HC Verma - Centre of Mass Linear Momentum Collision Solution For Class 11 Concepts Of Physics Part 1

Question 1:

Answer:

Yes, the centre of mass can be at a point outside the body.
For example, the centre of mass of a ring lies at its centre, which is not a part of the ring.

Question 2:

Yes, the centre of mass can be at a point outside the body.
For example, the centre of mass of a ring lies at its centre, which is not a part of the ring.

Answer:

Yes, if all the particles of a system lie in the X–Y plane, then it's necessary that its centre of mass lies in the X–Y plane.
$z_{cm} = \frac{m_{1} z_{1} + m_{2} z_{2} + . . . m_{n} z_{n}}{\sum_{} m}$

As all the particles lie in the X–Y plane, their z-coordinates are zero.
Therefore, for the whole system, z_cm = 0; i.e., its centre of mass lies in the X–Y plane.

Question 3:

Yes, if all the particles of a system lie in the X–Y plane, then it's necessary that its centre of mass lies in the X–Y plane.
$z_{cm} = \frac{m_{1} z_{1} + m_{2} z_{2} + . . . m_{n} z_{n}}{\sum_{} m}$

As all the particles lie in the X–Y plane, their z-coordinates are zero.
Therefore, for the whole system, z_cm = 0; i.e., its centre of mass lies in the X–Y plane.

Answer:

Yes. As a cube is a 3-dimensional body, all the particles of a system lying in a cube lie in the x,y and z plane.

Let the i^th element of mass âˆ†m_iis located at the point (x_i,y_i,z_i).
The co-ordinates of the centre of mass are given as:
$X = \frac{1}{M} \sum_{i = 1}^{i = n} (∆ m_{i}) x_{i} Y = \frac{1}{M} \sum_{i = 1}^{i = n} (∆ m_{i}) y_{i} Z = \frac{1}{M} \sum_{i = 1}^{i = n} (∆ m_{i}) z_{i}$

X, Y and Z lie inside the cube because it is a weighted mean.

Question 4:

Yes. As a cube is a 3-dimensional body, all the particles of a system lying in a cube lie in the x,y and z plane.

Let the i^th element of mass âˆ†m_iis located at the point (x_i,y_i,z_i).
The co-ordinates of the centre of mass are given as:
$X = \frac{1}{M} \sum_{i = 1}^{i = n} (∆ m_{i}) x_{i} Y = \frac{1}{M} \sum_{i = 1}^{i = n} (∆ m_{i}) y_{i} Z = \frac{1}{M} \sum_{i = 1}^{i = n} (∆ m_{i}) z_{i}$

X, Y and Z lie inside the cube because it is a weighted mean.

Answer:

(a) Yes

Consider a charge distributed in X-Y plane.

$X_{c m} = \frac{- 6 q \times 0 + q \times 5 a}{- 6 q + q} = - a$

(b) Yes. Because the z-coordinates of all the charges are zero, the centre of charge lies in X-Y plane.

(c) No, it is not necessary that the centre of charge lies in the cube because charge can be either negative or positive.

Question 5:

(a) Yes

Consider a charge distributed in X-Y plane.

$X_{c m} = \frac{- 6 q \times 0 + q \times 5 a}{- 6 q + q} = - a$

(b) Yes. Because the z-coordinates of all the charges are zero, the centre of charge lies in X-Y plane.

(c) No, it is not necessary that the centre of charge lies in the cube because charge can be either negative or positive.

Answer:

In order to simplify the situation, we consider that the weight Mg of an extended body acts through its centre of mass.
Although the earth attracts all the particles, the net effect can be assumed to be at the centre of mass.

Question 6:

In order to simplify the situation, we consider that the weight Mg of an extended body acts through its centre of mass.
Although the earth attracts all the particles, the net effect can be assumed to be at the centre of mass.

Answer:

There is no violation of conservation of momentum because in the earth's frame the component of tension is acting in the horizontal direction.

Question 7:

There is no violation of conservation of momentum because in the earth's frame the component of tension is acting in the horizontal direction.

Answer:

The trunk does not recoil as the girl jumps off on the platform because the force exerted by the girl is less than the limiting friction between the platform and the iron trunk.

Question 8:

The trunk does not recoil as the girl jumps off on the platform because the force exerted by the girl is less than the limiting friction between the platform and the iron trunk.

Answer:

Yes.
For example, consider particle-1 at a velocity of 4 ms^-1 and particle-2 at a velocity of 2 ms^-1 undergo a head-on collision.
The velocity of particle-1 decreases but particle-2 increases. Therefore, at an instant, their velocities will be equal.

Question 9:

Yes.
For example, consider particle-1 at a velocity of 4 ms^-1 and particle-2 at a velocity of 2 ms^-1 undergo a head-on collision.
The velocity of particle-1 decreases but particle-2 increases. Therefore, at an instant, their velocities will be equal.

Answer:

No. As the collision experiment is being done on a horizontal table in the elevator that is accelerating up or down in vertical direction, no extra force is experienced in the horizontal direction. Hence, the objects in the horizontal direction remain unaffected.

Question 10:

No. As the collision experiment is being done on a horizontal table in the elevator that is accelerating up or down in vertical direction, no extra force is experienced in the horizontal direction. Hence, the objects in the horizontal direction remain unaffected.

Answer:

No, due to the non-inertial character of the frame and the presence of a pseudo force, both the equations, i.e., Velocity of separation = Velocity of approach and Final momentum = Initial momentum, do not remain valid in the accelerating car.

Question 11:

No, due to the non-inertial character of the frame and the presence of a pseudo force, both the equations, i.e., Velocity of separation = Velocity of approach and Final momentum = Initial momentum, do not remain valid in the accelerating car.

Answer:

No. As the potential energy can have a negative value, the total energy of the system may sum up to zero.

For example:
Two masses A and B having masses 2 kg and 4 kg respectively move with a velocity of 4 ms^-1 in opposite directions.

Kinetic energy of system (A and B)
$= \frac{1}{2} \times 2 \times 4^{2} + \frac{1}{2} \times 4 \times 4^{2} = 48 J$

If the gravitational potential energy of the system is −48 J, the total energy of the system will be zero. However, the linear momentum will be non-zero.

Question 12:

No. As the potential energy can have a negative value, the total energy of the system may sum up to zero.

For example:
Two masses A and B having masses 2 kg and 4 kg respectively move with a velocity of 4 ms^-1 in opposite directions.

Kinetic energy of system (A and B)
$= \frac{1}{2} \times 2 \times 4^{2} + \frac{1}{2} \times 4 \times 4^{2} = 48 J$

If the gravitational potential energy of the system is −48 J, the total energy of the system will be zero. However, the linear momentum will be non-zero.

Answer:

Yes, the kinetic energy of the particle can be determined if the value of linear momentum is known.
The kinetic energy is calculated using the formula:
$K . E = \frac{1}{2} m v^{2} = \frac{p^{2}}{2 m} w here, p is the linear momemtum having value m v .$

But linear momentum cannot be determined even if the kinetic energy is known because linear momentum is a vector quantity, whereas kinetic energy is a scalar quantity. Thus, the direction of the linear momentum remains unknown, however its magnitude can be calculated.

Question 13:

Yes, the kinetic energy of the particle can be determined if the value of linear momentum is known.
The kinetic energy is calculated using the formula:
$K . E = \frac{1}{2} m v^{2} = \frac{p^{2}}{2 m} w here, p is the linear momemtum having value m v .$

But linear momentum cannot be determined even if the kinetic energy is known because linear momentum is a vector quantity, whereas kinetic energy is a scalar quantity. Thus, the direction of the linear momentum remains unknown, however its magnitude can be calculated.

Answer:

The distance of centre of mass of a uniform hemisphere from its centre will be less than r/2 because the portion of the hemisphere lying below r/2 from the diameter is heavier than the portion lying above r/2.

Question 14:

The distance of centre of mass of a uniform hemisphere from its centre will be less than r/2 because the portion of the hemisphere lying below r/2 from the diameter is heavier than the portion lying above r/2.

Answer:

More effort is needed when the cage is closed, while less effort is required when the cage has rods to let the air pass. When a bird flies from its position, it pushes the air downwards. Thus, when the bird is in a cage, the net downward force will be equal to the weight of the cage plus the downward force due to air (the weight of the bird).

However, if the cage has rods to let air pass, the downward force exerted by air become less. Therefore, less effort will be required to hold the cage.

Question 15:

More effort is needed when the cage is closed, while less effort is required when the cage has rods to let the air pass. When a bird flies from its position, it pushes the air downwards. Thus, when the bird is in a cage, the net downward force will be equal to the weight of the cage plus the downward force due to air (the weight of the bird).

However, if the cage has rods to let air pass, the downward force exerted by air become less. Therefore, less effort will be required to hold the cage.

Answer:

According to the question, the weight of plank is very less as compared to the fat person. Therefore, the centre of mass of the whole system effectively lies on the person. As the net external force on the system is zero, the centre of mass of the system does not move.

Question 16:

According to the question, the weight of plank is very less as compared to the fat person. Therefore, the centre of mass of the whole system effectively lies on the person. As the net external force on the system is zero, the centre of mass of the system does not move.

Answer:

From the figure, it can be seen that when a high-jumper successfully clears the bar, it is possible that her centre of mass crosses the bar from below it because the legs as well as the arms of the high-jumper are below the bar.
Hence, the point shown in the figure can be her centre of mass.

Question 17:

From the figure, it can be seen that when a high-jumper successfully clears the bar, it is possible that her centre of mass crosses the bar from below it because the legs as well as the arms of the high-jumper are below the bar.
Hence, the point shown in the figure can be her centre of mass.

Answer:

The person shown on the right hand side of the figure is more likely to fall down because in the given cart frame the pseudo force will be in backward direction.

Question 18:

The person shown on the right hand side of the figure is more likely to fall down because in the given cart frame the pseudo force will be in backward direction.

Answer:

It is not necessary that the linear momentum of a system remains constant even if no external force acts on it because during collision, the sum of magnitudes of momenta does not remain constant.

Question 19:

It is not necessary that the linear momentum of a system remains constant even if no external force acts on it because during collision, the sum of magnitudes of momenta does not remain constant.

Answer:

Yes, if the external force applied on the particle is zero, its speed does not change and hence, the momentum remains constant.

Question 20:

Yes, if the external force applied on the particle is zero, its speed does not change and hence, the momentum remains constant.

Answer:

Yes, it's proper to say that the force generated by the engine accelerates the car. When petrol burns inside the engine, the piston moves, which in turn rotates the wheel. As the wheel rotates, the frictional forces from the road moves the car.

Question 21:

Yes, it's proper to say that the force generated by the engine accelerates the car. When petrol burns inside the engine, the piston moves, which in turn rotates the wheel. As the wheel rotates, the frictional forces from the road moves the car.

Answer:

The frictional force acting between the surface of the table and the ball is responsible for the change in momentum of the ball. As the force opposes the motion of the ball, it stops after moving some distance.

Question 22:

The frictional force acting between the surface of the table and the ball is responsible for the change in momentum of the ball. As the force opposes the motion of the ball, it stops after moving some distance.

Answer:

Considering the table plus the ball as a system, it can be said that the frictional force is responsible for the change in the momentum. As the force acts between the surface of the table and ground, it opposes the motion of the table plus the ball. Hence, the ball slows down and the momentum of the system decreases.

Question 23:

Considering the table plus the ball as a system, it can be said that the frictional force is responsible for the change in the momentum. As the force acts between the surface of the table and ground, it opposes the motion of the table plus the ball. Hence, the ball slows down and the momentum of the system decreases.

Answer:

In view of the principle of conservation of momentum, the given situation is possible because as a beta particle is ejected, another particle called an antineutrino is also ejected.

Question 24:

In view of the principle of conservation of momentum, the given situation is possible because as a beta particle is ejected, another particle called an antineutrino is also ejected.

Answer:

According to the question, the van is standing on a frictionless surface. When throwing something in backward direction, the persons sitting inside the van sets the van in motion in the forward direction according to the principle of conservation of linear momentum.

Question 25:

According to the question, the van is standing on a frictionless surface. When throwing something in backward direction, the persons sitting inside the van sets the van in motion in the forward direction according to the principle of conservation of linear momentum.

Answer:

No. If the masses are different, the velocities in a one-dimensional collision cannot be interchanged because that would be violation of the principle of conservation of momentum.

Question 1:

No. If the masses are different, the velocities in a one-dimensional collision cannot be interchanged because that would be violation of the principle of conservation of momentum.

Answer:

(c) A is correct but B is wrong

In a non-inertial frame, the position of centre of mass of the particle does not change but an additional pseudo force acts on it.

Question 2:

(c) A is correct but B is wrong

In a non-inertial frame, the position of centre of mass of the particle does not change but an additional pseudo force acts on it.

Answer:

(d) B implies A but A does not imply B.

The centre of mass of a system is given by,
$\vec{R} = \frac{1}{M} \sum_{} m_{i} {\vec{r}}_{i}$

On differentiating the above equation with respect to time, we get:
$\frac{d \vec{R}}{d t} = \frac{1}{M} \sum_{} m_{i} \frac{d {\vec{r}}_{i}}{d t}$

As the centre of mass of the system remains at rest, we have:
$\frac{1}{M} \sum_{} m_{i} \frac{d {\vec{r}}_{i}}{d t} = 0 \sum_{} m_{i} {\vec{v}}_{i} = 0$

This implies that the linear momentum of the system remains constant.

Question 3:

(d) B implies A but A does not imply B.

The centre of mass of a system is given by,
$\vec{R} = \frac{1}{M} \sum_{} m_{i} {\vec{r}}_{i}$

On differentiating the above equation with respect to time, we get:
$\frac{d \vec{R}}{d t} = \frac{1}{M} \sum_{} m_{i} \frac{d {\vec{r}}_{i}}{d t}$

As the centre of mass of the system remains at rest, we have:
$\frac{1}{M} \sum_{} m_{i} \frac{d {\vec{r}}_{i}}{d t} = 0 \sum_{} m_{i} {\vec{v}}_{i} = 0$

This implies that the linear momentum of the system remains constant.

Answer:

(d) B implies A but A does not imply B.

If the linear momentum of a system is zero,
$\Rightarrow m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} + . . .$ =0

Thus, for a system of comprising two particles of same masses,
${\vec{v}}_{1} = - {\vec{v}}_{2}$ ...(1)

The kinetic energy of the system is given by,
$K . E . = \frac{1}{2} m {\vec{v}}_{1}^{2} + \frac{1}{2} m {\vec{v}}_{2}^{2}$

Using equation (1) to solve above equation, we can say:
$K . E . \neq 0$
i.e. A does not imply B.

Now,
If the kinetic energy of the system is zero,
$\Rightarrow \frac{1}{2} m {\vec{v}}_{1}^{2} + \frac{1}{2} m {\vec{v}}_{2}^{2} = 0$
$v_{1} = \pm v_{2}$

On calculating the linear momentum of the system, we get:
$\vec{P} = m {\vec{v}}_{1} + m {\vec{v}}_{2} taking v_{1} = - v_{2}, we can write : \vec{P} = 0$

Hence, we can say, B implies A but A does not imply B.

Question 4:

(d) B implies A but A does not imply B.

If the linear momentum of a system is zero,
$\Rightarrow m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} + . . .$ =0

Thus, for a system of comprising two particles of same masses,
${\vec{v}}_{1} = - {\vec{v}}_{2}$ ...(1)

The kinetic energy of the system is given by,
$K . E . = \frac{1}{2} m {\vec{v}}_{1}^{2} + \frac{1}{2} m {\vec{v}}_{2}^{2}$

Using equation (1) to solve above equation, we can say:
$K . E . \neq 0$
i.e. A does not imply B.

Now,
If the kinetic energy of the system is zero,
$\Rightarrow \frac{1}{2} m {\vec{v}}_{1}^{2} + \frac{1}{2} m {\vec{v}}_{2}^{2} = 0$
$v_{1} = \pm v_{2}$

On calculating the linear momentum of the system, we get:
$\vec{P} = m {\vec{v}}_{1} + m {\vec{v}}_{2} taking v_{1} = - v_{2}, we can write : \vec{P} = 0$

Hence, we can say, B implies A but A does not imply B.

Answer:

(d) both A and B are false.

As the velocity of the particle depends on the frame of reference, the linear momentum as well as the kinetic energy is dependent on the frame of reference.

Question 5:

(d) both A and B are false.

As the velocity of the particle depends on the frame of reference, the linear momentum as well as the kinetic energy is dependent on the frame of reference.

Answer:

(b) ≤ R

Distance of the centre of mass from the origin is given by,
$R' = \frac{1}{M} \sum_{i = 1}^{n} m_{i} {\vec{r}}_{i}$

Let half of the particles lie on +Y-axis and the rest of the particles lie on +X-axis.
$Y_{CM} = \frac{m_{3} R \vec{j} + m_{4} R \vec{j}}{m_{1} + m_{2} + m_{3} + m_{4} + . . .} = \frac{R \vec{j}}{2}$

Similarly, $X_{CM} = \frac{R \vec{i}}{2}$

Therefore, the coordinates of centre of mass are $(\frac{R}{2}, \frac{R}{2})$ .

Distance of the centre of mass from the origin $= \frac{R}{\sqrt{2}}$
For the given situation, $R' < R$ .

In general, R' ≤ R.

Question 6:

(b) ≤ R

Distance of the centre of mass from the origin is given by,
$R' = \frac{1}{M} \sum_{i = 1}^{n} m_{i} {\vec{r}}_{i}$

Let half of the particles lie on +Y-axis and the rest of the particles lie on +X-axis.
$Y_{CM} = \frac{m_{3} R \vec{j} + m_{4} R \vec{j}}{m_{1} + m_{2} + m_{3} + m_{4} + . . .} = \frac{R \vec{j}}{2}$

Similarly, $X_{CM} = \frac{R \vec{i}}{2}$

Therefore, the coordinates of centre of mass are $(\frac{R}{2}, \frac{R}{2})$ .

Distance of the centre of mass from the origin $= \frac{R}{\sqrt{2}}$
For the given situation, $R' < R$ .

In general, R' ≤ R.

Answer:

(b) inside the square plate

Let m₁ be the mass of circular plate and m₂ be the mass of square plate.
The thickness of both the plates is t.

$mass = densit y \times volume m_{1} = ρ π {(\frac{d}{2})}^{2} t m_{2} = ρ d^{2} t$

Centre of mass of the circular plate lies at its centre.
Let the centre of circular plate be the origin.
${\vec{r}}_{1} = 0$

Centre of mass of the square plate lies at its centre.
${\vec{r}}_{2} = 2 d Now, R = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}} = \frac{m_{1} \times 0 + ρ d^{2} t \times 2 d}{ρ π {(\frac{d}{2})}^{2} t + ρ d^{2} t} = \frac{2 d}{\frac{π}{4} + 1} = 1.12 d \Rightarrow R > d$

$∴$ Centre of mass of the system lies in the square plate.

Question 7:

(b) inside the square plate

Let m₁ be the mass of circular plate and m₂ be the mass of square plate.
The thickness of both the plates is t.

$mass = densit y \times volume m_{1} = ρ π {(\frac{d}{2})}^{2} t m_{2} = ρ d^{2} t$

Centre of mass of the circular plate lies at its centre.
Let the centre of circular plate be the origin.
${\vec{r}}_{1} = 0$

Centre of mass of the square plate lies at its centre.
${\vec{r}}_{2} = 2 d Now, R = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}} = \frac{m_{1} \times 0 + ρ d^{2} t \times 2 d}{ρ π {(\frac{d}{2})}^{2} t + ρ d^{2} t} = \frac{2 d}{\frac{π}{4} + 1} = 1.12 d \Rightarrow R > d$

$∴$ Centre of mass of the system lies in the square plate.

Answer:

(b) $\frac{1}{2} \vec{a}$

Acceleration of centre of mass of a two-particle system is given as,
${\vec{a}}_{c m} = \frac{m_{1} {\vec{a}}_{1} + m_{2} \vec{a_{2}}}{m_{1} + m_{2}}$ ...(1)

According to the question,

$m_{1} = m_{2} = m a_{1} = 0 a_{2} = a$

Substituting these values in equation (1), we get:
${\vec{a}}_{c m} = \frac{m \times 0 + m \vec{a}}{2 m} = \frac{1}{2} \vec{a}$

Question 8:

(b) $\frac{1}{2} \vec{a}$

Acceleration of centre of mass of a two-particle system is given as,
${\vec{a}}_{c m} = \frac{m_{1} {\vec{a}}_{1} + m_{2} \vec{a_{2}}}{m_{1} + m_{2}}$ ...(1)

According to the question,

$m_{1} = m_{2} = m a_{1} = 0 a_{2} = a$

Substituting these values in equation (1), we get:
${\vec{a}}_{c m} = \frac{m \times 0 + m \vec{a}}{2 m} = \frac{1}{2} \vec{a}$

Answer:

(b) the kinetic energy but not the linear momentum

Internal forces can not change the position of centre of mass of a system. Therefore, linear momentum of the system is constant, whereas kinetic energy of the system is not.

Question 9:

(b) the kinetic energy but not the linear momentum

Internal forces can not change the position of centre of mass of a system. Therefore, linear momentum of the system is constant, whereas kinetic energy of the system is not.

Answer:

(c) gravitational potential energy of the block

When the block kept at rest is hit by a bullet, the block acquires certain velocity by the conservation of linear momentum.
Therefore, the linear momentum and the kinetic energy of the block change.
As some of the kinetic energy carried by the bullet transforms into heat energy, its temperature also changes.
However, the gravitational potential energy of the block does not change, as the height of the block does not change in this process.

Question 10:

(c) gravitational potential energy of the block

When the block kept at rest is hit by a bullet, the block acquires certain velocity by the conservation of linear momentum.
Therefore, the linear momentum and the kinetic energy of the block change.
As some of the kinetic energy carried by the bullet transforms into heat energy, its temperature also changes.
However, the gravitational potential energy of the block does not change, as the height of the block does not change in this process.

Answer:

(d) is independent of h.

As the uniform sphere is placed on a smooth surface, the sphere only slips.
The acceleration of the centre of the sphere is given by,
$\frac{Applied force}{Mass of the sphere}$ ,which is independent of height h.

Question 11:

(d) is independent of h.

As the uniform sphere is placed on a smooth surface, the sphere only slips.
The acceleration of the centre of the sphere is given by,
$\frac{Applied force}{Mass of the sphere}$ ,which is independent of height h.

Answer:

(c) does not shift horizontally

As the body falls vertically downwards, no external force acts in the horizontal direction.
Hence, the centre of mass does not shift horizontally.

Question 12:

(c) does not shift horizontally

As the body falls vertically downwards, no external force acts in the horizontal direction.
Hence, the centre of mass does not shift horizontally.

Answer:

(b) of the box plus the ball system remains constant

Consider the box and the ball a system. As no external force acts on this system, the velocity of the centre of mass of the system remains constant.

Question 13:

(b) of the box plus the ball system remains constant

Consider the box and the ball a system. As no external force acts on this system, the velocity of the centre of mass of the system remains constant.

Answer:

(c) in opposite directions with equal speeds

Using the principle of conservation of linear momentum, we can write:
Initial momenta = Final momenta

$\Rightarrow M \times 0 = \frac{M}{2} v_{1} + \frac{M}{2} v_{2} \Rightarrow v_{1} = - v_{2}$
where M is the initial mass of the body, at rest.
The final mass of the two pieces moving with equal speeds in opposite direction is equal to $\frac{M}{2}$ .

Question 14:

(c) in opposite directions with equal speeds

Using the principle of conservation of linear momentum, we can write:
Initial momenta = Final momenta

$\Rightarrow M \times 0 = \frac{M}{2} v_{1} + \frac{M}{2} v_{2} \Rightarrow v_{1} = - v_{2}$
where M is the initial mass of the body, at rest.
The final mass of the two pieces moving with equal speeds in opposite direction is equal to $\frac{M}{2}$ .

Answer:

(c) $\frac{2 m v^{2}}{r}$ must be acting on the system

Total mass of the system = 2m
To move the centre of the system in a circle of radius r with a uniform speed v, the external force required is $\frac{2 m v^{2}}{R}$ .

Question 15:

(c) $\frac{2 m v^{2}}{r}$ must be acting on the system

Total mass of the system = 2m
To move the centre of the system in a circle of radius r with a uniform speed v, the external force required is $\frac{2 m v^{2}}{R}$ .

Answer:

(d) momentum, but neither kinetic energy nor temperature

Linear momentum of a system remains constant in a collision. However, the kinetic energy and temperature of the system may vary, as their values depend on the type of collision.

Question 16:

(d) momentum, but neither kinetic energy nor temperature

Linear momentum of a system remains constant in a collision. However, the kinetic energy and temperature of the system may vary, as their values depend on the type of collision.

Answer:

(d) $m_{1} \vec{v_{1}} + m_{2} \vec{v_{2}}$ must be parallel to $\vec{v}$

By the law of conservation of linear momentum, we can write:

$Initial momentum = Final momentum \Rightarrow m \vec{v} = m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} \Rightarrow (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}) must be parallel to \vec{v}$

Question 17:

(d) $m_{1} \vec{v_{1}} + m_{2} \vec{v_{2}}$ must be parallel to $\vec{v}$

By the law of conservation of linear momentum, we can write:

$Initial momentum = Final momentum \Rightarrow m \vec{v} = m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} \Rightarrow (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}) must be parallel to \vec{v}$

Answer:

(a) 3V cos θ

The linear momentum is conserved in horizontal direction.

$∴$ Initial momentum = Final momentum

$\Rightarrow m v \cos θ = - \frac{m}{2} v \cos θ + \frac{m}{2} v' \Rightarrow v' = m v \cos θ$

Question 18:

(a) 3V cos θ

The linear momentum is conserved in horizontal direction.

$∴$ Initial momentum = Final momentum

$\Rightarrow m v \cos θ = - \frac{m}{2} v \cos θ + \frac{m}{2} v' \Rightarrow v' = m v \cos θ$

Answer:

(a) the initial kinetic energy is equal to the final kinetic energy

As no energy is lost into heat in an elastic collision, the initial kinetic energy is equal to the final kinetic energy.

Question 19:

(a) the initial kinetic energy is equal to the final kinetic energy

As no energy is lost into heat in an elastic collision, the initial kinetic energy is equal to the final kinetic energy.

Answer:

(b) the final kinetic energy is less than the initial kinetic energy

As some energy is loss into heat in an inelastic collision, the final kinetic energy is less than the initial kinetic energy.

Question 1:

(b) the final kinetic energy is less than the initial kinetic energy

As some energy is loss into heat in an inelastic collision, the final kinetic energy is less than the initial kinetic energy.

Answer:

None.

The centre of mass of a system of particles depends on the product of individual masses and their distances from the origin.
Therefore, we may say about the given statements:
(a) Distance of particles from origin is not known.
(b) Masses are same but the distance of particles from the origin is not given.
(c) Distance of particles from origin is not given.
(d) It is not necessary that least one particle lies on the negative X-axis. The particles can be above the negative X-axis on X-Y plane.

Question 1:

None.

The centre of mass of a system of particles depends on the product of individual masses and their distances from the origin.
Therefore, we may say about the given statements:
(a) Distance of particles from origin is not known.
(b) Masses are same but the distance of particles from the origin is not given.
(c) Distance of particles from origin is not given.
(d) It is not necessary that least one particle lies on the negative X-axis. The particles can be above the negative X-axis on X-Y plane.

Answer:

Taking BC as the X-axis and point B as the origin, the positions of masses m₁= 1 kg, m₂ = 2 kg and m₃ = 3 kg are
$(0, 0), (1, 0) and (\frac{1}{2}, \frac{\sqrt{3}}{2})$ respectively.

The position ofcentre of mass is given by:

$X_{cm}, Y_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}, \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}}$
$= \frac{1 \times 0 + 2 \times 1 + 3 \times \frac{1}{2}}{1 + 2 + 3}, \frac{1 \times 0 + 2 \times 0 + 3 \times \frac{\sqrt{3}}{2}}{1 + 2 + 3} = \frac{7}{12} m, \frac{\sqrt{3}}{4} m$

Question 2:

Taking BC as the X-axis and point B as the origin, the positions of masses m₁= 1 kg, m₂ = 2 kg and m₃ = 3 kg are
$(0, 0), (1, 0) and (\frac{1}{2}, \frac{\sqrt{3}}{2})$ respectively.

The position ofcentre of mass is given by:

$X_{cm}, Y_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}, \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}}$
$= \frac{1 \times 0 + 2 \times 1 + 3 \times \frac{1}{2}}{1 + 2 + 3}, \frac{1 \times 0 + 2 \times 0 + 3 \times \frac{\sqrt{3}}{2}}{1 + 2 + 3} = \frac{7}{12} m, \frac{\sqrt{3}}{4} m$

Answer:

Let OX be the x-axis, OY be the Y-axis and O be the origin.

$Mass of O atom, m_{1} = 16 unit$
Let the position of oxygen atom be origin.
$\Rightarrow x_{1} = y_{1} = 0 Mass of H atom, m_{2} = 1 unit x_{2} = - 0.96 \times 10^{- 10} \sin 52 ° m y_{2} = - 0.96 \times 10^{- 10} \cos 52 ° Now, m_{3} = 1 unit x_{3} = 0.96 \times 10^{- 10} \sin 52 ° m y_{3} = - 0.96 \times 10^{- 10} \cos 52 ° The X coordinate of the center of mass is given by : x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{16 \times 0 + 1 \times (- 0.96 \times 10^{- 10} \sin 52 °) + 1 \times 0.96 \times {10^{-}}^{10} \sin 52 °}{1 + 1 + 16} = 0 The Y coordinate of the center of mass is given by : y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}} = \frac{16 \times 0 + 2 \times 0.96 \times 10^{- 10} \cos 52 °}{1 + 1 + 16} = \frac{2 \times 0.96 \times 10^{- 10} \cos 52 °}{18} = 6.4 \times 10^{- 12} m$

Hence, the distance of centre of mass of the molecule from the centre of the oxygen atom is $(0, 6.4 \times 10^{- 12} m)$ .

Question 3:

Let OX be the x-axis, OY be the Y-axis and O be the origin.

$Mass of O atom, m_{1} = 16 unit$
Let the position of oxygen atom be origin.
$\Rightarrow x_{1} = y_{1} = 0 Mass of H atom, m_{2} = 1 unit x_{2} = - 0.96 \times 10^{- 10} \sin 52 ° m y_{2} = - 0.96 \times 10^{- 10} \cos 52 ° Now, m_{3} = 1 unit x_{3} = 0.96 \times 10^{- 10} \sin 52 ° m y_{3} = - 0.96 \times 10^{- 10} \cos 52 ° The X coordinate of the center of mass is given by : x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{16 \times 0 + 1 \times (- 0.96 \times 10^{- 10} \sin 52 °) + 1 \times 0.96 \times {10^{-}}^{10} \sin 52 °}{1 + 1 + 16} = 0 The Y coordinate of the center of mass is given by : y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}} = \frac{16 \times 0 + 2 \times 0.96 \times 10^{- 10} \cos 52 °}{1 + 1 + 16} = \frac{2 \times 0.96 \times 10^{- 10} \cos 52 °}{18} = 6.4 \times 10^{- 12} m$

Hence, the distance of centre of mass of the molecule from the centre of the oxygen atom is $(0, 6.4 \times 10^{- 12} m)$ .

Answer:

Let OX be the X-axis and point O (0, 0) be the origin of the system.

The mass of each brick is M.
The length of each brick is L.
Each brick is displaced with respect to another in contact by a distance $\frac{L}{10}$ .

∴ The X-coordinate of the centre of mass is given as,
$X_{cm} = {\frac{1}{7 m \frac{L}{10}}}^{\{\frac{m L}{2} + m (\frac{L}{2} + \frac{L}{10}) + . . . . . . . . . . + m (\frac{L}{2})\}} X_{cm} = \frac{\frac{L}{2} + \frac{L}{2} + \frac{L}{10} + \frac{L}{5} + \frac{L}{2} + \frac{3 L}{10} + \frac{L}{2} + \frac{L}{5} + \frac{L}{2} + \frac{L}{10} + \frac{L}{2}}{7} = \frac{7 \frac{L}{2} + 5 \frac{L}{10} + 2 \frac{L}{5}}{7} = \frac{22 L}{35}$

The X-coordinate of the centre of mass relative to the origin is $\frac{22 L}{35}$ .

Question 4:

Let OX be the X-axis and point O (0, 0) be the origin of the system.

The mass of each brick is M.
The length of each brick is L.
Each brick is displaced with respect to another in contact by a distance $\frac{L}{10}$ .

∴ The X-coordinate of the centre of mass is given as,
$X_{cm} = {\frac{1}{7 m \frac{L}{10}}}^{\{\frac{m L}{2} + m (\frac{L}{2} + \frac{L}{10}) + . . . . . . . . . . + m (\frac{L}{2})\}} X_{cm} = \frac{\frac{L}{2} + \frac{L}{2} + \frac{L}{10} + \frac{L}{5} + \frac{L}{2} + \frac{3 L}{10} + \frac{L}{2} + \frac{L}{5} + \frac{L}{2} + \frac{L}{10} + \frac{L}{2}}{7} = \frac{7 \frac{L}{2} + 5 \frac{L}{10} + 2 \frac{L}{5}}{7} = \frac{22 L}{35}$

The X-coordinate of the centre of mass relative to the origin is $\frac{22 L}{35}$ .

Answer:

Let the centre O (0, 0) of the bigger disc be the origin.

Radius of bigger disc = 2R
Radius of smaller disc = R

Now,
m₁ = μR × T × ρ
m2 = μ (2R)² × T × ρ

where T is the thickness of the two discs, and
ρ is the density of the two discs.

Position of the centre of mass is calculated as:

$x_{1} = R, y_{1} = 0 x_{2} = 0, y_{2} = 0$

$= (\frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}) = (\frac{{πR}^{2} TρR + 0}{{μR}^{2} TρR + μ (2 R)^{2} Tρ}, \frac{0}{m_{1} + m_{2}}) = (\frac{{πR}^{2} TρR}{5 {μR}^{2} Tρ}, 0) = (\frac{R}{5}, 0)$

$\Rightarrow$ The centre of mass lies at distance $\frac{R}{5}$ from the centre of bigger disc, towards the centre of smaller disc.

Hence, the centre of mass of the system is $(\frac{R}{5}, 0)$ .

Question 2:

Let the centre O (0, 0) of the bigger disc be the origin.

Radius of bigger disc = 2R
Radius of smaller disc = R

Now,
m₁ = μR × T × ρ
m2 = μ (2R)² × T × ρ

where T is the thickness of the two discs, and
ρ is the density of the two discs.

Position of the centre of mass is calculated as:

$x_{1} = R, y_{1} = 0 x_{2} = 0, y_{2} = 0$

$= (\frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}) = (\frac{{πR}^{2} TρR + 0}{{μR}^{2} TρR + μ (2 R)^{2} Tρ}, \frac{0}{m_{1} + m_{2}}) = (\frac{{πR}^{2} TρR}{5 {μR}^{2} Tρ}, 0) = (\frac{R}{5}, 0)$

$\Rightarrow$ The centre of mass lies at distance $\frac{R}{5}$ from the centre of bigger disc, towards the centre of smaller disc.

Hence, the centre of mass of the system is $(\frac{R}{5}, 0)$ .

Answer:

(c) may be all non-negative
(d) may be positive for some cases and negative for other cases

According to the question, the centre of mass is at origin.

$∴$ $X = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + . . .}{m_{1} + m_{2} + m_{3} + . . . . . . .} = 0$
$\Rightarrow m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + . . . = 0$

From the above equation, it can be concluded that all the x-coordinates may be non-negative.

In other words, they may be positive for some cases and negative for others.

Question 3:

(c) may be all non-negative
(d) may be positive for some cases and negative for other cases

According to the question, the centre of mass is at origin.

$∴$ $X = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + . . .}{m_{1} + m_{2} + m_{3} + . . . . . . .} = 0$
$\Rightarrow m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + . . . = 0$

From the above equation, it can be concluded that all the x-coordinates may be non-negative.

In other words, they may be positive for some cases and negative for others.

Answer:

(a) the density continuously increases from left to right
(b) the density continuously decreases from left to right

As the density continuously increases/decreases from left to right, there will be difference in the masses of rod that lie on either sides of the centre of mass. Thus, the centre of mass of a rod in such a case will certainly not be at its centre.

Question 4:

(a) the density continuously increases from left to right
(b) the density continuously decreases from left to right

As the density continuously increases/decreases from left to right, there will be difference in the masses of rod that lie on either sides of the centre of mass. Thus, the centre of mass of a rod in such a case will certainly not be at its centre.

Answer:

(b) must not accelerate
(c) may move

If the external force acting on a system has zero resultant,
then, acceleration of centre of mass $= \frac{{\vec{F}}_{net}}{M} = 0$ .
However, it may move uniformly with constant velocity.

Question 5:

(b) must not accelerate
(c) may move

If the external force acting on a system has zero resultant,
then, acceleration of centre of mass $= \frac{{\vec{F}}_{net}}{M} = 0$ .
However, it may move uniformly with constant velocity.

Answer:

(b) v₀ = 0, a₀ ≠ 0
(d) v₀ ≠ 0, a₀ ≠ 0

If a non-zero external force acts on a system of particles, it causes the centre of mass of the system to accelerate with acceleration a₀ at any instant t. In such a case, the velocity of centre of mass of the system of particles is either v₀ or zero.

Question 6:

(b) v₀ = 0, a₀ ≠ 0
(d) v₀ ≠ 0, a₀ ≠ 0

If a non-zero external force acts on a system of particles, it causes the centre of mass of the system to accelerate with acceleration a₀ at any instant t. In such a case, the velocity of centre of mass of the system of particles is either v₀ or zero.

Answer:

(d) is equal to g

The acceleration of the centre of mass of two balls having masses m₁ and m₂ is given by
$a_{cm} = \frac{m_{1} {\vec{a}}_{1} + m_{2} {\vec{a}}_{2}}{m_{1} + m_{2}} = \frac{m_{1} g + m_{2} g}{m_{1} + m_{2}} = g$

Question 7:

(d) is equal to g

The acceleration of the centre of mass of two balls having masses m₁ and m₂ is given by
$a_{cm} = \frac{m_{1} {\vec{a}}_{1} + m_{2} {\vec{a}}_{2}}{m_{1} + m_{2}} = \frac{m_{1} g + m_{2} g}{m_{1} + m_{2}} = g$

Answer:

(a) the total momentum must be conserved
(d) the total kinetic energy must change

As no external force acts on the block, the linear momentum is conserved.
Some energy is used to break the block, thus the total kinetic energy must change.

Question 8:

(a) the total momentum must be conserved
(d) the total kinetic energy must change

As no external force acts on the block, the linear momentum is conserved.
Some energy is used to break the block, thus the total kinetic energy must change.

Answer:

(b) the linear momentum remains constant
(c) the final kinetic energy is equal to the initial kinetic energy
(d) the final linear momentum is equal to the initial linear momentum.

During an elastic collision, all of the above statements are valid.

Question 9:

(b) the linear momentum remains constant
(c) the final kinetic energy is equal to the initial kinetic energy
(d) the final linear momentum is equal to the initial linear momentum.

During an elastic collision, all of the above statements are valid.

Answer:

(c) the total momentum of the ball and the earth is conserved
(d) the total energy of the ball and the earth remains the same

As the ball rebounds after hitting the floor, its velocity changes.
i.e. Velocity of ball before collision ≠ Velocity of ball after collision

Therefore, the momentum of the ball just after the collision is not same as that just before the collision.
The mechanical energy of the ball also changes during the collision.

However, the total momentum of the system (earth plus ball) and the total energy of the system remain conserved.

Question 10:

(c) the total momentum of the ball and the earth is conserved
(d) the total energy of the ball and the earth remains the same

As the ball rebounds after hitting the floor, its velocity changes.
i.e. Velocity of ball before collision ≠ Velocity of ball after collision

Therefore, the momentum of the ball just after the collision is not same as that just before the collision.
The mechanical energy of the ball also changes during the collision.

However, the total momentum of the system (earth plus ball) and the total energy of the system remain conserved.

Answer:

(b) both the bodies move after collision
(c) the moving body comes to rest and the stationary body starts moving

By using the law of conservation of linear momentum we can write:
Initial momentum of the two bodies = Final momentum of the two bodies
$\Rightarrow p_{i} = p_{f} \Rightarrow m_{1} u = m_{1} v_{1} + m_{2} v_{2} \Rightarrow v_{1} = v_{2} \neq 0 However, it may be possible that v_{1} becomes 0 and v_{2} becomes \frac{m_{1} u}{m_{2}} .$

Question 11:

(b) both the bodies move after collision
(c) the moving body comes to rest and the stationary body starts moving

By using the law of conservation of linear momentum we can write:
Initial momentum of the two bodies = Final momentum of the two bodies
$\Rightarrow p_{i} = p_{f} \Rightarrow m_{1} u = m_{1} v_{1} + m_{2} v_{2} \Rightarrow v_{1} = v_{2} \neq 0 However, it may be possible that v_{1} becomes 0 and v_{2} becomes \frac{m_{1} u}{m_{2}} .$

Answer:

All of the above.

(a) the velocities are interchanged
(b) the speeds are interchanged
(c) the momenta are interchanged
(d) the faster body slows down and the slower body speeds up

Question 5:

All of the above.

(a) the velocities are interchanged
(b) the speeds are interchanged
(c) the momenta are interchanged
(d) the faster body slows down and the slower body speeds up

Answer:

Let O be the origin of the system (smaller disc plus bigger disc).

The density of the rods is ρ.
The thickness of rods is T.

Let m₁ be the mass and R be the radius of the smaller disc, and
m₂ be the mass and 2R be the radius of the bigger disc.

According to the question, the smaller disc is cut out from the bigger disc.

From the figure given below, we can write:

$m_{1} = π R^{2} T ρ Position of centre of mass of smaller disc, x_{1} = R, y_{1} = 0 m_{2} = π (2 R)^{2} T ρ Position of centre of mass of bigger disc, x_{2} = 0, y_{2} = 0 Therefore, the position of center of mass of the system, X_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, Y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} Since the smaller disc is removed from the bigger disc, the mass of the smaller disc is taken as negative (- m_{1}) . X_{cm} = \frac{- π R^{2} T ρ R + 0 m_{2}}{- π R^{2} T ρ + π (2 R)^{2} T ρ}, Y_{cm} = \frac{0 + 0}{m_{1} + m_{2}} X_{cm} = \frac{- π R^{2} T ρ R}{3 π R^{2} T ρ}, Y_{cm} = 0 X_{cm} = \frac{- R}{3}, Y_{cm} = 0$
Hence, the centre of mass of the system lies at distance $\frac{R}{3}$ from the centre of bigger disc, away from centre of the hole.

Question 6:

Let O be the origin of the system (smaller disc plus bigger disc).

The density of the rods is ρ.
The thickness of rods is T.

Let m₁ be the mass and R be the radius of the smaller disc, and
m₂ be the mass and 2R be the radius of the bigger disc.

According to the question, the smaller disc is cut out from the bigger disc.

From the figure given below, we can write:

$m_{1} = π R^{2} T ρ Position of centre of mass of smaller disc, x_{1} = R, y_{1} = 0 m_{2} = π (2 R)^{2} T ρ Position of centre of mass of bigger disc, x_{2} = 0, y_{2} = 0 Therefore, the position of center of mass of the system, X_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, Y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} Since the smaller disc is removed from the bigger disc, the mass of the smaller disc is taken as negative (- m_{1}) . X_{cm} = \frac{- π R^{2} T ρ R + 0 m_{2}}{- π R^{2} T ρ + π (2 R)^{2} T ρ}, Y_{cm} = \frac{0 + 0}{m_{1} + m_{2}} X_{cm} = \frac{- π R^{2} T ρ R}{3 π R^{2} T ρ}, Y_{cm} = 0 X_{cm} = \frac{- R}{3}, Y_{cm} = 0$
Hence, the centre of mass of the system lies at distance $\frac{R}{3}$ from the centre of bigger disc, away from centre of the hole.

Answer:

Let m be the mass per unit area of the square plate and the circular disc.

$\Rightarrow$ Mass of the square plate, M₁ = d²m
Mass of the circular disc, M₂ = $\frac{π d^{2}}{4} m$

Let the centre of the circular disc be the origin of the system.
$\Rightarrow$ x₁ = d, y₁ = 0
x₂ = 0, y₂ = 0

$\Rightarrow$ Position of the centre of mass of circular disc and square plate:
$= (\frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}) = (\frac{d^{2} m d + π (d^{2} / 4) m \times 0}{d^{2} m + π (d^{2} / 4) m}, \frac{0 + 0}{m_{1} + m_{2}}) = (\frac{d^{3} m}{d^{2} m (1 + π / 4)}, 0) = (\frac{4 d}{π + 4}, 0)$
Hence, the new centre of mass of the system (circular disc plus square plate) lies at distance $\frac{4 d}{(π + 4)}$ from the centre of circular disc, towards right.

Question 7:

Let m be the mass per unit area of the square plate and the circular disc.

$\Rightarrow$ Mass of the square plate, M₁ = d²m
Mass of the circular disc, M₂ = $\frac{π d^{2}}{4} m$

Let the centre of the circular disc be the origin of the system.
$\Rightarrow$ x₁ = d, y₁ = 0
x₂ = 0, y₂ = 0

$\Rightarrow$ Position of the centre of mass of circular disc and square plate:
$= (\frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}) = (\frac{d^{2} m d + π (d^{2} / 4) m \times 0}{d^{2} m + π (d^{2} / 4) m}, \frac{0 + 0}{m_{1} + m_{2}}) = (\frac{d^{3} m}{d^{2} m (1 + π / 4)}, 0) = (\frac{4 d}{π + 4}, 0)$
Hence, the new centre of mass of the system (circular disc plus square plate) lies at distance $\frac{4 d}{(π + 4)}$ from the centre of circular disc, towards right.

Answer:

From the figure, the velocities of different masses can be written as:

$For m_{1} = 1.0 kg, Velocity, {\vec{v}}_{1} = (- 1.5 \cos 37 ° \hat{i} - 1.5 \sin 37 ° \hat{j}) = - 1.2 \hat{i} - 0.9 \hat{j} For m_{2} = 1.2 kg, Velocity, {\vec{v}}_{2} = 0.4 \vec{j} For m_{3} = 1.5 kg, Velocity, {\vec{v}}_{3} = - 1.0 \cos 37 ° \hat{i} + 1.0 \sin 37 ° \hat{j} For m_{4} = 0.50 kg, Velocity, {\vec{v}}_{4} = 3.0 \hat{i} For m_{5} = 1.0 kg, Velocity, {\vec{v}}_{5} = 2.0 \cos 37 ° \hat{i} - 2.0 \sin 37 ° \hat{j} (\cos 37 ° = \frac{4}{5} and \sin 37 ° = \frac{3}{5}) V_{cm} = \frac{m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} + m_{3} {\vec{v}}_{3} + m_{4} {\vec{v}}_{4} + m_{5} {\vec{v}}_{5}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5}} = {\frac{1}{1.0 + 1.2 + 1.5 + 1.0 + 0.50}}^{[1.0 (- 1.5 \times \frac{4}{5} \vec{i} - 1.5 \times \frac{3}{5} \vec{j}) + . . . - 2.0 \times \frac{3}{5} \vec{j}]}$

On solving the above equation, we get:
V_cm is 0.20 m/s , at 45° below the direction, towards right.

Question 8:

From the figure, the velocities of different masses can be written as:

$For m_{1} = 1.0 kg, Velocity, {\vec{v}}_{1} = (- 1.5 \cos 37 ° \hat{i} - 1.5 \sin 37 ° \hat{j}) = - 1.2 \hat{i} - 0.9 \hat{j} For m_{2} = 1.2 kg, Velocity, {\vec{v}}_{2} = 0.4 \vec{j} For m_{3} = 1.5 kg, Velocity, {\vec{v}}_{3} = - 1.0 \cos 37 ° \hat{i} + 1.0 \sin 37 ° \hat{j} For m_{4} = 0.50 kg, Velocity, {\vec{v}}_{4} = 3.0 \hat{i} For m_{5} = 1.0 kg, Velocity, {\vec{v}}_{5} = 2.0 \cos 37 ° \hat{i} - 2.0 \sin 37 ° \hat{j} (\cos 37 ° = \frac{4}{5} and \sin 37 ° = \frac{3}{5}) V_{cm} = \frac{m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} + m_{3} {\vec{v}}_{3} + m_{4} {\vec{v}}_{4} + m_{5} {\vec{v}}_{5}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5}} = {\frac{1}{1.0 + 1.2 + 1.5 + 1.0 + 0.50}}^{[1.0 (- 1.5 \times \frac{4}{5} \vec{i} - 1.5 \times \frac{3}{5} \vec{j}) + . . . - 2.0 \times \frac{3}{5} \vec{j}]}$

On solving the above equation, we get:
V_cm is 0.20 m/s , at 45° below the direction, towards right.

Answer:

Let the two masses m₁ and m₂ be placed on the X-axis.
It is given that:
m₁ = 10 kg
m₂ = 20 kg

The first mass is displaced by a distance of 2 cm.
$∴ X_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} \Rightarrow X_{cm} = \frac{10 \times 2 + 20 x_{2}}{30}$

As the position of the centre of mass remains unchanged,
X_cm = 0

$\Rightarrow 0 = \frac{20 + 20 x_{2}}{30} \Rightarrow 20 + 20 x_{2} = 0 \Rightarrow 20 = - 20 x_{2} \Rightarrow x_{2} = - 1$

Therefore, to keep the position of centre of mass unchanged, the block of mass 20 kg should be moved by a distance of 1 cm, towards left.

Question 9:

Let the two masses m₁ and m₂ be placed on the X-axis.
It is given that:
m₁ = 10 kg
m₂ = 20 kg

The first mass is displaced by a distance of 2 cm.
$∴ X_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} \Rightarrow X_{cm} = \frac{10 \times 2 + 20 x_{2}}{30}$

As the position of the centre of mass remains unchanged,
X_cm = 0

$\Rightarrow 0 = \frac{20 + 20 x_{2}}{30} \Rightarrow 20 + 20 x_{2} = 0 \Rightarrow 20 = - 20 x_{2} \Rightarrow x_{2} = - 1$

Therefore, to keep the position of centre of mass unchanged, the block of mass 20 kg should be moved by a distance of 1 cm, towards left.

Answer:

Let the two masses m₁ and m₂ are kept along a vertical line.

It is given that:
m₁ = 10 kg
m₂ = 30 kg

First block is raised through a height of 7 cm.

Thus, to raise the centre of mass by 1 cm:

$Y cm = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} \Rightarrow 1 = \frac{10 \times 7 + 30 \times y_{2}}{40} \Rightarrow 70 + 30 y_{2} = 40 \Rightarrow 30 y_{2} = - 30 \Rightarrow y_{2} = - 1$

Therefore, the 30 kg block (m₂₎ should be moved by 1 cm downwards in order to raise the centre of mass by 1 cm.

Question 10:

Let the two masses m₁ and m₂ are kept along a vertical line.

It is given that:
m₁ = 10 kg
m₂ = 30 kg

First block is raised through a height of 7 cm.

Thus, to raise the centre of mass by 1 cm:

$Y cm = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} \Rightarrow 1 = \frac{10 \times 7 + 30 \times y_{2}}{40} \Rightarrow 70 + 30 y_{2} = 40 \Rightarrow 30 y_{2} = - 30 \Rightarrow y_{2} = - 1$

Therefore, the 30 kg block (m₂₎ should be moved by 1 cm downwards in order to raise the centre of mass by 1 cm.

Answer:

As there is no gravity or other external forces acting on the system, the melting ice tends to acquire a spherical shape. Therefore, the centre of mass of the system does not move.

Question 11:

As there is no gravity or other external forces acting on the system, the melting ice tends to acquire a spherical shape. Therefore, the centre of mass of the system does not move.

Answer:

Let the mass of the plate be M.
Consider a small semicircular portion of mass dm and radius r, as shown in fig.

$d m = \frac{M π r d r}{\frac{π (R_{2}^{2} - R_{1}^{2})}{2}} = \frac{M}{\frac{(R_{2}^{2} - R_{1}^{2})}{2}} r d r$
The centre of mass is given as:
$y_{c m} = \frac{\int y d m}{M}$

$y_{c m} = \int_{R_{1}}^{R_{2}} (\frac{2 r}{π}) . \frac{M}{\frac{(R_{2}^{2} - R_{1}^{2})}{2}} \times \frac{r d r}{M} = \frac{2}{π \frac{(R_{2}^{2} - R_{1}^{2})}{2}} \int_{R_{1}}^{R_{2}} r^{2} d r = \frac{2}{π \frac{(R_{2}^{2} - R_{1}^{2})}{2}} [\frac{(R_{2}^{3} - R_{1}^{3})}{3}] = \frac{4 (R_{1}^{2} + R_{1} R_{2} + R_{2}^{2})}{3 π (R_{1} + R_{2})}$

Question 12:

Let the mass of the plate be M.
Consider a small semicircular portion of mass dm and radius r, as shown in fig.

$d m = \frac{M π r d r}{\frac{π (R_{2}^{2} - R_{1}^{2})}{2}} = \frac{M}{\frac{(R_{2}^{2} - R_{1}^{2})}{2}} r d r$
The centre of mass is given as:
$y_{c m} = \frac{\int y d m}{M}$

$y_{c m} = \int_{R_{1}}^{R_{2}} (\frac{2 r}{π}) . \frac{M}{\frac{(R_{2}^{2} - R_{1}^{2})}{2}} \times \frac{r d r}{M} = \frac{2}{π \frac{(R_{2}^{2} - R_{1}^{2})}{2}} \int_{R_{1}}^{R_{2}} r^{2} d r = \frac{2}{π \frac{(R_{2}^{2} - R_{1}^{2})}{2}} [\frac{(R_{2}^{3} - R_{1}^{3})}{3}] = \frac{4 (R_{1}^{2} + R_{1} R_{2} + R_{2}^{2})}{3 π (R_{1} + R_{2})}$

Answer:

It is given that:
Mass of Mr. Verma, m₁ = 50 kg
Mass of Mr. Mathur, m₂ = 60 kg
Mass of boat, m₃ = 40 kg

Let A be the origin of the system (boat plus two men).

Initially, Mr. Verma and Mr. Mathur were at two extremes of the boat.

∴ Distance of the centre of mass:

$X_{cm} = \frac{m_{1} \times x_{1} + m_{2} \times x_{2} + m_{3} \times x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{60 \times 0 + 50 \times 4 + 40 \times 2}{60 + 50 + 40} = \frac{280}{150} = 1.87 m from A$

As no external force is experienced in longitudinal direction, the centre of mass would not shift.

Initially, the centre of mass lies at a distance of 2 m from A.
When the men move towards the middle of the boat, the centre of mass shifts and lies at 1.87 m from A.

Therefore, the shift in centre of mass = 2 − 1.87 = 0.13 m, towards right
Hence, the boat moves 13 cm or 0.13 m towards right.

Question 13:

It is given that:
Mass of Mr. Verma, m₁ = 50 kg
Mass of Mr. Mathur, m₂ = 60 kg
Mass of boat, m₃ = 40 kg

Let A be the origin of the system (boat plus two men).

Initially, Mr. Verma and Mr. Mathur were at two extremes of the boat.

∴ Distance of the centre of mass:

$X_{cm} = \frac{m_{1} \times x_{1} + m_{2} \times x_{2} + m_{3} \times x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{60 \times 0 + 50 \times 4 + 40 \times 2}{60 + 50 + 40} = \frac{280}{150} = 1.87 m from A$

As no external force is experienced in longitudinal direction, the centre of mass would not shift.

Initially, the centre of mass lies at a distance of 2 m from A.
When the men move towards the middle of the boat, the centre of mass shifts and lies at 1.87 m from A.

Therefore, the shift in centre of mass = 2 − 1.87 = 0.13 m, towards right
Hence, the boat moves 13 cm or 0.13 m towards right.

Answer:

The mass of the bob is m.
The mass of the cart is M.

Considering the bob falls at point A.

Initial distance of centre of mass of the system from P is given as
$x = \frac{m \times L + M \times 0}{M + m} = \frac{m}{M + m} L$

When the bob falls in the slot, the distance of centre of mass of the system from P becomes zero.
$∴ Shift in centre of mass = 0 - \frac{m L}{M + m} = - \frac{m L}{M + m} towards left = \frac{m L}{M + m} towards right$

Therefore, the cart moves a distance of $\frac{m L}{M + m}$ towards right.

Question 14:

The mass of the bob is m.
The mass of the cart is M.

Considering the bob falls at point A.

Initial distance of centre of mass of the system from P is given as
$x = \frac{m \times L + M \times 0}{M + m} = \frac{m}{M + m} L$

When the bob falls in the slot, the distance of centre of mass of the system from P becomes zero.
$∴ Shift in centre of mass = 0 - \frac{m L}{M + m} = - \frac{m L}{M + m} towards left = \frac{m L}{M + m} towards right$

Therefore, the cart moves a distance of $\frac{m L}{M + m}$ towards right.

Answer:

Given:
The mass of monkey is m.
The mass of balloon is M.

Initially, the monkey, balloon and the rope are at rest.

Let the centre of mass is at a point P.

When the monkey descends through a distance L,

The centre of mass shifts.
$l_{0} = \frac{m \times L + M \times 0}{M + m} = \frac{m L}{M + m} from P$

Therefore, the balloon descends through a distance $\frac{m L}{M + m}$ .

Question 15:

Given:
The mass of monkey is m.
The mass of balloon is M.

Initially, the monkey, balloon and the rope are at rest.

Let the centre of mass is at a point P.

When the monkey descends through a distance L,

The centre of mass shifts.
$l_{0} = \frac{m \times L + M \times 0}{M + m} = \frac{m L}{M + m} from P$

Therefore, the balloon descends through a distance $\frac{m L}{M + m}$ .

Answer:

Let the masses of the two particles be m₁ and m₂.

Given:
m₁ = 1 kg
m₂ = 4 kg

Now,
Kinetic energy of the first particle = Kinetic energy of the second particle
$(\frac{1}{2}) m_{1} v_{1}^{2} = (\frac{1}{2}) m_{2} v_{2}^{2} \Rightarrow \frac{m_{1}}{m_{2}} = \frac{v_{2}^{2}}{v_{1}^{2}} \Rightarrow \frac{v_{2}}{v_{1}} = \sqrt{\frac{m_{1}}{m_{2}}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{m_{2}}{m_{1}}} The ratio of linear momenta (m v) of the two particles, \frac{P_{1}}{P_{2}} = \frac{m_{1} v_{1}}{m_{2} v_{2}} = \frac{m_{1}}{m_{2}} \sqrt{\frac{m_{2}}{m_{1}}} = \sqrt{\frac{m_{1}}{m_{2}}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \Rightarrow P_{1} : P_{2} = 1 : 2$

Therefore, the ratio of linear momenta is 1:2.

Question 16:

Let the masses of the two particles be m₁ and m₂.

Given:
m₁ = 1 kg
m₂ = 4 kg

Now,
Kinetic energy of the first particle = Kinetic energy of the second particle
$(\frac{1}{2}) m_{1} v_{1}^{2} = (\frac{1}{2}) m_{2} v_{2}^{2} \Rightarrow \frac{m_{1}}{m_{2}} = \frac{v_{2}^{2}}{v_{1}^{2}} \Rightarrow \frac{v_{2}}{v_{1}} = \sqrt{\frac{m_{1}}{m_{2}}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{m_{2}}{m_{1}}} The ratio of linear momenta (m v) of the two particles, \frac{P_{1}}{P_{2}} = \frac{m_{1} v_{1}}{m_{2} v_{2}} = \frac{m_{1}}{m_{2}} \sqrt{\frac{m_{2}}{m_{1}}} = \sqrt{\frac{m_{1}}{m_{2}}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \Rightarrow P_{1} : P_{2} = 1 : 2$

Therefore, the ratio of linear momenta is 1:2.

Answer:

According to the question, uranium 238 nucleus emits an alpha-particle with a speed of 1.4 × 10⁷ m/s.

Let the speed of the residual nucleus thorium 234 be v₂.

By the law of conservation of linear momentum, we have:
$m_{1} v_{1} = m_{2} v_{2}$
Here, m₁ and v₁ are the mass and velocity of the alpha-particle repectively, and m₂ is the mass of the residual nucleus.

$\Rightarrow 4 \times 1.4 \times 10^{7} = 234 \times v_{2} v_{2} = \frac{4 \times 1.4 \times 10^{7}}{234} = 2.4 \times 10^{5} m / s$

Therefore, the speed of the residual nucleus is $2.4 \times 10^{5} m / s$ .

Question 17:

According to the question, uranium 238 nucleus emits an alpha-particle with a speed of 1.4 × 10⁷ m/s.

Let the speed of the residual nucleus thorium 234 be v₂.

By the law of conservation of linear momentum, we have:
$m_{1} v_{1} = m_{2} v_{2}$
Here, m₁ and v₁ are the mass and velocity of the alpha-particle repectively, and m₂ is the mass of the residual nucleus.

$\Rightarrow 4 \times 1.4 \times 10^{7} = 234 \times v_{2} v_{2} = \frac{4 \times 1.4 \times 10^{7}}{234} = 2.4 \times 10^{5} m / s$

Therefore, the speed of the residual nucleus is $2.4 \times 10^{5} m / s$ .

Answer:

By the law of conservation of linear momentum, we have:
$m_{1} v_{1} = m_{2} v_{2}$
Here, m₁ and v₁ are the mass and velocity of the man respectively, and m₂and v₂ are the mass and velocity of the Earth respectively.

$\Rightarrow 50 \times 1.8 = 6 \times 10^{24} \times v_{2} ∴ v_{2} = \frac{50 \times 1.8}{6 \times 10^{24}} = 15 \times 10^{- 24} m / s \Rightarrow v_{2} = 1.5 \times 10^{- 23} m / s$
Hence, the earth recoils with a speed of 1.5 × 10⁻²³ m/s.

Question 18:

By the law of conservation of linear momentum, we have:
$m_{1} v_{1} = m_{2} v_{2}$
Here, m₁ and v₁ are the mass and velocity of the man respectively, and m₂and v₂ are the mass and velocity of the Earth respectively.

$\Rightarrow 50 \times 1.8 = 6 \times 10^{24} \times v_{2} ∴ v_{2} = \frac{50 \times 1.8}{6 \times 10^{24}} = 15 \times 10^{- 24} m / s \Rightarrow v_{2} = 1.5 \times 10^{- 23} m / s$
Hence, the earth recoils with a speed of 1.5 × 10⁻²³ m/s.

Answer:

It is given that:
Mass of proton, m_p = 1.67 × 10⁻²⁷ kg
Momentum of electron = 1.4 × 10⁻²⁶ kg m/s
Momentum of antineutrino = 6.4 × 10⁻²⁷ kg m/s

Let the recoil speed of the proton be V_p.

(a) When the electron and the antineutrino are ejected in the same direction,

Applying the law of conservation of momentum, we get:

$m_{p} V_{p} = p_{electron} + p_{antineutrino} 1.67 \times 10^{- 27} \times V_{P} = 1.4 \times 10^{- 26} + 6.4 \times 10^{- 27} \Rightarrow V_{P} = (\frac{20.4}{1.67}) = 12.2 m / s, in the opposite direction$

(b) When the electron and the antineutrino are ejected perpendicular to each other,

Total momentum of electron and antineutrino is given as,

$p = \sqrt{p_{electron} + p_{antineutrino}} = \sqrt{(14)^{2} + (6.4)^{2}} \times 10^{- 27} = 15.4 \times 10^{- 27} kg m / s \Rightarrow V_{p} = \frac{15.4 \times 10^{- 27} kgm / s}{1.67 \times 10^{- 27} kg} \Rightarrow V_{p} = 9.2 m / s$

Question 19:

It is given that:
Mass of proton, m_p = 1.67 × 10⁻²⁷ kg
Momentum of electron = 1.4 × 10⁻²⁶ kg m/s
Momentum of antineutrino = 6.4 × 10⁻²⁷ kg m/s

Let the recoil speed of the proton be V_p.

(a) When the electron and the antineutrino are ejected in the same direction,

Applying the law of conservation of momentum, we get:

$m_{p} V_{p} = p_{electron} + p_{antineutrino} 1.67 \times 10^{- 27} \times V_{P} = 1.4 \times 10^{- 26} + 6.4 \times 10^{- 27} \Rightarrow V_{P} = (\frac{20.4}{1.67}) = 12.2 m / s, in the opposite direction$

(b) When the electron and the antineutrino are ejected perpendicular to each other,

Total momentum of electron and antineutrino is given as,

$p = \sqrt{p_{electron} + p_{antineutrino}} = \sqrt{(14)^{2} + (6.4)^{2}} \times 10^{- 27} = 15.4 \times 10^{- 27} kg m / s \Rightarrow V_{p} = \frac{15.4 \times 10^{- 27} kgm / s}{1.67 \times 10^{- 27} kg} \Rightarrow V_{p} = 9.2 m / s$

Answer:

Mass of man = M
Initial velocity of the man = 0
Mass of bag = m

Let the man throws the bag towards left with a velocity v and himself moves towards right with a velocity V.

Using the law of conservation of momentum,
$m v = M V \Rightarrow v = \frac{M V}{m} . . . (1) Let the total time he takes to reach ground be t_{1} . \Rightarrow t_{1} = \sqrt{\frac{2 H}{g}} Let the total time he takes to reach the height h be t_{2} . \Rightarrow t_{2} = \sqrt{\frac{2 (H - h)}{g}} ∴ The time of flying in covering the remaining height h is, t = t_{1} - t_{2} \Rightarrow t = \sqrt{\frac{2 H}{g}} - \sqrt{\frac{2 (H - h)}{g}} = \sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{H - h})$

During this time, the man covers a horizontal distance x and lands in the water.
$\Rightarrow x = V \times t \Rightarrow V = \frac{x}{t} ∴ v = \frac{M}{m} \frac{x}{t} [using equation (1)] = \frac{M}{m} \frac{x}{\sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{H - h})}$

Thus, the minimum horizontal velocity imparted to the bag, such that the man lands in the water is $\frac{M}{m} \frac{x}{\sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{H - h})}$ .

Let the bag lands at a distance x' towards left from actual line of fall.
As there is no external force in horizontal direction, the x-coordinate of the centre of mass will remain same.

$\Rightarrow 0 = \frac{M \times (x) + m \times x'}{M + m} \Rightarrow x' = - \frac{M}{m} x$
Therefore, the bag will land at a distance $\frac{M}{m} x$ .

Question 20:

Mass of man = M
Initial velocity of the man = 0
Mass of bag = m

Let the man throws the bag towards left with a velocity v and himself moves towards right with a velocity V.

Using the law of conservation of momentum,
$m v = M V \Rightarrow v = \frac{M V}{m} . . . (1) Let the total time he takes to reach ground be t_{1} . \Rightarrow t_{1} = \sqrt{\frac{2 H}{g}} Let the total time he takes to reach the height h be t_{2} . \Rightarrow t_{2} = \sqrt{\frac{2 (H - h)}{g}} ∴ The time of flying in covering the remaining height h is, t = t_{1} - t_{2} \Rightarrow t = \sqrt{\frac{2 H}{g}} - \sqrt{\frac{2 (H - h)}{g}} = \sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{H - h})$

During this time, the man covers a horizontal distance x and lands in the water.
$\Rightarrow x = V \times t \Rightarrow V = \frac{x}{t} ∴ v = \frac{M}{m} \frac{x}{t} [using equation (1)] = \frac{M}{m} \frac{x}{\sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{H - h})}$

Thus, the minimum horizontal velocity imparted to the bag, such that the man lands in the water is $\frac{M}{m} \frac{x}{\sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{H - h})}$ .

Let the bag lands at a distance x' towards left from actual line of fall.
As there is no external force in horizontal direction, the x-coordinate of the centre of mass will remain same.

$\Rightarrow 0 = \frac{M \times (x) + m \times x'}{M + m} \Rightarrow x' = - \frac{M}{m} x$
Therefore, the bag will land at a distance $\frac{M}{m} x$ .

Answer:

It is given that:
Mass of the ball = 50 g = $0.05 Kg$
Speed of the ball, v = 2.0 m/s
Incident angle = 45Ëš

$v = 2 \cos 45 ° \hat{i} - 2 \sin 45 ° \hat{j} v' = - 2 \cos 45 ° \hat{i} - 2 \sin 45 ° \hat{j} (a) Change in momentum = m \vec{v} - m \vec{v}' = 0.05 (2 \cos 45 ° \hat{i} - 2 \sin 45 ° \hat{j}) - 0.05 (- 2 \cos 45 ° \hat{i} - 2 \sin 45 ° \hat{j}) = 0.1 \cos 45 ° \hat{i} - 0.1 \sin 45 ° \hat{j} + 0.1 \cos 45 ° \hat{i} + 0.1 \sin 45 ° \hat{j} = 0.2 \cos 45 ° \hat{i} ∴ Magnitude = \frac{0.2}{\sqrt{2}} = 0.14 Kg m / s$

(b) The change in magnitude of the momentum of the ball,

$|{\vec{P}}_{2}| - |{\vec{P}}_{1}| = 2 \times 0.5 - 2 \times 0.5 = 0$

i.e. There is no change in magnitude of the momentum of the ball.

Question 21:

It is given that:
Mass of the ball = 50 g = $0.05 Kg$
Speed of the ball, v = 2.0 m/s
Incident angle = 45Ëš

$v = 2 \cos 45 ° \hat{i} - 2 \sin 45 ° \hat{j} v' = - 2 \cos 45 ° \hat{i} - 2 \sin 45 ° \hat{j} (a) Change in momentum = m \vec{v} - m \vec{v}' = 0.05 (2 \cos 45 ° \hat{i} - 2 \sin 45 ° \hat{j}) - 0.05 (- 2 \cos 45 ° \hat{i} - 2 \sin 45 ° \hat{j}) = 0.1 \cos 45 ° \hat{i} - 0.1 \sin 45 ° \hat{j} + 0.1 \cos 45 ° \hat{i} + 0.1 \sin 45 ° \hat{j} = 0.2 \cos 45 ° \hat{i} ∴ Magnitude = \frac{0.2}{\sqrt{2}} = 0.14 Kg m / s$

(b) The change in magnitude of the momentum of the ball,

$|{\vec{P}}_{2}| - |{\vec{P}}_{1}| = 2 \times 0.5 - 2 \times 0.5 = 0$

i.e. There is no change in magnitude of the momentum of the ball.

Answer:

It is given that:
Wavelength of light = λ
Momentum of each photon = h/λ
Angle of incidence = θ

${\vec{P}}_{I n c i d e n c e} = (\frac{h}{λ}) \cos θ \hat{i} - (\frac{h}{λ}) \sin θ \hat{j} {\vec{P}}_{Reflected} = - (\frac{h}{λ}) \cos θ \hat{i} - (\frac{h}{λ}) \sin θ \hat{j} The change in momentum will only be in the direction of x - axi s i . e ., |ΔP| = (\frac{h}{λ}) \cos θ - (- \frac{h}{λ} \cos θ) = (\frac{2 h}{λ}) \cos θ$

Question 22:

It is given that:
Wavelength of light = λ
Momentum of each photon = h/λ
Angle of incidence = θ

${\vec{P}}_{I n c i d e n c e} = (\frac{h}{λ}) \cos θ \hat{i} - (\frac{h}{λ}) \sin θ \hat{j} {\vec{P}}_{Reflected} = - (\frac{h}{λ}) \cos θ \hat{i} - (\frac{h}{λ}) \sin θ \hat{j} The change in momentum will only be in the direction of x - axi s i . e ., |ΔP| = (\frac{h}{λ}) \cos θ - (- \frac{h}{λ} \cos θ) = (\frac{2 h}{λ}) \cos θ$

Answer:

As the block is exploded only because of its internal energy, the net external force on the system is zero.
Thus, the centre of mass of does not change.

Let the body was at the origin of the co-ordinate system during explosion.

Resultant velocity of two bodies of equal mass moving at a speed of 10 m/s in + x-axis and + y-axis direction, is given as:
$v = \sqrt{10^{2} + 10^{2} + 2.10.10 \cos 90 °} = 10 \sqrt{2} m / s, 45 ° w . r . t x - axis$

If the centre of mass is at rest, the third part having equal mass as that of the other two masses will move in the opposite direction (i.e. $135 °$ w.r.t. +x-axis) at the same velocity of $10 \sqrt{2} m / s .$ .

Question 23:

As the block is exploded only because of its internal energy, the net external force on the system is zero.
Thus, the centre of mass of does not change.

Let the body was at the origin of the co-ordinate system during explosion.

Resultant velocity of two bodies of equal mass moving at a speed of 10 m/s in + x-axis and + y-axis direction, is given as:
$v = \sqrt{10^{2} + 10^{2} + 2.10.10 \cos 90 °} = 10 \sqrt{2} m / s, 45 ° w . r . t x - axis$

If the centre of mass is at rest, the third part having equal mass as that of the other two masses will move in the opposite direction (i.e. $135 °$ w.r.t. +x-axis) at the same velocity of $10 \sqrt{2} m / s .$ .

Answer:

According to the question, the spaceship is removed from all other material objects and kept totally isolated from the surroundings. Thus, the mass loss by astronauts couldn't slip away from the spaceship. Therefore, the total mass of the spaceship remains unchanged and so does its velocity.

Hence, the spaceship moves with the speed of 15 km/s.

Question 24:

According to the question, the spaceship is removed from all other material objects and kept totally isolated from the surroundings. Thus, the mass loss by astronauts couldn't slip away from the spaceship. Therefore, the total mass of the spaceship remains unchanged and so does its velocity.

Hence, the spaceship moves with the speed of 15 km/s.

Answer:

It is given that:
Diameter of hailstone = 1 cm = 0.01 m
⇒ Radius of hailstone, r = 0.005 m

Average speed of hailstone = 20 m/s
Density of hailstone = 900 kg/m³ = 0.9 g/cm³

$Volume of the hailstones is given as, V = \frac{4}{3} π r^{3} \Rightarrow V = \frac{4}{3} π (0.005)^{3} = 5.235 \times 10^{- 7} m^{3} Mass = v olume \times density = 5.235 \times 10^{- 7} \times 900 = 4.711 \times 10^{- 4} kg ∴ Mass of 2000 hailstone = 2000 \times 4.711 \times 10^{- 4} = 0.9422 Rate of change of momentum = 0.9422 \times 20 \approx 19 N / m^{2}$

$∴$ The total force exerted on the roof = $19 \times 100 = 1900 N$

Question 25:

It is given that:
Diameter of hailstone = 1 cm = 0.01 m
⇒ Radius of hailstone, r = 0.005 m

Average speed of hailstone = 20 m/s
Density of hailstone = 900 kg/m³ = 0.9 g/cm³

$Volume of the hailstones is given as, V = \frac{4}{3} π r^{3} \Rightarrow V = \frac{4}{3} π (0.005)^{3} = 5.235 \times 10^{- 7} m^{3} Mass = v olume \times density = 5.235 \times 10^{- 7} \times 900 = 4.711 \times 10^{- 4} kg ∴ Mass of 2000 hailstone = 2000 \times 4.711 \times 10^{- 4} = 0.9422 Rate of change of momentum = 0.9422 \times 20 \approx 19 N / m^{2}$

$∴$ The total force exerted on the roof = $19 \times 100 = 1900 N$

Answer:

It is given that the mass of the ball is m.

Let the ball be dropped from a height h.

The speed of ball before the collision is v₁.
$∴ v_{1} = \sqrt{2 g h}$
The speed of ball after the collision is v₂.
$v_{2} = - \sqrt{2 g h} Rate of change of velocity = acceleration \Rightarrow a = \frac{2 \sqrt{2 g h}}{t} ∴ Force, F = \frac{m \times 2 \sqrt{2 g h}}{t} \dots (1)$

Using Newton's laws of motion, we can write:
$v = \sqrt{2 g h}, s = h, u = 0 \Rightarrow \sqrt{2 g h} = g t$
$\Rightarrow t = \sqrt{\frac{2 h}{g}} ∴ Total time = 2 \sqrt{\frac{2 h}{g}}$
Substituting this value of time t in equation (1), we get:
F = mg

Question 26:

It is given that the mass of the ball is m.

Let the ball be dropped from a height h.

The speed of ball before the collision is v₁.
$∴ v_{1} = \sqrt{2 g h}$
The speed of ball after the collision is v₂.
$v_{2} = - \sqrt{2 g h} Rate of change of velocity = acceleration \Rightarrow a = \frac{2 \sqrt{2 g h}}{t} ∴ Force, F = \frac{m \times 2 \sqrt{2 g h}}{t} \dots (1)$

Using Newton's laws of motion, we can write:
$v = \sqrt{2 g h}, s = h, u = 0 \Rightarrow \sqrt{2 g h} = g t$
$\Rightarrow t = \sqrt{\frac{2 h}{g}} ∴ Total time = 2 \sqrt{\frac{2 h}{g}}$
Substituting this value of time t in equation (1), we get:
F = mg

Answer:

Given:
The mass of the railroad car is M.
The mass of the man is m.

The car recoils with a speed v, backwards on the rails.

Let the man of mass m approaches towards the engine with a velocity v' w.r.t the engine.

∴ The velocity of man w.r.t earth is v' − v, towards right.

$V_{centre of mass} = 0 (Initially at rest) ∴ 0 = - M v + m (v' - v) \Rightarrow M v = m (v' - v) \Rightarrow m v' = M v + m v \Rightarrow v' = (\frac{M + m}{m}) v \Rightarrow v' = (1 + \frac{M}{m}) v$

Question 27:

Given:
The mass of the railroad car is M.
The mass of the man is m.

The car recoils with a speed v, backwards on the rails.

Let the man of mass m approaches towards the engine with a velocity v' w.r.t the engine.

∴ The velocity of man w.r.t earth is v' − v, towards right.

$V_{centre of mass} = 0 (Initially at rest) ∴ 0 = - M v + m (v' - v) \Rightarrow M v = m (v' - v) \Rightarrow m v' = M v + m v \Rightarrow v' = (\frac{M + m}{m}) v \Rightarrow v' = (1 + \frac{M}{m}) v$

Answer:

It is given that:
Mass of the car, the gun, the shells and the operator = 50m
Mass of one shell = m
Muzzle velocity of the shells, v = 200 m/s

Let the speed of car be v.

On applying the law of conservation of linear momentum, we get:

$49 m V + m v = 0 \Rightarrow 49 m \times V + m \times 200 = 0 \Rightarrow V = - \frac{200}{49} m / s (' -' sign indicates direction towards left)$

Thus, when another shell is fired, the velocity of the car with respect to the platform is,
$V = \frac{200}{49} m / s, towards left$

When one more shell is fired, the velocity of the car with respect to the platform is,
$V_{1} = \frac{200}{48} m / s, towards left$
∴ Velocity of the car w.r.t the earth $= 200 (\frac{1}{49} + \frac{1}{48}) m / s$

Question 28:

It is given that:
Mass of the car, the gun, the shells and the operator = 50m
Mass of one shell = m
Muzzle velocity of the shells, v = 200 m/s

Let the speed of car be v.

On applying the law of conservation of linear momentum, we get:

$49 m V + m v = 0 \Rightarrow 49 m \times V + m \times 200 = 0 \Rightarrow V = - \frac{200}{49} m / s (' -' sign indicates direction towards left)$

Thus, when another shell is fired, the velocity of the car with respect to the platform is,
$V = \frac{200}{49} m / s, towards left$

When one more shell is fired, the velocity of the car with respect to the platform is,
$V_{1} = \frac{200}{48} m / s, towards left$
∴ Velocity of the car w.r.t the earth $= 200 (\frac{1}{49} + \frac{1}{48}) m / s$

Answer:

It is given that:
Mass of each persons = m
Mass of railroad car = M

Let the velocity of the railroad w.r.t. earth, when the man on the left jumps off be V.

By the law of conservation of momentum:

$0 = - m u + (M + m) V \Rightarrow V = (\frac{m u}{M + m}), towards right$

When the man on the right jumps, his velocity w.r.t. the car is u.

$0 = m u - M V' \Rightarrow V' = \frac{m u}{M}$

(V is the change in velocity of the platform when the platform itself is taken as reference, assuming the car to be at rest.)

∴ Net velocity towards left, (i.e. the velocity of the car)

$V' - V = \frac{m u}{M} - \frac{m u}{(M + m)} = \frac{m M u + m^{2} u - M m u}{M (M + m)} \Rightarrow V' - V = \frac{m^{2} u}{M (M + m)}$

Question 29:

It is given that:
Mass of each persons = m
Mass of railroad car = M

Let the velocity of the railroad w.r.t. earth, when the man on the left jumps off be V.

By the law of conservation of momentum:

$0 = - m u + (M + m) V \Rightarrow V = (\frac{m u}{M + m}), towards right$

When the man on the right jumps, his velocity w.r.t. the car is u.

$0 = m u - M V' \Rightarrow V' = \frac{m u}{M}$

(V is the change in velocity of the platform when the platform itself is taken as reference, assuming the car to be at rest.)

∴ Net velocity towards left, (i.e. the velocity of the car)

$V' - V = \frac{m u}{M} - \frac{m u}{(M + m)} = \frac{m M u + m^{2} u - M m u}{M (M + m)} \Rightarrow V' - V = \frac{m^{2} u}{M (M + m)}$

Answer:

Given:
The mass of the small block is m.
Initial speed of this block is v.

The mass of the bigger block is M.
Initial speed of this block is zero.

At point A, as the small block transfers its momentum to the bigger block, both the blocks move with a common velocity V (say) in the same direction as v.

Using the law of conservation of linear momentum, we can write:
Initial momentum = final momentum

$m v + M \times O = (m + M) V$
$\Rightarrow V = \frac{m v}{m + M}$

Therefore, the speed of the bigger block when the smaller block reaches point A of the surface is $\frac{m v}{m + M}$ .

Question 30:

Given:
The mass of the small block is m.
Initial speed of this block is v.

The mass of the bigger block is M.
Initial speed of this block is zero.

At point A, as the small block transfers its momentum to the bigger block, both the blocks move with a common velocity V (say) in the same direction as v.

Using the law of conservation of linear momentum, we can write:
Initial momentum = final momentum

$m v + M \times O = (m + M) V$
$\Rightarrow V = \frac{m v}{m + M}$

Therefore, the speed of the bigger block when the smaller block reaches point A of the surface is $\frac{m v}{m + M}$ .

Answer:

It is given that:
Mass of the bugghi, m_b= 200 kg
Velocity of the bugghi, V_b= 10 km/h
Mass of the boy, m_boy = 25 kg
Velocity of the boy, V_Boy = 4 km/h

Consider the boy and the bugghi as a system.

The total momentum before the process of sitting remains same after the process of sitting.

Using the law of conservation of momentum, we can write:

$m_{b} V_{b} + m_{boy} V_{boy} = (m_{b} + m_{boy}) V \Rightarrow 200 \times 10 + 25 \times 4 = (200 + 25) \times V \Rightarrow V = \frac{2100}{225} = \frac{28}{3} Km / h$

Question 31:

It is given that:
Mass of the bugghi, m_b= 200 kg
Velocity of the bugghi, V_b= 10 km/h
Mass of the boy, m_boy = 25 kg
Velocity of the boy, V_Boy = 4 km/h

Consider the boy and the bugghi as a system.

The total momentum before the process of sitting remains same after the process of sitting.

Using the law of conservation of momentum, we can write:

$m_{b} V_{b} + m_{boy} V_{boy} = (m_{b} + m_{boy}) V \Rightarrow 200 \times 10 + 25 \times 4 = (200 + 25) \times V \Rightarrow V = \frac{2100}{225} = \frac{28}{3} Km / h$

Answer:

It is given that:
Speed of the ball, v₁ = 5.0 m/s
Mass of the ball, m₁ = 0.5 kg
Mass of another ball, m₂ = 1 kg
Let the velocity of this ball be v₂ m/s.

On applying the law of conservation of momentum, we get:

$m_{1} v_{1} + m_{2} v_{2} = 0 0.5 \times 5 + 1 \times v_{2} = 0 \Rightarrow v_{2} = - 2.5 m / s$
Hence, the velocity of second ball is 2.5 m/s, opposite to the direction of motion of the first ball.

Question 32:

It is given that:
Speed of the ball, v₁ = 5.0 m/s
Mass of the ball, m₁ = 0.5 kg
Mass of another ball, m₂ = 1 kg
Let the velocity of this ball be v₂ m/s.

On applying the law of conservation of momentum, we get:

$m_{1} v_{1} + m_{2} v_{2} = 0 0.5 \times 5 + 1 \times v_{2} = 0 \Rightarrow v_{2} = - 2.5 m / s$
Hence, the velocity of second ball is 2.5 m/s, opposite to the direction of motion of the first ball.

Answer:

It is given that:
Mass of the skater who is skating, m₁ = 60 kg
Initial speed of this man, v₁ = 10 m/s
Mass of the skater at rest, m₂ = 40 kg
Initial speedof this man, v₂ = 0

Let the velocity of both men after collision be v.

Using the law of conservation of momentum, we can write:

$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v \Rightarrow 60 \times 10 + 0 = 100 \times v \Rightarrow v = 6 m / s The loss in kinetic energy during collision is given as, ∆ K E = \frac{1}{2} m_{1} v_{1}^{2} - \frac{1}{2} (m_{1} + m_{2}) v^{2} \Rightarrow ∆ K E = \frac{1}{2} \times 60 \times (10)^{2} - \frac{1}{2} \times 100 \times 36 \Rightarrow ∆ K E = 1200 J$

Question 33:

It is given that:
Mass of the skater who is skating, m₁ = 60 kg
Initial speed of this man, v₁ = 10 m/s
Mass of the skater at rest, m₂ = 40 kg
Initial speedof this man, v₂ = 0

Let the velocity of both men after collision be v.

Using the law of conservation of momentum, we can write:

$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v \Rightarrow 60 \times 10 + 0 = 100 \times v \Rightarrow v = 6 m / s The loss in kinetic energy during collision is given as, ∆ K E = \frac{1}{2} m_{1} v_{1}^{2} - \frac{1}{2} (m_{1} + m_{2}) v^{2} \Rightarrow ∆ K E = \frac{1}{2} \times 60 \times (10)^{2} - \frac{1}{2} \times 100 \times 36 \Rightarrow ∆ K E = 1200 J$

Answer:

It is given that:
Speed of the first particle during collision, $v (t) = u_{1} + \frac{t}{∆ t} (v_{1} - u_{1})$

Let v' be the speed of the second particle, during collision.

On applying the law of conservation of linear momentum on both particles, we get:

m₁u₁ + m₂u₂ = m₁v(t) + m₂v'
$\Rightarrow m_{1} u_{1} + m_{2} u_{2} = m_{1} u_{1} + m_{1} \times (\frac{t}{∆ t}) (v_{1} - u_{1}) + m_{2} v' On dividing both the sides by m_{2}, we get : u_{2} = \frac{m_{1}}{m_{2}} (\frac{t}{∆ t}) (v_{1} - u_{1}) + v' \Rightarrow v' = u_{2} - \frac{m_{1}}{m_{2}} (\frac{t}{∆ t}) (v_{1} - u_{1})$

The speed of the second particle during collision can be written as a function of time and is given by the expression, $u_{2} - \frac{m_{1}}{m_{2}} (\frac{t}{∆ t}) (v_{1} - u_{1})$ .

Question 34:

It is given that:
Speed of the first particle during collision, $v (t) = u_{1} + \frac{t}{∆ t} (v_{1} - u_{1})$

Let v' be the speed of the second particle, during collision.

On applying the law of conservation of linear momentum on both particles, we get:

m₁u₁ + m₂u₂ = m₁v(t) + m₂v'
$\Rightarrow m_{1} u_{1} + m_{2} u_{2} = m_{1} u_{1} + m_{1} \times (\frac{t}{∆ t}) (v_{1} - u_{1}) + m_{2} v' On dividing both the sides by m_{2}, we get : u_{2} = \frac{m_{1}}{m_{2}} (\frac{t}{∆ t}) (v_{1} - u_{1}) + v' \Rightarrow v' = u_{2} - \frac{m_{1}}{m_{2}} (\frac{t}{∆ t}) (v_{1} - u_{1})$

The speed of the second particle during collision can be written as a function of time and is given by the expression, $u_{2} - \frac{m_{1}}{m_{2}} (\frac{t}{∆ t}) (v_{1} - u_{1})$ .

Answer:

Given:
The mass of bullet moving with speed v is m.
The mass of the ball is M and it is at rest.

m' is the fractional mass of the ball that sticks with the bullet.
The remaining mass of the ball moves with the velocity v₁.

Let v₂ be the final velocity of the bullet plus fractional mass system.

On applying the law of conservation of momentum, we get:

mv + 0 = (m' + m)v₂ + (M − m') v₁

$\Rightarrow v_{2} = \frac{m v - (M - m') v_{1}}{m + m'}$
Therefore, the velocity of the bullet after the collision is $\frac{m v - (M - m') v_{1}}{m + m'}$ .

Question 35:

Given:
The mass of bullet moving with speed v is m.
The mass of the ball is M and it is at rest.

m' is the fractional mass of the ball that sticks with the bullet.
The remaining mass of the ball moves with the velocity v₁.

Let v₂ be the final velocity of the bullet plus fractional mass system.

On applying the law of conservation of momentum, we get:

mv + 0 = (m' + m)v₂ + (M − m') v₁

$\Rightarrow v_{2} = \frac{m v - (M - m') v_{1}}{m + m'}$
Therefore, the velocity of the bullet after the collision is $\frac{m v - (M - m') v_{1}}{m + m'}$ .

Answer:

Given:
The mass of the both balls is m.
Initial speed of first ball = v
Initial speed of second ball = 0

Let the final velocities of balls be v₁ and v₂ respectively.

$e = \frac{velocity of separation}{velocity of approach} \Rightarrow e = \frac{v_{1} - v_{2}}{v} \Rightarrow v_{1} - v_{2} = e v . . . (1)$

On applying the law of conservation of linear momentum, we get:

$m (v_{1} + v_{2}) = m v \Rightarrow v_{1} + v_{2} = v . . . (2) According to the given condition, Final K . E . = \frac{3}{4} Initial K . E . \Rightarrow \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} = \frac{3}{4} \times \frac{1}{2} m v^{2} \Rightarrow v_{1}^{2} + v_{2}^{2} = \frac{3}{4} v^{2} \Rightarrow \frac{(v_{1} + v_{2})^{2} + (v_{1} - v_{2})^{2}}{2} = \frac{3}{4} v^{2} \Rightarrow \frac{(1 + e^{2}) v^{2}}{2} = \frac{3}{4} v^{2} [using the equations (1) and (2)] \Rightarrow 1 + e^{2} = (\frac{3}{2}) \Rightarrow e^{2} = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$
Hence, the coefficient of restitution is found to be $\frac{1}{\sqrt{2}}$ .

Question 36:

Given:
The mass of the both balls is m.
Initial speed of first ball = v
Initial speed of second ball = 0

Let the final velocities of balls be v₁ and v₂ respectively.

$e = \frac{velocity of separation}{velocity of approach} \Rightarrow e = \frac{v_{1} - v_{2}}{v} \Rightarrow v_{1} - v_{2} = e v . . . (1)$

On applying the law of conservation of linear momentum, we get:

$m (v_{1} + v_{2}) = m v \Rightarrow v_{1} + v_{2} = v . . . (2) According to the given condition, Final K . E . = \frac{3}{4} Initial K . E . \Rightarrow \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} = \frac{3}{4} \times \frac{1}{2} m v^{2} \Rightarrow v_{1}^{2} + v_{2}^{2} = \frac{3}{4} v^{2} \Rightarrow \frac{(v_{1} + v_{2})^{2} + (v_{1} - v_{2})^{2}}{2} = \frac{3}{4} v^{2} \Rightarrow \frac{(1 + e^{2}) v^{2}}{2} = \frac{3}{4} v^{2} [using the equations (1) and (2)] \Rightarrow 1 + e^{2} = (\frac{3}{2}) \Rightarrow e^{2} = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$
Hence, the coefficient of restitution is found to be $\frac{1}{\sqrt{2}}$ .

Answer:

It is given that:
Mass of first block, m₁= 2 kg
Initial speed,v₁ = 2.0 m/s
Mass of second block, m₂ = 2 kg
Initial speed of this block = 0

For maximum possible loss in kinetic energy, we assume that the collision is elastic and both the blocks move with same final velocity v (say).

On applying the law of conservation of linear momentum, we get:

m₁v₁ + m₂ ×0 = (m₁+m₂)v
2 × 2 = (2 + 2)v
⇒ v = 1 m/s

Loss in K.E. in elastic collision is give by,
$= \frac{1}{2} m_{1} v_{1}^{2} - \frac{1}{2} (m_{1} + m_{2}) v = (\frac{1}{2}) \times 2 \times 2^{2} - (\frac{1}{2}) (2 + 2) \times (1)^{2} = 4 - 2 = 2 J$

$(b) The actual loss in K . E . = \frac{Maximum loss in K . E .}{2} = 1 J$

Let the final velocities of the blocks be v₁ and v₂ respectively.
The coefficient of restitution is e.

$∴$ The loss in K.E. is given by,

$(\frac{1}{2}) \times 2 \times (2)^{2} - (\frac{1}{2}) 2 \times v_{1}^{2} - (\frac{1}{2}) \times 2 v_{2}^{2} = 1 \Rightarrow 4 - (v_{1}^{2} + v_{2}^{2}) = 1 \Rightarrow 4 - \frac{(1 + e^{2}) \times 4}{2} = 1 \Rightarrow 2 (1 + e^{2}) = 3 \Rightarrow 1 + e^{2} = \frac{3}{2} \Rightarrow e^{2} = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$

Question 37:

It is given that:
Mass of first block, m₁= 2 kg
Initial speed,v₁ = 2.0 m/s
Mass of second block, m₂ = 2 kg
Initial speed of this block = 0

For maximum possible loss in kinetic energy, we assume that the collision is elastic and both the blocks move with same final velocity v (say).

On applying the law of conservation of linear momentum, we get:

m₁v₁ + m₂ ×0 = (m₁+m₂)v
2 × 2 = (2 + 2)v
⇒ v = 1 m/s

Loss in K.E. in elastic collision is give by,
$= \frac{1}{2} m_{1} v_{1}^{2} - \frac{1}{2} (m_{1} + m_{2}) v = (\frac{1}{2}) \times 2 \times 2^{2} - (\frac{1}{2}) (2 + 2) \times (1)^{2} = 4 - 2 = 2 J$

$(b) The actual loss in K . E . = \frac{Maximum loss in K . E .}{2} = 1 J$

Let the final velocities of the blocks be v₁ and v₂ respectively.
The coefficient of restitution is e.

$∴$ The loss in K.E. is given by,

$(\frac{1}{2}) \times 2 \times (2)^{2} - (\frac{1}{2}) 2 \times v_{1}^{2} - (\frac{1}{2}) \times 2 v_{2}^{2} = 1 \Rightarrow 4 - (v_{1}^{2} + v_{2}^{2}) = 1 \Rightarrow 4 - \frac{(1 + e^{2}) \times 4}{2} = 1 \Rightarrow 2 (1 + e^{2}) = 3 \Rightarrow 1 + e^{2} = \frac{3}{2} \Rightarrow e^{2} = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$

Answer:

It is given that:
Mass of particles = 100 g
Initial speed of the first particle = u
Final K.E. of the system after collision = 0.2J

Initial K.E. of the system, before collision = $\frac{1}{2} m u^{2} + 0$
i.e. Initial K.E. = $\frac{1}{2} \times 0.1 \times u^{2} = 0.05 u^{2}$

Let v₁ and v₂ be the final velocities of the first and second block respectively.

By law of conservation of momentum, we know:

$m v_{1} + m v_{2} = m u$
$\Rightarrow v_{1} + v_{2} = u . . . (1) (v_{1} - v_{2}) + e (u_{1} - u_{2}) = 0 \Rightarrow e u = v_{2} - v_{1} . . . (2) [Putting u_{2} = 0, u_{1} = u] Adding the equations (1) and (2), we get : 2 v_{2} = (1 + e) u \Rightarrow v_{2} = (\frac{u}{2}) (1 + e) ∴ v_{1} = u - \frac{u}{2} (1 + e) v_{1} = \frac{u}{2} (1 - e) According to given condition, \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} = 0.2 \Rightarrow v_{1}^{2} + v_{2}^{2} = 4 \Rightarrow \frac{u^{2}}{2} (1 + e^{2}) = 4 \Rightarrow u^{2} = \frac{8}{1 + e^{2}}$

For maximum value of u, denominator should be minimum in the above equation.
i.e. e = 0
⇒ u² = 8

$\Rightarrow u = 2 \sqrt{2} m / s$

For minimum value of u, denominator should have maximum value.
i.e. e = 1
⇒ u²= 4

⇒ u = 2 m/s

Question 38:

It is given that:
Mass of particles = 100 g
Initial speed of the first particle = u
Final K.E. of the system after collision = 0.2J

Initial K.E. of the system, before collision = $\frac{1}{2} m u^{2} + 0$
i.e. Initial K.E. = $\frac{1}{2} \times 0.1 \times u^{2} = 0.05 u^{2}$

Let v₁ and v₂ be the final velocities of the first and second block respectively.

By law of conservation of momentum, we know:

$m v_{1} + m v_{2} = m u$
$\Rightarrow v_{1} + v_{2} = u . . . (1) (v_{1} - v_{2}) + e (u_{1} - u_{2}) = 0 \Rightarrow e u = v_{2} - v_{1} . . . (2) [Putting u_{2} = 0, u_{1} = u] Adding the equations (1) and (2), we get : 2 v_{2} = (1 + e) u \Rightarrow v_{2} = (\frac{u}{2}) (1 + e) ∴ v_{1} = u - \frac{u}{2} (1 + e) v_{1} = \frac{u}{2} (1 - e) According to given condition, \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} = 0.2 \Rightarrow v_{1}^{2} + v_{2}^{2} = 4 \Rightarrow \frac{u^{2}}{2} (1 + e^{2}) = 4 \Rightarrow u^{2} = \frac{8}{1 + e^{2}}$

For maximum value of u, denominator should be minimum in the above equation.
i.e. e = 0
⇒ u² = 8

$\Rightarrow u = 2 \sqrt{2} m / s$

For minimum value of u, denominator should have maximum value.
i.e. e = 1
⇒ u²= 4

⇒ u = 2 m/s

Answer:

It is given that:
Weight of A = Weight of B = 40 kg
Velocity of ball = 5 m/s

(a) Case-1: Total momentum of the man A and ball remains constant.

∴ 0 = 4 × 5 − 40 × v
⇒ v = 0.5 m/s, towards left

(b) Case-2: When B catches the ball, the momentum between B and the ball remains constant.

⇒ 4 × 5 = 44 v
⇒ $v = (\frac{20}{44}) m / s$

Case-3: When B throws the ball,

On applying the law of conservation of linear momentum, we get:

$\Rightarrow 44 \times (\frac{20}{44}) = - 4 \times 5 + 40 \times v \Rightarrow v = 1 m / s, (towards right)$

Case-4: When A catches the ball,

Applying the law of conservation of liner momentum, we get:

$- 4 \times 5 + (- 0.5) \times 40 = 44 v \Rightarrow v = \frac{40}{44} = \frac{10}{11} m / s, towards left$

(c) Case-5: When A throws the ball,

Applying the law of conservation of linear momentum, we get:

$44 \times (\frac{10}{11}) = 4 \times 5 + 40 \times v \Rightarrow v = \frac{60}{40} = \frac{3}{2} m / s (towards left)$

Case-6: When B receives the ball,

Applying the law of conservation of linear momentum, we get:

$40 \times 1 + 4 \times 5 = 44 \times v \Rightarrow v = \frac{60}{44} m / s, towards right$

Case-7: When B throws the ball,

On applying the law of conservation of linear momentum, we get:

$\Rightarrow v = 44 \times (\frac{60}{44}) m / s, towards right$

Case-8: When A catches the ball,

On applying the law of conservation of linear momentum, we get:
$- 4 \times 5 + 40 (\frac{3}{2}) = - 44 v \Rightarrow V = - \frac{40}{44} = (\frac{10}{11}) m / s, towards left$

Similarly, after 5 round trips,
The velocity of A will be $(\frac{50}{11})$ m/s and the velocity of B will be 5 m/s.

(d) As after 6 round trips, the velocity of A becomes $\frac{60}{11}$ i.e. > 5 m/s, it cannot catch the ball. Thus, A can only roll the ball six times.

(e) Let the ball and the body A be at origin, in the initial position.

$∴ X_{c} = \frac{40 \times 0 + 4 \times 0 + 40 \times d}{40 + 40 + 4} = \frac{10}{21} d$

Question 39:

It is given that:
Weight of A = Weight of B = 40 kg
Velocity of ball = 5 m/s

(a) Case-1: Total momentum of the man A and ball remains constant.

∴ 0 = 4 × 5 − 40 × v
⇒ v = 0.5 m/s, towards left

(b) Case-2: When B catches the ball, the momentum between B and the ball remains constant.

⇒ 4 × 5 = 44 v
⇒ $v = (\frac{20}{44}) m / s$

Case-3: When B throws the ball,

On applying the law of conservation of linear momentum, we get:

$\Rightarrow 44 \times (\frac{20}{44}) = - 4 \times 5 + 40 \times v \Rightarrow v = 1 m / s, (towards right)$

Case-4: When A catches the ball,

Applying the law of conservation of liner momentum, we get:

$- 4 \times 5 + (- 0.5) \times 40 = 44 v \Rightarrow v = \frac{40}{44} = \frac{10}{11} m / s, towards left$

(c) Case-5: When A throws the ball,

Applying the law of conservation of linear momentum, we get:

$44 \times (\frac{10}{11}) = 4 \times 5 + 40 \times v \Rightarrow v = \frac{60}{40} = \frac{3}{2} m / s (towards left)$

Case-6: When B receives the ball,

Applying the law of conservation of linear momentum, we get:

$40 \times 1 + 4 \times 5 = 44 \times v \Rightarrow v = \frac{60}{44} m / s, towards right$

Case-7: When B throws the ball,

On applying the law of conservation of linear momentum, we get:

$\Rightarrow v = 44 \times (\frac{60}{44}) m / s, towards right$

Case-8: When A catches the ball,

On applying the law of conservation of linear momentum, we get:
$- 4 \times 5 + 40 (\frac{3}{2}) = - 44 v \Rightarrow V = - \frac{40}{44} = (\frac{10}{11}) m / s, towards left$

Similarly, after 5 round trips,
The velocity of A will be $(\frac{50}{11})$ m/s and the velocity of B will be 5 m/s.

(d) As after 6 round trips, the velocity of A becomes $\frac{60}{11}$ i.e. > 5 m/s, it cannot catch the ball. Thus, A can only roll the ball six times.

(e) Let the ball and the body A be at origin, in the initial position.

$∴ X_{c} = \frac{40 \times 0 + 4 \times 0 + 40 \times d}{40 + 40 + 4} = \frac{10}{21} d$

Answer:

Let the velocity of the ball falling from height h₁ be u (when it approaches the ground).

Velocity on the ground, $u = \sqrt{2 g h_{1}}$
$\Rightarrow u = \sqrt{2 \times 9.8 \times 2}$

Let the velocity of ball when it separates from the ground be v. (Assuming it goes up to height h₂)
$\Rightarrow v = \sqrt{2 g h_{2}} = \sqrt{2 \times 9.8 \times 1.5}$

Let the coefficient of restitution be e.
We know, v = eu

$\Rightarrow e = \frac{\sqrt{2 \times 9.8 \times 1.5}}{\sqrt{2 \times 9.8 \times 2}} = \frac{\sqrt{3}}{2}$
Hence, the coefficient of restitution is $\frac{\sqrt{3}}{2}$ .

Question 40:

Let the velocity of the ball falling from height h₁ be u (when it approaches the ground).

Velocity on the ground, $u = \sqrt{2 g h_{1}}$
$\Rightarrow u = \sqrt{2 \times 9.8 \times 2}$

Let the velocity of ball when it separates from the ground be v. (Assuming it goes up to height h₂)
$\Rightarrow v = \sqrt{2 g h_{2}} = \sqrt{2 \times 9.8 \times 1.5}$

Let the coefficient of restitution be e.
We know, v = eu

$\Rightarrow e = \frac{\sqrt{2 \times 9.8 \times 1.5}}{\sqrt{2 \times 9.8 \times 2}} = \frac{\sqrt{3}}{2}$
Hence, the coefficient of restitution is $\frac{\sqrt{3}}{2}$ .

Answer:

Let the nucleus recoils with a velocity v.

Applying the law of conservation of linear momentum, we get:
Linear momentum of recoiled nucleus = Linear momentum of gamma photon
⇒ mv = $\frac{E}{c}$

$∴$ $v = \frac{E}{m c}$

Kinetic energy of the recoiled nucleus = $\frac{1}{2} M v^{2}$
$\Rightarrow K . E . = \frac{1}{2} m {(\frac{E}{m c})}^{2} = \frac{1}{2} \frac{E^{2}}{m c^{2}}$

Decrease in the internal energy = photon energy + the kinetic energy of the recoiled nucleus
⇒ Decrease in the internal energy = $E + \frac{E^{2}}{2 m c^{2}}$

Question 41:

Let the nucleus recoils with a velocity v.

Applying the law of conservation of linear momentum, we get:
Linear momentum of recoiled nucleus = Linear momentum of gamma photon
⇒ mv = $\frac{E}{c}$

$∴$ $v = \frac{E}{m c}$

Kinetic energy of the recoiled nucleus = $\frac{1}{2} M v^{2}$
$\Rightarrow K . E . = \frac{1}{2} m {(\frac{E}{m c})}^{2} = \frac{1}{2} \frac{E^{2}}{m c^{2}}$

Decrease in the internal energy = photon energy + the kinetic energy of the recoiled nucleus
⇒ Decrease in the internal energy = $E + \frac{E^{2}}{2 m c^{2}}$

Answer:

Given,
Mass of each block, M_A = M_B = 2 kg
Initial velocities of block A, V_a = 1 m/s
Initial velocity of block B, V_b = 0
Spring constant of the spring = 100 N/m

Block A strikes the spring with a velocity of 1 m/s.
After the collision, it's velocity decreases continuously. At an instant the whole system (Block A + the compound spring + Block B) moves together with a common velocity V (say).

Using the law of conservation of energy, we get:

$(\frac{1}{2}) M_{A} V_{A}^{2} + (\frac{1}{2}) M_{B} V_{B}^{2} = (\frac{1}{2}) M_{A} V^{2} + (\frac{1}{2}) M_{B} V^{2} + (\frac{1}{2}) k x^{2} (\frac{1}{2}) \times 2 (1)^{2} + 0 = (\frac{1}{2}) + (\frac{1}{2}) \times v^{2} + (\frac{1}{2}) x^{2} \times 100$
(where x is the maximum compression of the spring)

⇒ 1 − 2v² = 50x² ...(1)

As there is no external force acting in the horizontal direction, the momentum is conserved.

$\Rightarrow M_{A} V_{A} + M_{B} V_{B} = (M_{A} + M_{B}) V \Rightarrow 2 \times 1 = 4 \times V \Rightarrow V = (\frac{1}{2}) m / s . . . (2) Susbstituting this value of V in equation (1), we get : 1 = 2 \times (\frac{1}{4}) + 50 x^{2} \Rightarrow \frac{1}{4} = 50 x^{2} \Rightarrow x^{2} = \frac{1}{100} \Rightarrow x = \frac{1}{10} m \Rightarrow x = 10 cm$

Question 42:

Given,
Mass of each block, M_A = M_B = 2 kg
Initial velocities of block A, V_a = 1 m/s
Initial velocity of block B, V_b = 0
Spring constant of the spring = 100 N/m

Block A strikes the spring with a velocity of 1 m/s.
After the collision, it's velocity decreases continuously. At an instant the whole system (Block A + the compound spring + Block B) moves together with a common velocity V (say).

Using the law of conservation of energy, we get:

$(\frac{1}{2}) M_{A} V_{A}^{2} + (\frac{1}{2}) M_{B} V_{B}^{2} = (\frac{1}{2}) M_{A} V^{2} + (\frac{1}{2}) M_{B} V^{2} + (\frac{1}{2}) k x^{2} (\frac{1}{2}) \times 2 (1)^{2} + 0 = (\frac{1}{2}) + (\frac{1}{2}) \times v^{2} + (\frac{1}{2}) x^{2} \times 100$
(where x is the maximum compression of the spring)

⇒ 1 − 2v² = 50x² ...(1)

As there is no external force acting in the horizontal direction, the momentum is conserved.

$\Rightarrow M_{A} V_{A} + M_{B} V_{B} = (M_{A} + M_{B}) V \Rightarrow 2 \times 1 = 4 \times V \Rightarrow V = (\frac{1}{2}) m / s . . . (2) Susbstituting this value of V in equation (1), we get : 1 = 2 \times (\frac{1}{4}) + 50 x^{2} \Rightarrow \frac{1}{4} = 50 x^{2} \Rightarrow x^{2} = \frac{1}{100} \Rightarrow x = \frac{1}{10} m \Rightarrow x = 10 cm$

Answer:

It is given that:
Mass of bullet, m = 20 g =0.02 kg
The initial speed, v₁ = 500 m/s
Mass of block, M = 10 kg
The initial speed of block = 0
Final velocity of bullet, v₂= 100 m/s
Let the final velocity of block when the bullet emerges out = v'

Applying conservation of linear momentum,
mv₁ + M× 0 = mv₂ + Mv'
⇒ 0.02 × 500 = 0.02 × 100 + 10 × v'
⇒ v' = 0.8 m/s

Distance covered by the block, d = 20 cm = 0.02 m .
Let friction coefficient between the block and the surface = μ
Thus, the value of friction force, $F = μ m g$ $F = μ m g$

Change in K.E. of block = Work done by the friction force
$\Rightarrow \frac{1}{2} \times M \times 0 - \frac{1}{2} \times M \times {(v')}^{2} = μ m g d \Rightarrow 0 - (\frac{1}{2}) \times 10 \times (0.8)^{2} = μ \times 10 \times 10 \times 0.2 \Rightarrow μ = 0.16$

Question 43:

It is given that:
Mass of bullet, m = 20 g =0.02 kg
The initial speed, v₁ = 500 m/s
Mass of block, M = 10 kg
The initial speed of block = 0
Final velocity of bullet, v₂= 100 m/s
Let the final velocity of block when the bullet emerges out = v'

Applying conservation of linear momentum,
mv₁ + M× 0 = mv₂ + Mv'
⇒ 0.02 × 500 = 0.02 × 100 + 10 × v'
⇒ v' = 0.8 m/s

Distance covered by the block, d = 20 cm = 0.02 m .
Let friction coefficient between the block and the surface = μ
Thus, the value of friction force, $F = μ m g$ $F = μ m g$

Change in K.E. of block = Work done by the friction force
$\Rightarrow \frac{1}{2} \times M \times 0 - \frac{1}{2} \times M \times {(v')}^{2} = μ m g d \Rightarrow 0 - (\frac{1}{2}) \times 10 \times (0.8)^{2} = μ \times 10 \times 10 \times 0.2 \Rightarrow μ = 0.16$

Answer:

Given:
Initial velocity of the projectile = u
Angle of projection of the projectile with respect to ground = θ

When the projectile hits the ground for the first time, the velocity remains same i.e. u.
The component of velocity parallel to ground, u cos θ should remain constant.
However, the vertical component of the projectile undergoes a change after the collision.

If the coefficient of restitution of collision between the projectile and the field is e,
The velocity of separation is given by,
⇒ v = eu sin θ

Therefore, for the second projectile motion,
Velocity of projection (u') will be,
$u' = \sqrt{(u \cos θ)^{2} + (e u \sin θ)^{2}} Angle of projection, α = \tan^{- 1} (\frac{e u \sin θ}{u \cos θ}) \Rightarrow α = \tan^{- 1} (e \tan θ) or, \tan α = e \tan θ . . . (2) Also, y = x \tan α - (\frac{g x^{2} \sec^{2} α}{2 u'^{2}}) . . . (3) Here, y = 0 ∴ \tan α = e \tan θ \Rightarrow \sec^{2} α = 1 + e^{2} \tan^{2} θ and u'^{2} = u^{2} \cos^{2} θ + e^{2} u^{2} \sin^{2} θ$

Putting the above calculated values in equation (3), we get:
$x e tanθ = \frac{g x^{2} (1 + e^{2} \tan^{2} θ)}{2 u^{2} (\cos^{2} θ + e^{2} \sin^{2} θ)} or, x = \frac{2 e u^{2} \tan θ (\cos^{2} θ + e^{2} \sin^{2} θ)}{g (1 + e^{2} \tan^{2} θ)} \Rightarrow x = \frac{2 e u^{2} \tan θ . \cos^{2} θ}{g} \Rightarrow x = \frac{e u^{2} \sin 2 θ}{g}$

Thus, from the starting point the projectile makes its second collision with the field at a distance,
$x' = \frac{u^{2} \sin 2 θ}{g} + \frac{e u^{2} \sin 2 θ}{g} \Rightarrow x' = \frac{u^{2} \sin 2 θ}{g} (1 + e)$

Question 44:

Given:
Initial velocity of the projectile = u
Angle of projection of the projectile with respect to ground = θ

When the projectile hits the ground for the first time, the velocity remains same i.e. u.
The component of velocity parallel to ground, u cos θ should remain constant.
However, the vertical component of the projectile undergoes a change after the collision.

If the coefficient of restitution of collision between the projectile and the field is e,
The velocity of separation is given by,
⇒ v = eu sin θ

Therefore, for the second projectile motion,
Velocity of projection (u') will be,
$u' = \sqrt{(u \cos θ)^{2} + (e u \sin θ)^{2}} Angle of projection, α = \tan^{- 1} (\frac{e u \sin θ}{u \cos θ}) \Rightarrow α = \tan^{- 1} (e \tan θ) or, \tan α = e \tan θ . . . (2) Also, y = x \tan α - (\frac{g x^{2} \sec^{2} α}{2 u'^{2}}) . . . (3) Here, y = 0 ∴ \tan α = e \tan θ \Rightarrow \sec^{2} α = 1 + e^{2} \tan^{2} θ and u'^{2} = u^{2} \cos^{2} θ + e^{2} u^{2} \sin^{2} θ$

Putting the above calculated values in equation (3), we get:
$x e tanθ = \frac{g x^{2} (1 + e^{2} \tan^{2} θ)}{2 u^{2} (\cos^{2} θ + e^{2} \sin^{2} θ)} or, x = \frac{2 e u^{2} \tan θ (\cos^{2} θ + e^{2} \sin^{2} θ)}{g (1 + e^{2} \tan^{2} θ)} \Rightarrow x = \frac{2 e u^{2} \tan θ . \cos^{2} θ}{g} \Rightarrow x = \frac{e u^{2} \sin 2 θ}{g}$

Thus, from the starting point the projectile makes its second collision with the field at a distance,
$x' = \frac{u^{2} \sin 2 θ}{g} + \frac{e u^{2} \sin 2 θ}{g} \Rightarrow x' = \frac{u^{2} \sin 2 θ}{g} (1 + e)$

Answer:

Given:
The angle of inclination of the inclined plane is θ.
A ball falls on the inclined plane from height h.
The coefficient of restitution is e.

Let the ball strikes the inclined plane at origin with velocity $v_{0} = \sqrt{2 g h}$ .
As the ball elastically rebounds, it recalls with the same velocity v₀ at the same angle θ from the normal or y-axis.
Let the ball strikes the incline second time at any point P, which is at a distance l from the origin along the incline.
From the equation,
$y = v_{o y} t + \frac{1}{2} a_{y} t^{2} we may write : 0 = v_{0} \cos θ t - \frac{1}{2} g \cos θ t^{2}$
where t is time of motion of the ball in air, as the ball moves from origin to point P.

$As t \neq 0, the value of t is \frac{2 v_{0}}{g} .$

Now from the equation,
$x = v_{0} t + \frac{1}{2} a_{x} t^{2} we can write, l = v_{0} \sin θ t + \frac{1}{2} g \sin θ t^{2} ∴ l = v_{0} \sin θ (\frac{2 v_{0}}{g}) + \frac{1}{2} g \sin θ {(\frac{2 v_{0}}{g})}^{2} \Rightarrow l = \frac{4 v_{0}^{2} \sin θ}{g}$

Substituting the value of v₀ in the above equation, we get:
l = 8h sin θ

Therefore, the ball again hits the plain at a distance, l = 8h sin θ

Question 45:

Given:
The angle of inclination of the inclined plane is θ.
A ball falls on the inclined plane from height h.
The coefficient of restitution is e.

Let the ball strikes the inclined plane at origin with velocity $v_{0} = \sqrt{2 g h}$ .
As the ball elastically rebounds, it recalls with the same velocity v₀ at the same angle θ from the normal or y-axis.
Let the ball strikes the incline second time at any point P, which is at a distance l from the origin along the incline.
From the equation,
$y = v_{o y} t + \frac{1}{2} a_{y} t^{2} we may write : 0 = v_{0} \cos θ t - \frac{1}{2} g \cos θ t^{2}$
where t is time of motion of the ball in air, as the ball moves from origin to point P.

$As t \neq 0, the value of t is \frac{2 v_{0}}{g} .$

Now from the equation,
$x = v_{0} t + \frac{1}{2} a_{x} t^{2} we can write, l = v_{0} \sin θ t + \frac{1}{2} g \sin θ t^{2} ∴ l = v_{0} \sin θ (\frac{2 v_{0}}{g}) + \frac{1}{2} g \sin θ {(\frac{2 v_{0}}{g})}^{2} \Rightarrow l = \frac{4 v_{0}^{2} \sin θ}{g}$

Substituting the value of v₀ in the above equation, we get:
l = 8h sin θ

Therefore, the ball again hits the plain at a distance, l = 8h sin θ

Answer:

Given:
The angle of inclination of the plane, θ = 45°
A ball falls on the inclined plane from a height (h) of 5 m.

The coefficient of restitution, $e = (\frac{3}{4})$
The velocity with which ball strikes the inclined plane is given as,
$v = \sqrt{2 g \times 5} = 10 m / s$

The ball makes an angle β with the horizontal, after the collision.
The horizontal component of velocity, 10cos45° remains unchanged.
However, the velocity in perpendicular direction to the plane after the collision will now be:
$v_{1} = e \times 10 \sin 45 ° = (\frac{3}{4}) \times 10 \times \frac{1}{\sqrt{2}} = (3.75) \sqrt{2} m / s Similarly, v_{2} = 5 \sqrt{2} m / s Now, u = \sqrt{v_{2}^{2} + v_{1}^{2}} = \sqrt{50 + 28.125} = \sqrt{78.125} = 8.83 m / s$

Angle of reflection from the wall is given as,
$β = \tan^{- 1} \frac{(3.75 \sqrt{2})}{5 \sqrt{2}} = \tan^{- 1} (\frac{3}{4}) = 37 °$

Angle of projection α = 90 − (θ + β)
$\Rightarrow$ α = 90° − (45° + 37°) = 8°

Let the distance where the ball falls after the collision be L.
⇒ x = L cos θ

Angle of projection (α) = −8°

Now,
Y = $x \tan α - \frac{g x^{2} s e c^{2} α}{2 u^{2}}$
$\Rightarrow$ −L sin θ = L cos θ × tan 8° $- \frac{g}{2} \frac{L^{2} \cos^{2} θ \sec^{2} 80 °}{(u)^{2}}$
$\Rightarrow - \sin 45 ° = \cos 45 ° \times \tan 8 ° - \frac{10 \cos^{2} 45 \sec^{2} 8 °}{(8.83)^{2}}$

On solving the above equation, we get:
L = 18.5 m

Question 46:

Given:
The angle of inclination of the plane, θ = 45°
A ball falls on the inclined plane from a height (h) of 5 m.

The coefficient of restitution, $e = (\frac{3}{4})$
The velocity with which ball strikes the inclined plane is given as,
$v = \sqrt{2 g \times 5} = 10 m / s$

The ball makes an angle β with the horizontal, after the collision.
The horizontal component of velocity, 10cos45° remains unchanged.
However, the velocity in perpendicular direction to the plane after the collision will now be:
$v_{1} = e \times 10 \sin 45 ° = (\frac{3}{4}) \times 10 \times \frac{1}{\sqrt{2}} = (3.75) \sqrt{2} m / s Similarly, v_{2} = 5 \sqrt{2} m / s Now, u = \sqrt{v_{2}^{2} + v_{1}^{2}} = \sqrt{50 + 28.125} = \sqrt{78.125} = 8.83 m / s$

Angle of reflection from the wall is given as,
$β = \tan^{- 1} \frac{(3.75 \sqrt{2})}{5 \sqrt{2}} = \tan^{- 1} (\frac{3}{4}) = 37 °$

Angle of projection α = 90 − (θ + β)
$\Rightarrow$ α = 90° − (45° + 37°) = 8°

Let the distance where the ball falls after the collision be L.
⇒ x = L cos θ

Angle of projection (α) = −8°

Now,
Y = $x \tan α - \frac{g x^{2} s e c^{2} α}{2 u^{2}}$
$\Rightarrow$ −L sin θ = L cos θ × tan 8° $- \frac{g}{2} \frac{L^{2} \cos^{2} θ \sec^{2} 80 °}{(u)^{2}}$
$\Rightarrow - \sin 45 ° = \cos 45 ° \times \tan 8 ° - \frac{10 \cos^{2} 45 \sec^{2} 8 °}{(8.83)^{2}}$

On solving the above equation, we get:
L = 18.5 m

Answer:

It is given that:
Mass of block, M = 200 g = 0.20 kg
Mass of the particle, m = 120 gm = 0.12 kg
Height of the particle, h = 45 cm = 0.45 m

According to question, as the block attains equilibrium, the spring is stretched by a distance, x = 1.00 cm = 0.01 m.

i.e. M × g = K × x

⇒ 0.2 × g = K × x
⇒ 2 = K × 0.01
⇒ K = 200 N/m

The velocity with which the particle m strikes M is given by,
$u = \sqrt{2 g h} u = \sqrt{2 \times 10 \times 0.45} = \sqrt{9} = 3 m / s$

After the collision, let the velocity of the particle and the block be V.
According to law of conservation of momentum, we can write:
mu = (m + M)V

Solving for V , we get:
$V = \frac{0.12 \times 3}{0.32} = \frac{9}{8} m / s$

Let the spring be stretched through an extra deflection of δ.

On applying the law of conservation of energy, we can write:
Initial energy of the system before collision = Final energy of the system

$\Rightarrow \frac{1}{2} m u^{2} + \frac{1}{2} K x^{2} = \frac{1}{2} (m + M) V^{2} + \frac{1}{2} K {(x + δ)}^{2}$

Substituting appropriate values in the above equation, we get:
$(\frac{1}{2}) \times 0.12 \times 9 + (\frac{1}{2}) \times 200 \times (0.01)^{2} = (\frac{1}{2}) 0.32 \times (\frac{81}{64}) + (\frac{1}{2}) \times 200 \times (δ + 0.1)^{2}$

On solving the above equation, we get:
δ = 0.061
m = 6.1 cm

Question 47:

It is given that:
Mass of block, M = 200 g = 0.20 kg
Mass of the particle, m = 120 gm = 0.12 kg
Height of the particle, h = 45 cm = 0.45 m

According to question, as the block attains equilibrium, the spring is stretched by a distance, x = 1.00 cm = 0.01 m.

i.e. M × g = K × x

⇒ 0.2 × g = K × x
⇒ 2 = K × 0.01
⇒ K = 200 N/m

The velocity with which the particle m strikes M is given by,
$u = \sqrt{2 g h} u = \sqrt{2 \times 10 \times 0.45} = \sqrt{9} = 3 m / s$

After the collision, let the velocity of the particle and the block be V.
According to law of conservation of momentum, we can write:
mu = (m + M)V

Solving for V , we get:
$V = \frac{0.12 \times 3}{0.32} = \frac{9}{8} m / s$

Let the spring be stretched through an extra deflection of δ.

On applying the law of conservation of energy, we can write:
Initial energy of the system before collision = Final energy of the system

$\Rightarrow \frac{1}{2} m u^{2} + \frac{1}{2} K x^{2} = \frac{1}{2} (m + M) V^{2} + \frac{1}{2} K {(x + δ)}^{2}$

Substituting appropriate values in the above equation, we get:
$(\frac{1}{2}) \times 0.12 \times 9 + (\frac{1}{2}) \times 200 \times (0.01)^{2} = (\frac{1}{2}) 0.32 \times (\frac{81}{64}) + (\frac{1}{2}) \times 200 \times (δ + 0.1)^{2}$

On solving the above equation, we get:
δ = 0.061
m = 6.1 cm

Answer:

Given:
Mass of bullet, m = 25 g = 0.025 kg
Mass of ballistic pendulum, M = 5 kg
Vertical displacement, h = 10 cm = 0.1 m

Let the bullet strikes the pendulum with a velocity u.
Let the final velocity be v.

Using the law of conservation of linear momentum, we can write:
$m u = (M + m) v \Rightarrow v = \frac{m}{(M + m)} u \Rightarrow v = \frac{0.25}{5.025} \times u = \frac{u}{201}$

Applying the law of conservation of energy, we get:
$(\frac{1}{2}) (M + m) v^{2} = (M + m) g h \Rightarrow \frac{u^{2}}{(201)^{2}} = 2 \times 10 \times 0.1 \Rightarrow u = 201 \times \sqrt{2} = 280 m / s$

The bullet strikes the pendulum with a velocity of 280 m/s.

Question 48:

Given:
Mass of bullet, m = 25 g = 0.025 kg
Mass of ballistic pendulum, M = 5 kg
Vertical displacement, h = 10 cm = 0.1 m

Let the bullet strikes the pendulum with a velocity u.
Let the final velocity be v.

Using the law of conservation of linear momentum, we can write:
$m u = (M + m) v \Rightarrow v = \frac{m}{(M + m)} u \Rightarrow v = \frac{0.25}{5.025} \times u = \frac{u}{201}$

Applying the law of conservation of energy, we get:
$(\frac{1}{2}) (M + m) v^{2} = (M + m) g h \Rightarrow \frac{u^{2}}{(201)^{2}} = 2 \times 10 \times 0.1 \Rightarrow u = 201 \times \sqrt{2} = 280 m / s$

The bullet strikes the pendulum with a velocity of 280 m/s.

Answer:

Given:
Mass of bullet, m = 20 gm = 0.02 kg
Horizontal speed of the bullet, u = 300 m/s
Mass of wooden block, M = 500 gm = 0.5 kg

Let the bullet emerges out with velocity v.
Let the velocity of the block be v'.

Using the law of conservation of momentum, we get:
mu = Mv' + mv ...(1)

Now, applying the work-energy principle for the block after the collision, we get:
$0 - (\frac{1}{2}) M \times {(v')}^{2} = - M g h \Rightarrow (v')^{2} = 2 g h v' = \sqrt{2 g h} = \sqrt{20 \times 10 \times 0.2} = 2 m / s$

On substituting the value of v' in equation (1), we get:

$0.02 \times 300 = 0.5 \times 2 + 0.02 \times v \Rightarrow v = \frac{6 - 1}{0.02} = \frac{5}{0.02} \Rightarrow v = 250 m / s$

Hence, the speed of the bullet as it emerges out from the block is 250 m/s.

Question 49:

Given:
Mass of bullet, m = 20 gm = 0.02 kg
Horizontal speed of the bullet, u = 300 m/s
Mass of wooden block, M = 500 gm = 0.5 kg

Let the bullet emerges out with velocity v.
Let the velocity of the block be v'.

Using the law of conservation of momentum, we get:
mu = Mv' + mv ...(1)

Now, applying the work-energy principle for the block after the collision, we get:
$0 - (\frac{1}{2}) M \times {(v')}^{2} = - M g h \Rightarrow (v')^{2} = 2 g h v' = \sqrt{2 g h} = \sqrt{20 \times 10 \times 0.2} = 2 m / s$

On substituting the value of v' in equation (1), we get:

$0.02 \times 300 = 0.5 \times 2 + 0.02 \times v \Rightarrow v = \frac{6 - 1}{0.02} = \frac{5}{0.02} \Rightarrow v = 250 m / s$

Hence, the speed of the bullet as it emerges out from the block is 250 m/s.

Answer:

It is given that two blocks of masses m₁ and m₂ are connected with a spring having spring constant k.
Initially the spring is stretched by a distance x₀.

For the block to come to rest again,
Let the distance travelled by m₁ be x₁(towards right), and that travelled by m₂ be x₂ towards left.

As no external force acts in horizontal direction, we can write:
m₁x₁ = m₂x₂ ...(1)

As the energy is conserved in the spring, we get:
$(\frac{1}{2}) k x_{0}^{2} = (\frac{1}{2}) k (x_{1} + x_{2} - x_{0})^{2} \Rightarrow x_{0} = x_{1} + x_{2} - x_{0} \Rightarrow x_{1} + x_{2} = 2 x_{0} . . . (2)$

$∴ x_{1} = 2 x_{0} - x_{2} Putting this value in equation (1), we get : m_{1} (2 x_{0} - x_{0}) = m_{2} x_{2} \Rightarrow 2 m_{1} x_{0} - m_{1} x_{2} = m_{2} x_{2} \Rightarrow x_{2} = \frac{2 m_{2}}{m_{1} + m_{2}} x_{0} Similarly, x_{1} = (\frac{2 m_{2}}{m_{1} + m_{2}}) x_{0}$

Question 50:

It is given that two blocks of masses m₁ and m₂ are connected with a spring having spring constant k.
Initially the spring is stretched by a distance x₀.

For the block to come to rest again,
Let the distance travelled by m₁ be x₁(towards right), and that travelled by m₂ be x₂ towards left.

As no external force acts in horizontal direction, we can write:
m₁x₁ = m₂x₂ ...(1)

As the energy is conserved in the spring, we get:
$(\frac{1}{2}) k x_{0}^{2} = (\frac{1}{2}) k (x_{1} + x_{2} - x_{0})^{2} \Rightarrow x_{0} = x_{1} + x_{2} - x_{0} \Rightarrow x_{1} + x_{2} = 2 x_{0} . . . (2)$

$∴ x_{1} = 2 x_{0} - x_{2} Putting this value in equation (1), we get : m_{1} (2 x_{0} - x_{0}) = m_{2} x_{2} \Rightarrow 2 m_{1} x_{0} - m_{1} x_{2} = m_{2} x_{2} \Rightarrow x_{2} = \frac{2 m_{2}}{m_{1} + m_{2}} x_{0} Similarly, x_{1} = (\frac{2 m_{2}}{m_{1} + m_{2}}) x_{0}$

Answer:

Given,
Velocity of mass, m₂ = v₀
Velocity of mass, m₁= 0

(a) Velocity of centre of mass is given by,
$v_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$\Rightarrow v_{c m} = \frac{m_{1} \times 0 + m_{2} \times v_{0}}{m_{1} + m_{2}} \Rightarrow v_{c m} = \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

(b) Let the maximum elongation in spring be x.

The spring attains maximum elongation when velocities of both the blocks become equal to the velocity of centre of mass.
i.e. v₁ = v₂ = v_cm

On applying the law of conservation of energy, we can write:
Change in kinetic energy = Potential energy stored in spring

$\Rightarrow \frac{1}{2} m_{2} v_{0}^{2} - \frac{1}{2} (m_{1} + m_{2}) {(\frac{m_{2} v_{0}}{m_{1} + m_{2}})}^{2} = \frac{1}{2} k x^{2} \Rightarrow m_{2} v_{0}^{2} (1 - \frac{m_{2}}{m_{1} + m_{2}}) = k x^{2}$

$\Rightarrow x = v_{0} {[\frac{m_{1} m_{2}}{(m_{1} + m_{2}) k}]}^{1 / 2}$

Question 51:

Given,
Velocity of mass, m₂ = v₀
Velocity of mass, m₁= 0

(a) Velocity of centre of mass is given by,
$v_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$\Rightarrow v_{c m} = \frac{m_{1} \times 0 + m_{2} \times v_{0}}{m_{1} + m_{2}} \Rightarrow v_{c m} = \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

(b) Let the maximum elongation in spring be x.

The spring attains maximum elongation when velocities of both the blocks become equal to the velocity of centre of mass.
i.e. v₁ = v₂ = v_cm

On applying the law of conservation of energy, we can write:
Change in kinetic energy = Potential energy stored in spring

$\Rightarrow \frac{1}{2} m_{2} v_{0}^{2} - \frac{1}{2} (m_{1} + m_{2}) {(\frac{m_{2} v_{0}}{m_{1} + m_{2}})}^{2} = \frac{1}{2} k x^{2} \Rightarrow m_{2} v_{0}^{2} (1 - \frac{m_{2}}{m_{1} + m_{2}}) = k x^{2}$

$\Rightarrow x = v_{0} {[\frac{m_{1} m_{2}}{(m_{1} + m_{2}) k}]}^{1 / 2}$

Answer:

It is given that the force on both the blocks is F.
â€‹

Let x₁ and x₂ be the extensions of blocks m₁ and m₂ respectively.

Total work done by the forces on the blocks = Fx₁ + Fx₂ ...(1)
∴ Increase in the potential energy of spring = $(\frac{1}{2}) K (x_{1} + x_{2})^{2} . . . (2)$
$Equating the equations (1) and (2), we get : \Rightarrow (x_{1} + x_{2}) = (\frac{2 F}{k}) \dots (3)$

As the net external force on the system is zero, the centre of mass does not shift.
$∴ m_{1} x_{1} = m_{2} x_{2} . . . (4) Using the equations (3) and (4), we get : \frac{m_{1}}{m_{2}} x_{1} + x_{1} = \frac{2 F}{k} \Rightarrow x_{1} (1 + \frac{m_{1}}{m_{2}}) = \frac{2 F}{k} \Rightarrow x_{1} = \frac{2 F m_{2}}{k (m_{1} + m_{2})}$

Similarly,
$x_{2} = \frac{2 F m_{1}}{k (m_{1} + m_{2})}$

Question 52:

It is given that the force on both the blocks is F.
â€‹

Let x₁ and x₂ be the extensions of blocks m₁ and m₂ respectively.

Total work done by the forces on the blocks = Fx₁ + Fx₂ ...(1)
∴ Increase in the potential energy of spring = $(\frac{1}{2}) K (x_{1} + x_{2})^{2} . . . (2)$
$Equating the equations (1) and (2), we get : \Rightarrow (x_{1} + x_{2}) = (\frac{2 F}{k}) \dots (3)$

As the net external force on the system is zero, the centre of mass does not shift.
$∴ m_{1} x_{1} = m_{2} x_{2} . . . (4) Using the equations (3) and (4), we get : \frac{m_{1}}{m_{2}} x_{1} + x_{1} = \frac{2 F}{k} \Rightarrow x_{1} (1 + \frac{m_{1}}{m_{2}}) = \frac{2 F}{k} \Rightarrow x_{1} = \frac{2 F m_{2}}{k (m_{1} + m_{2})}$

Similarly,
$x_{2} = \frac{2 F m_{1}}{k (m_{1} + m_{2})}$

Answer:

Given:
Force on block of mass, m₁ = F₁
Force on block of mass, m₂ =â€‹ F₂

Let the acceleration produced in mass m₁ be a₁.
$a_{1} = \frac{F_{1} - F_{2}}{m_{1} + m_{2}}$

Let the acceleration of mass m₂be a₂.
â€‹ $a_{2} = \frac{F_{2} - F_{1}}{m_{1} + m_{2}}$

Due to the force F₂, the mass m₁ experiences a pseudo forceâ€‹.

$∴ Net force on m_{1} = F_{1} + m_{1} a_{2} F' = F_{1} + m_{1} \times \frac{(F_{2} - F_{1})}{m_{1} + m_{2}} = \frac{m_{1} F_{1} + m_{2} F_{1} + m_{1} F_{2} - m_{1} F_{1}}{m_{1} + m_{2}} = \frac{m_{2} F_{1} + m_{1} F_{2}}{m_{1} + m_{2}}$

Similarly, mass m₂ experiences a pseudo force due to force F₁.

$∴ Net force on m_{2} = F_{2} + m_{2} a_{1}$
$F " = F_{2} + m_{2} \times \frac{(F_{1} - F_{2})}{m_{1} + m_{2}} = \frac{m_{1} F_{2} + m_{2} F_{2} + m_{2} F_{1} - m_{2} F_{2}}{m_{1} + m_{2}} = \frac{m_{1} F_{2} + m_{2} F_{1}}{m_{1} + m_{2}}$

Let m₁ be displaced by a distance x₁ and m₂ be displaced by a distance â€‹x₂.

Therefore, the maximum elongation of the spring = x₁ + x₂

Work done by the blocks = Energy stored in the spring

$\Rightarrow \frac{m_{2} F_{1} + m_{1} F_{2}}{m_{1} + m_{2}} \times x_{1} \times \frac{m_{2} F_{1} + m_{1} F_{2}}{m_{1} + m_{2}} \times x_{2} = (\frac{1}{2}) k (x_{1} + x_{2})^{2} \Rightarrow x_{1} + x_{2} = \frac{2}{k} (\frac{m_{1} F_{2} + m_{2} F_{1}}{m_{1} + m_{2}})$

Question 53:

Given:
Force on block of mass, m₁ = F₁
Force on block of mass, m₂ =â€‹ F₂

Let the acceleration produced in mass m₁ be a₁.
$a_{1} = \frac{F_{1} - F_{2}}{m_{1} + m_{2}}$

Let the acceleration of mass m₂be a₂.
â€‹ $a_{2} = \frac{F_{2} - F_{1}}{m_{1} + m_{2}}$

Due to the force F₂, the mass m₁ experiences a pseudo forceâ€‹.

$∴ Net force on m_{1} = F_{1} + m_{1} a_{2} F' = F_{1} + m_{1} \times \frac{(F_{2} - F_{1})}{m_{1} + m_{2}} = \frac{m_{1} F_{1} + m_{2} F_{1} + m_{1} F_{2} - m_{1} F_{1}}{m_{1} + m_{2}} = \frac{m_{2} F_{1} + m_{1} F_{2}}{m_{1} + m_{2}}$

Similarly, mass m₂ experiences a pseudo force due to force F₁.

$∴ Net force on m_{2} = F_{2} + m_{2} a_{1}$
$F " = F_{2} + m_{2} \times \frac{(F_{1} - F_{2})}{m_{1} + m_{2}} = \frac{m_{1} F_{2} + m_{2} F_{2} + m_{2} F_{1} - m_{2} F_{2}}{m_{1} + m_{2}} = \frac{m_{1} F_{2} + m_{2} F_{1}}{m_{1} + m_{2}}$

Let m₁ be displaced by a distance x₁ and m₂ be displaced by a distance â€‹x₂.

Therefore, the maximum elongation of the spring = x₁ + x₂

Work done by the blocks = Energy stored in the spring

$\Rightarrow \frac{m_{2} F_{1} + m_{1} F_{2}}{m_{1} + m_{2}} \times x_{1} \times \frac{m_{2} F_{1} + m_{1} F_{2}}{m_{1} + m_{2}} \times x_{2} = (\frac{1}{2}) k (x_{1} + x_{2})^{2} \Rightarrow x_{1} + x_{2} = \frac{2}{k} (\frac{m_{1} F_{2} + m_{2} F_{1}}{m_{1} + m_{2}})$

Answer:

Given,
Mass of the man, M_m = 50 kg
Mass of the pillow, M_P = 5 kg
Velocity of pillow w.r.t. man, ${\vec{V}}_{p m}$ = 8 ft/s

Let the velocity of man be ${\vec{V}}_{m}$ .

${\vec{V}}_{pm} = {\vec{V}}_{p} - (- {\vec{V}}_{m}) = {\vec{V}}_{p} + {\vec{V}}_{m} \Rightarrow {\vec{V}}_{p} = {\vec{V}}_{pm} - {\vec{V}}_{m}$

As no external force acts on the system, the acceleration of centre of mass is zero and the velocity of centre of mass remains constant.

Using the law of conservation of linear momentum:
M_m × V_m = M_p × V_p ...(1)

$M_{m} \times V_{m} = M_{p} \times (V_{pm} - V_{m}) \Rightarrow 50 \times V_{m} = 5 \times (8 - V_{m}) \Rightarrow V_{m} = \frac{8}{11} = 0.727 ft / s$

∴ Velocity of pillow, ${\vec{V}}_{p}$ = 8 − 0.727 = 7.2 ft/s

The time taken to reach the floor is given by,
$t = \frac{s}{v} = \frac{8}{7.2} = 1.1 s$

As the mass of wall is much greater than the mass of the pillow,
velocity of block before the collision = velocity after the collision

⇒ time of ascent = 1.11 s

Hence, total time taken = 1.11 + 1.11 = 2.22 s

Question 54:

Given,
Mass of the man, M_m = 50 kg
Mass of the pillow, M_P = 5 kg
Velocity of pillow w.r.t. man, ${\vec{V}}_{p m}$ = 8 ft/s

Let the velocity of man be ${\vec{V}}_{m}$ .

${\vec{V}}_{pm} = {\vec{V}}_{p} - (- {\vec{V}}_{m}) = {\vec{V}}_{p} + {\vec{V}}_{m} \Rightarrow {\vec{V}}_{p} = {\vec{V}}_{pm} - {\vec{V}}_{m}$

As no external force acts on the system, the acceleration of centre of mass is zero and the velocity of centre of mass remains constant.

Using the law of conservation of linear momentum:
M_m × V_m = M_p × V_p ...(1)

$M_{m} \times V_{m} = M_{p} \times (V_{pm} - V_{m}) \Rightarrow 50 \times V_{m} = 5 \times (8 - V_{m}) \Rightarrow V_{m} = \frac{8}{11} = 0.727 ft / s$

∴ Velocity of pillow, ${\vec{V}}_{p}$ = 8 − 0.727 = 7.2 ft/s

The time taken to reach the floor is given by,
$t = \frac{s}{v} = \frac{8}{7.2} = 1.1 s$

As the mass of wall is much greater than the mass of the pillow,
velocity of block before the collision = velocity after the collision

⇒ time of ascent = 1.11 s

Hence, total time taken = 1.11 + 1.11 = 2.22 s

Answer:

Given:
Mass of the block, A = m
Mass of the block, B = 2m

Let the initial velocity of block A be u₁ and the final velocity of block A,when it reaches the block B be v₁.

Using the work-energy theorem for block A, we can write:
Gain in kinetic energy = Loss in potential energy

$∴ (\frac{1}{2}) m v_{1}^{2} - (\frac{1}{2}) m u_{1}^{2} = m g h \Rightarrow v_{1}^{2} - u_{1}^{2} = 2 gh \Rightarrow v_{1} = \sqrt{2 g h + u_{1}^{2}} . . . (1)$

Let the block B just manages to reach the man's head.
i.e. the velocity of block B is zero at that point.

Again, applying the work-energy theorem for block B, we get:
$(\frac{1}{2}) \times 2 m \times (0)^{2} - (\frac{1}{2}) \times 2 m \times v^{2} = m g h \Rightarrow v = \sqrt{2 gh} Therefore, before the collision : Velocity of A, u_{A} = v_{1} Velocity of B, u_{B} = 0 After the collision : Velocity of A, v_{A} = v (say) Velocity of B, v_{B} = \sqrt{2 gh}$

As the collision is elastic, K.E. and momentum are conserved.

$m v_{1} + 2 m \times 0 = m v + 2 m \sqrt{2 g h} \Rightarrow v_{1} - v = 2 \sqrt{2 g h} . . . (2)$
$\Rightarrow (\frac{1}{2}) m v_{1}^{2} + (\frac{1}{2}) 2 m \times (0)^{2} = (\frac{1}{2}) m v^{2} + (\frac{1}{2}) 2 m {(\sqrt{2 g h})}^{2} \Rightarrow v_{1}^{2} - v^{2} = 2 \times \sqrt{2 g h} \times \sqrt{2 g h} . . . (3) Dividing equation (3) by equation (2), we get : v_{1} + v = \sqrt{2 g h} . . . (4) Adding the equations (4) and (2), we get : 2 v_{1} = 3 \sqrt{2 g h} Now using equation (1) to substitue the value of v_{1}, we get : \sqrt{2 g h + u^{2}} = (\frac{3}{2}) \sqrt{2 g h} \Rightarrow 2 g h + u^{2} = (\frac{9}{4}) (2 g h) \Rightarrow u = \sqrt{2.5 g h}$

Block a should be started with a minimum velocity of $\sqrt{2.5 g h}$ to get the sleeping man awakened.

Question 55:

Given:
Mass of the block, A = m
Mass of the block, B = 2m

Let the initial velocity of block A be u₁ and the final velocity of block A,when it reaches the block B be v₁.

Using the work-energy theorem for block A, we can write:
Gain in kinetic energy = Loss in potential energy

$∴ (\frac{1}{2}) m v_{1}^{2} - (\frac{1}{2}) m u_{1}^{2} = m g h \Rightarrow v_{1}^{2} - u_{1}^{2} = 2 gh \Rightarrow v_{1} = \sqrt{2 g h + u_{1}^{2}} . . . (1)$

Let the block B just manages to reach the man's head.
i.e. the velocity of block B is zero at that point.

Again, applying the work-energy theorem for block B, we get:
$(\frac{1}{2}) \times 2 m \times (0)^{2} - (\frac{1}{2}) \times 2 m \times v^{2} = m g h \Rightarrow v = \sqrt{2 gh} Therefore, before the collision : Velocity of A, u_{A} = v_{1} Velocity of B, u_{B} = 0 After the collision : Velocity of A, v_{A} = v (say) Velocity of B, v_{B} = \sqrt{2 gh}$

As the collision is elastic, K.E. and momentum are conserved.

$m v_{1} + 2 m \times 0 = m v + 2 m \sqrt{2 g h} \Rightarrow v_{1} - v = 2 \sqrt{2 g h} . . . (2)$
$\Rightarrow (\frac{1}{2}) m v_{1}^{2} + (\frac{1}{2}) 2 m \times (0)^{2} = (\frac{1}{2}) m v^{2} + (\frac{1}{2}) 2 m {(\sqrt{2 g h})}^{2} \Rightarrow v_{1}^{2} - v^{2} = 2 \times \sqrt{2 g h} \times \sqrt{2 g h} . . . (3) Dividing equation (3) by equation (2), we get : v_{1} + v = \sqrt{2 g h} . . . (4) Adding the equations (4) and (2), we get : 2 v_{1} = 3 \sqrt{2 g h} Now using equation (1) to substitue the value of v_{1}, we get : \sqrt{2 g h + u^{2}} = (\frac{3}{2}) \sqrt{2 g h} \Rightarrow 2 g h + u^{2} = (\frac{9}{4}) (2 g h) \Rightarrow u = \sqrt{2.5 g h}$

Block a should be started with a minimum velocity of $\sqrt{2.5 g h}$ to get the sleeping man awakened.

Answer:

Given:
Mass of block = 490 gm
Initial speed of the block = 0
Mass of bullet = 10 gm
Initial speed of the bullet, v_{1 =} $50 \sqrt{7} m / s$

As the bullet gets embedded inside the block, this is an example of a perfectly inelastic collision.

Let the final velocity of the system (block and bullet) be V_A.

Using the law of conservation of linear momentum, we get:
m₁v₁ + m₂ × 0 = (m_₁ + m₂)v_A
⇒10 × 10 - 3 × 507 + 490 × 10 - 3 × 0 = (490 + 10) × 10 - 3 × v_A
∴v_A= $\sqrt{7}$ m/s

When the block loses contact at D, the component mg acts on it.
Let the velocity at D be v_B.

$\frac{m (v_{B})^{2}}{r} = m g \sin θ (v_{B})^{2} = g r \sin θ . . . (1)$
Using work energy theorem
$(\frac{1}{2}) m v_{B}^{2} - (\frac{1}{2}) m v_{A}^{2} = - m g (0.2 + 0.2 \sin θ) (\frac{1}{2}) g r \sin θ - (\frac{1}{2}) {(\sqrt{7})}^{2} = g (0.2 + 0.2 \sin θ)$

$3.5 - \frac{1}{2} + 9.8 + 0.2 + \sin θ = 9.8 + 0.2 (1 + \sin θ) 3.5 - 0.98 \sin θ = 1.96 + 1.96 \sin θ \sin θ = \frac{1}{2} \Rightarrow θ = 30^{o}$

∴ Angle of projection = 90° − 30° = 60°

Time of reaching the ground,

$\Rightarrow t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times (0.2 + 0.2 \sin 30^{o})}{9.8}} \Rightarrow t = 0.247 s$

Distance travelled in the horizontal direction is given by,
S = v cos θ × t

S = gr sin θ cos θ×t
S = 9.8 × 2 × 1232 × 0.247
S = 0.196 m

Total distance = (0.2 − 0.2 cos 30° + 0.196)
                        = 0.22 m

Question 56:

Given:
Mass of block = 490 gm
Initial speed of the block = 0
Mass of bullet = 10 gm
Initial speed of the bullet, v_{1 =} $50 \sqrt{7} m / s$

As the bullet gets embedded inside the block, this is an example of a perfectly inelastic collision.

Let the final velocity of the system (block and bullet) be V_A.

Using the law of conservation of linear momentum, we get:
m₁v₁ + m₂ × 0 = (m_₁ + m₂)v_A
⇒10 × 10 - 3 × 507 + 490 × 10 - 3 × 0 = (490 + 10) × 10 - 3 × v_A
∴v_A= $\sqrt{7}$ m/s

When the block loses contact at D, the component mg acts on it.
Let the velocity at D be v_B.

$\frac{m (v_{B})^{2}}{r} = m g \sin θ (v_{B})^{2} = g r \sin θ . . . (1)$
Using work energy theorem
$(\frac{1}{2}) m v_{B}^{2} - (\frac{1}{2}) m v_{A}^{2} = - m g (0.2 + 0.2 \sin θ) (\frac{1}{2}) g r \sin θ - (\frac{1}{2}) {(\sqrt{7})}^{2} = g (0.2 + 0.2 \sin θ)$

$3.5 - \frac{1}{2} + 9.8 + 0.2 + \sin θ = 9.8 + 0.2 (1 + \sin θ) 3.5 - 0.98 \sin θ = 1.96 + 1.96 \sin θ \sin θ = \frac{1}{2} \Rightarrow θ = 30^{o}$

∴ Angle of projection = 90° − 30° = 60°

Time of reaching the ground,

$\Rightarrow t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times (0.2 + 0.2 \sin 30^{o})}{9.8}} \Rightarrow t = 0.247 s$

Distance travelled in the horizontal direction is given by,
S = v cos θ × t

S = gr sin θ cos θ×t
S = 9.8 × 2 × 1232 × 0.247
S = 0.196 m

Total distance = (0.2 − 0.2 cos 30° + 0.196)
                        = 0.22 m

Answer:

Given:
Mass of the 1st ball = m
Mass of the 2nd ball = 2m

When the balls reach the lower end,
Let the velocity of m be v; and the velocity of 2m be v_'.

Using the work-energy theorem, we can write:
$(\frac{1}{2}) \times m v^{2} - (\frac{1}{2}) \times m (0)^{2} = m g l \Rightarrow v = \sqrt{2 g l} Similarly, velocity of block having mass 2 m will be, v' = \sqrt{2 g l} ∴ v = v' = u (say)$

Let the velocities of m and 2m after the collision be v₁ and v₂ respectively.

Using the law of conservation of momentum, we can write:
$m \times u - 2 m u = m v_{1} + 2 m v_{2} \Rightarrow v_{1} + 2 v_{2} = - u . . . (1) On applying collision formula, we get : v_{1} - v_{2} = [(u - v)] = - 2 u . . . (2) Substracting equation (1) from (2), we get : 3 v_{2} = u \Rightarrow v_{2} = \frac{u}{3} = \sqrt{\frac{2 g l}{3}} Substituting this value in equation (1), we get : v_{1} - v_{2} = - 2 u \Rightarrow v_{1} = - 2 u + v_{2} = - 2 u + (\frac{u}{3}) = \frac{- 5}{3} u = \frac{- 5}{3} \times \sqrt{2 g l} = \frac{- \sqrt{50 g l}}{3}$

(b) Let the heights reached by balls 2m and m be h₁ and h respectively.

Using the work energy principle, we get:
$(\frac{1}{2}) \times 2 m \times (0)^{2} - (\frac{1}{2}) \times 2 m \times {(v_{2})}^{2} = - 2 m \times g \times h_{1} \Rightarrow h_{1} = (\frac{l}{9}) Similarly, (\frac{1}{2}) \times m \times (0)^{2} - (\frac{1}{2}) \times m v_{1}^{2} = m \times g \times h_{2} \Rightarrow (\frac{1}{2}) \times \frac{50 g l}{9} = g \times h_{2} \Rightarrow h_{2} = \frac{25 l}{9}$

As h₂ is more than 2l, the velocity at the highest point will not be zero.
Also, the mass m will rise by a distance 2l.

Question 57:

Given:
Mass of the 1st ball = m
Mass of the 2nd ball = 2m

When the balls reach the lower end,
Let the velocity of m be v; and the velocity of 2m be v_'.

Using the work-energy theorem, we can write:
$(\frac{1}{2}) \times m v^{2} - (\frac{1}{2}) \times m (0)^{2} = m g l \Rightarrow v = \sqrt{2 g l} Similarly, velocity of block having mass 2 m will be, v' = \sqrt{2 g l} ∴ v = v' = u (say)$

Let the velocities of m and 2m after the collision be v₁ and v₂ respectively.

Using the law of conservation of momentum, we can write:
$m \times u - 2 m u = m v_{1} + 2 m v_{2} \Rightarrow v_{1} + 2 v_{2} = - u . . . (1) On applying collision formula, we get : v_{1} - v_{2} = [(u - v)] = - 2 u . . . (2) Substracting equation (1) from (2), we get : 3 v_{2} = u \Rightarrow v_{2} = \frac{u}{3} = \sqrt{\frac{2 g l}{3}} Substituting this value in equation (1), we get : v_{1} - v_{2} = - 2 u \Rightarrow v_{1} = - 2 u + v_{2} = - 2 u + (\frac{u}{3}) = \frac{- 5}{3} u = \frac{- 5}{3} \times \sqrt{2 g l} = \frac{- \sqrt{50 g l}}{3}$

(b) Let the heights reached by balls 2m and m be h₁ and h respectively.

Using the work energy principle, we get:
$(\frac{1}{2}) \times 2 m \times (0)^{2} - (\frac{1}{2}) \times 2 m \times {(v_{2})}^{2} = - 2 m \times g \times h_{1} \Rightarrow h_{1} = (\frac{l}{9}) Similarly, (\frac{1}{2}) \times m \times (0)^{2} - (\frac{1}{2}) \times m v_{1}^{2} = m \times g \times h_{2} \Rightarrow (\frac{1}{2}) \times \frac{50 g l}{9} = g \times h_{2} \Rightarrow h_{2} = \frac{25 l}{9}$

As h₂ is more than 2l, the velocity at the highest point will not be zero.
Also, the mass m will rise by a distance 2l.

Answer:

Given is a uniform chain of mass M and length L.

Let us consider a small element of chain at a distance x from the floor having length 'dx'.

Therefore, mass of element, $d m = \frac{M}{L} d x$     ...(1)
The velocity with which the element strikes the floor is,
$v = \sqrt{2 g x}$     ...(2)

∴ The momentum transferred to the floor is,
$P = (d m) v = \frac{M}{L} . d x \sqrt{2 g x}$

According to the given condition, the element comes to rest.

Thus, the force (say F₁) exerted on the floor = Change in momentum
$F_{1} = \frac{d P}{d t} = \frac{M}{L} \times \frac{d x}{d t} \sqrt{2 g x} (for the chain element) A s \frac{d x}{d t} = v, F_{1} = \frac{M}{L} \times {(\sqrt{2 g x})}^{2} [using equation (2)] \Rightarrow F_{1} = \frac{M}{L} 2 g x = \frac{2 M g x}{L}$

Again, the force exerted due to length x of the chain on the floor due to its own weight is given by,
$W = \frac{M}{L} (x) \times g = \frac{M g x}{L}$

Thus, the total force exerted is given by.
F = F₁ + W
where W is the weight of chain below the element dx up to the floor.

$\Rightarrow F = \frac{2 M g x}{L} + \frac{M g x}{L} = \frac{3 M g x}{L}$

Question 58:

Given is a uniform chain of mass M and length L.

Let us consider a small element of chain at a distance x from the floor having length 'dx'.

Therefore, mass of element, $d m = \frac{M}{L} d x$     ...(1)
The velocity with which the element strikes the floor is,
$v = \sqrt{2 g x}$     ...(2)

∴ The momentum transferred to the floor is,
$P = (d m) v = \frac{M}{L} . d x \sqrt{2 g x}$

According to the given condition, the element comes to rest.

Thus, the force (say F₁) exerted on the floor = Change in momentum
$F_{1} = \frac{d P}{d t} = \frac{M}{L} \times \frac{d x}{d t} \sqrt{2 g x} (for the chain element) A s \frac{d x}{d t} = v, F_{1} = \frac{M}{L} \times {(\sqrt{2 g x})}^{2} [using equation (2)] \Rightarrow F_{1} = \frac{M}{L} 2 g x = \frac{2 M g x}{L}$

Again, the force exerted due to length x of the chain on the floor due to its own weight is given by,
$W = \frac{M}{L} (x) \times g = \frac{M g x}{L}$

Thus, the total force exerted is given by.
F = F₁ + W
where W is the weight of chain below the element dx up to the floor.

$\Rightarrow F = \frac{2 M g x}{L} + \frac{M g x}{L} = \frac{3 M g x}{L}$

Answer:

Given,
Speed of the block A = 10 m/s
The block B is kept at rest.
Coefficient of friction between floor and block B, μ = 0.10

Lets v₁ and v₂ be the velocities of A and B after collision respectively.

(a) If the collision is perfectly elastic, linear momentum is conserved.

Using the law of conservation of linear momentum, we can write:
$m u_{1} + m u_{2} = m v_{1} + m v_{2} \Rightarrow 10 + 0 = v_{1} + v_{2} v_{1} + v_{2} = 10 . . . (1) We know, Velocity of separation (after collision) = Velocity of approach (before collision) v_{1} - v_{2} = - (u_{1} - v_{2}) \Rightarrow v_{1} - v_{2} = - 10 . . . (2) Substracting equation (2) from (1), we get : 2 v_{2} = 20 \Rightarrow v_{2} = 10 m / s$

The deceleration of block B is calculated as follows:

Applying the work energy principle, we get:

$(\frac{1}{2}) \times m \times (0)^{2} - (\frac{1}{2}) \times m \times v^{2} = - m \times a \times s_{1} \Rightarrow - (\frac{1}{2}) \times (10)^{2} = - μ g \times s_{1} \Rightarrow s_{1} = \frac{100}{2 \times 1 \times 10} = 50 m$

(b) If the collision is perfectly inelastic, we can write:

$m \times u_{1} + m \times u_{2} = (m + m) \times v \Rightarrow m \times 10 + m \times 0 = 2 m \times v \Rightarrow v = (\frac{10}{2}) = 5 m / s$

The two blocks move together, sticking to each other.
∴ Applying the work-energy principle again, we get:

$(\frac{1}{2}) \times 2 m \times (0)^{2} - (\frac{1}{2}) \times 2 m \times (v)^{2} = 2 m \times μ g \times s_{2} \Rightarrow \frac{(5)^{2}}{0.1 \times 10 \times 2} = s_{2} \Rightarrow s_{2} = 12.5 m$

Question 59:

Given,
Speed of the block A = 10 m/s
The block B is kept at rest.
Coefficient of friction between floor and block B, μ = 0.10

Lets v₁ and v₂ be the velocities of A and B after collision respectively.

(a) If the collision is perfectly elastic, linear momentum is conserved.

Using the law of conservation of linear momentum, we can write:
$m u_{1} + m u_{2} = m v_{1} + m v_{2} \Rightarrow 10 + 0 = v_{1} + v_{2} v_{1} + v_{2} = 10 . . . (1) We know, Velocity of separation (after collision) = Velocity of approach (before collision) v_{1} - v_{2} = - (u_{1} - v_{2}) \Rightarrow v_{1} - v_{2} = - 10 . . . (2) Substracting equation (2) from (1), we get : 2 v_{2} = 20 \Rightarrow v_{2} = 10 m / s$

The deceleration of block B is calculated as follows:

Applying the work energy principle, we get:

$(\frac{1}{2}) \times m \times (0)^{2} - (\frac{1}{2}) \times m \times v^{2} = - m \times a \times s_{1} \Rightarrow - (\frac{1}{2}) \times (10)^{2} = - μ g \times s_{1} \Rightarrow s_{1} = \frac{100}{2 \times 1 \times 10} = 50 m$

(b) If the collision is perfectly inelastic, we can write:

$m \times u_{1} + m \times u_{2} = (m + m) \times v \Rightarrow m \times 10 + m \times 0 = 2 m \times v \Rightarrow v = (\frac{10}{2}) = 5 m / s$

The two blocks move together, sticking to each other.
∴ Applying the work-energy principle again, we get:

$(\frac{1}{2}) \times 2 m \times (0)^{2} - (\frac{1}{2}) \times 2 m \times (v)^{2} = 2 m \times μ g \times s_{2} \Rightarrow \frac{(5)^{2}}{0.1 \times 10 \times 2} = s_{2} \Rightarrow s_{2} = 12.5 m$

Answer:

Given:
Initial velocity of 2 kg block, v₁= 1.0 m/s
Initial velocity of the 4 kg block, v₂ = 0

Let the velocity of 2 kg block, just before the collision be u₁.

Using the work-energy theorem on the block of 2 kg mass:

The separation between two blocks, s = 16 cm = 0.16 m
$∴ (\frac{1}{2}) m \times u_{1}^{2} - (\frac{1}{2}) m \times (1)^{2} = - μ \times m g \times s \Rightarrow u_{1} = \sqrt{(1)^{2} - 2 \times 0.20 \times 10 \times 0.16} \Rightarrow u_{1} = 0.6 m / s$

As the collision is perfectly elastic, linear momentum is conserved.

Let v₁, v₂ be the velocities of 2 kg and 4 kg blocks, just after collision.

Using the law of conservation of linear momentum, we can write:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \Rightarrow 2 \times 0.6 + 4 \times 0 = 2 v_{1} + 4 v_{2} \Rightarrow 2 v_{1} + 4 v_{2} = 1.2 . . . (1)$

For elastic collision,
Velocity of separation (after collision) = Velocity of approach (before collision)
$i . e . v_{1} - v_{2} = + (u_{1} - u_{2}) = + (0.6 - 0) \Rightarrow v_{1} - v_{2} = - 0.6 . . . (2) Substracting equation (2) from (1), we get : 3 v_{2} = 1.2 \Rightarrow v_{2} = 0.4 m / s ∴ v_{1} = - 0.6 + 0.4 = - 0.2 m / s$

Let the 2 kg block covers a distance of S₁.

∴ Applying work-energy theorem for this block, when it comes to rest:

$(\frac{1}{2}) \times 2 \times (0)^{2} + (\frac{1}{2}) \times 2 \times (0.2)^{2} = - 2 \times 0.2 \times 10 \times S_{1} \Rightarrow S_{1} = 1 cm .$

Let the 4 kg block covers a distance of S₂.

Applying work energy principle for this block:

$(\frac{1}{2}) \times 4 \times (0)^{2} - (\frac{1}{2}) \times 4 \times (0.4)^{2} = - 4 \times 0.2 \times 10 \times S_{2} \Rightarrow 2 \times 0.4 \times 0.4 = 4 \times 0.2 \times 10 \times S_{2} \Rightarrow S_{2} = 4 cm$

Therefore, the distance between the 2 kg and 4 kg block is given as,
S₁ + S₂ = 1 + 4 = 5 cm

Question 60:

Given:
Initial velocity of 2 kg block, v₁= 1.0 m/s
Initial velocity of the 4 kg block, v₂ = 0

Let the velocity of 2 kg block, just before the collision be u₁.

Using the work-energy theorem on the block of 2 kg mass:

The separation between two blocks, s = 16 cm = 0.16 m
$∴ (\frac{1}{2}) m \times u_{1}^{2} - (\frac{1}{2}) m \times (1)^{2} = - μ \times m g \times s \Rightarrow u_{1} = \sqrt{(1)^{2} - 2 \times 0.20 \times 10 \times 0.16} \Rightarrow u_{1} = 0.6 m / s$

As the collision is perfectly elastic, linear momentum is conserved.

Let v₁, v₂ be the velocities of 2 kg and 4 kg blocks, just after collision.

Using the law of conservation of linear momentum, we can write:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \Rightarrow 2 \times 0.6 + 4 \times 0 = 2 v_{1} + 4 v_{2} \Rightarrow 2 v_{1} + 4 v_{2} = 1.2 . . . (1)$

For elastic collision,
Velocity of separation (after collision) = Velocity of approach (before collision)
$i . e . v_{1} - v_{2} = + (u_{1} - u_{2}) = + (0.6 - 0) \Rightarrow v_{1} - v_{2} = - 0.6 . . . (2) Substracting equation (2) from (1), we get : 3 v_{2} = 1.2 \Rightarrow v_{2} = 0.4 m / s ∴ v_{1} = - 0.6 + 0.4 = - 0.2 m / s$

Let the 2 kg block covers a distance of S₁.

∴ Applying work-energy theorem for this block, when it comes to rest:

$(\frac{1}{2}) \times 2 \times (0)^{2} + (\frac{1}{2}) \times 2 \times (0.2)^{2} = - 2 \times 0.2 \times 10 \times S_{1} \Rightarrow S_{1} = 1 cm .$

Let the 4 kg block covers a distance of S₂.

Applying work energy principle for this block:

$(\frac{1}{2}) \times 4 \times (0)^{2} - (\frac{1}{2}) \times 4 \times (0.4)^{2} = - 4 \times 0.2 \times 10 \times S_{2} \Rightarrow 2 \times 0.4 \times 0.4 = 4 \times 0.2 \times 10 \times S_{2} \Rightarrow S_{2} = 4 cm$

Therefore, the distance between the 2 kg and 4 kg block is given as,
S₁ + S₂ = 1 + 4 = 5 cm

Answer:

According to the question, the surface is frictionless. Thus, the block m will slide down the inclined plane of mass M.
Acceleration, a₁ = g sin α (Relative to the inclined plane)

The horizontal component of acceleration a₁ is given by a_x = g sin α cos α, for which the block M accelerates towards left.
Let the left acceleration be a₂.

By the concept of centre of mass, we can say that the external force is zero in the horizontal direction.

$m a_{x} = (M + m) a_{2}$

Absolute (resultant) acceleration of m on the plane M, along the direction of the incline will be = $a = g \sin α - a_{2} \cos α$

Let the time taken by the block m to reach the bottom end be t.

Now,
$s = u t + (\frac{1}{2}) a t^{2} \Rightarrow \frac{h}{\sin α} = (\frac{1}{2}) a t^{2} \Rightarrow t = \sqrt{\frac{2 h}{a \sin α}}$

Thus, the velocity of the bigger block after time t will be,

$v_{m} = u = a_{2} t = \frac{m g \sin α \cos α}{M + m} \sqrt{\frac{2 h}{a \sin α}} = {[\frac{2 m^{2} g^{2} h \sin^{2} α \cos^{2} α}{(M + m)^{2} a \sin α}]}^{1 / 2}$

Subtracting the value of a from equation (2), we get:

$v_{M} = [\frac{2 m^{2} g^{2} h \sin^{2} α}{(M + m)^{2} \sin α} \times \frac{\cos^{2} α}{g \sin α} \frac{(M + m)}{(M + m \sin^{2} α)}]$

Question 61:

According to the question, the surface is frictionless. Thus, the block m will slide down the inclined plane of mass M.
Acceleration, a₁ = g sin α (Relative to the inclined plane)

The horizontal component of acceleration a₁ is given by a_x = g sin α cos α, for which the block M accelerates towards left.
Let the left acceleration be a₂.

By the concept of centre of mass, we can say that the external force is zero in the horizontal direction.

$m a_{x} = (M + m) a_{2}$

Absolute (resultant) acceleration of m on the plane M, along the direction of the incline will be = $a = g \sin α - a_{2} \cos α$

Let the time taken by the block m to reach the bottom end be t.

Now,
$s = u t + (\frac{1}{2}) a t^{2} \Rightarrow \frac{h}{\sin α} = (\frac{1}{2}) a t^{2} \Rightarrow t = \sqrt{\frac{2 h}{a \sin α}}$

Thus, the velocity of the bigger block after time t will be,

$v_{m} = u = a_{2} t = \frac{m g \sin α \cos α}{M + m} \sqrt{\frac{2 h}{a \sin α}} = {[\frac{2 m^{2} g^{2} h \sin^{2} α \cos^{2} α}{(M + m)^{2} a \sin α}]}^{1 / 2}$

Subtracting the value of a from equation (2), we get:

$v_{M} = [\frac{2 m^{2} g^{2} h \sin^{2} α}{(M + m)^{2} \sin α} \times \frac{\cos^{2} α}{g \sin α} \frac{(M + m)}{(M + m \sin^{2} α)}]$

Answer:

,

According to the question, mass m is given with a speed v over the larger mass M.

(a) When the smaller block travels on the vertical part, let the velocity of the bigger block be v₁, towards left.

From the law of conservation of momentum (in the horizontal direction), we get:
$m v = (M + m) v_{1} \Rightarrow v_{1} = \frac{m v}{M + m}$

(b) When the smaller block breaks off, let its resultant velocity be v₂.

Using the law a of conservation of energy, we get:

$(\frac{1}{2}) m v^{2} = (\frac{1}{2}) M v_{1}^{2} + (\frac{1}{2}) m v_{2}^{2} + mgh$

(c) Vertical component of the velocity v₂ of mass m is given by,

${v_{y}}^{2} = {v_{2}}^{2} - {v_{1}}^{2}$

To determine the maximum height (from the ground),
Let us assume that the body rises to a height h₁ over and above h.

$Now, (\frac{1}{2}) m v_{y}^{2} = m g h_{1}$

$\Rightarrow H = \frac{M v^{2}}{(M + m) 2 g}$

(d) Because the smaller mass also has a horizontal component of velocity V₁ at the time it breaks off from M (that has a velocity v₁), the block m will again land on the block M.

The time of flight of block m after it breaks off is calculated as:

During the upward motion (BC),
$0 = v_{y} - g t$

Thus, the time for which the smaller block is in flight is given by,
$T = 2 t_{1} = \frac{2}{g} [\frac{M v^{2} - 2 (M + m) g h}{(M + m)}]$

The distance travelled by the bigger block during this time is,
$S = v_{1} T$

Question 62:

,

According to the question, mass m is given with a speed v over the larger mass M.

(a) When the smaller block travels on the vertical part, let the velocity of the bigger block be v₁, towards left.

From the law of conservation of momentum (in the horizontal direction), we get:
$m v = (M + m) v_{1} \Rightarrow v_{1} = \frac{m v}{M + m}$

(b) When the smaller block breaks off, let its resultant velocity be v₂.

Using the law a of conservation of energy, we get:

$(\frac{1}{2}) m v^{2} = (\frac{1}{2}) M v_{1}^{2} + (\frac{1}{2}) m v_{2}^{2} + mgh$

(c) Vertical component of the velocity v₂ of mass m is given by,

${v_{y}}^{2} = {v_{2}}^{2} - {v_{1}}^{2}$

To determine the maximum height (from the ground),
Let us assume that the body rises to a height h₁ over and above h.

$Now, (\frac{1}{2}) m v_{y}^{2} = m g h_{1}$

$\Rightarrow H = \frac{M v^{2}}{(M + m) 2 g}$

(d) Because the smaller mass also has a horizontal component of velocity V₁ at the time it breaks off from M (that has a velocity v₁), the block m will again land on the block M.

The time of flight of block m after it breaks off is calculated as:

During the upward motion (BC),
$0 = v_{y} - g t$

Thus, the time for which the smaller block is in flight is given by,
$T = 2 t_{1} = \frac{2}{g} [\frac{M v^{2} - 2 (M + m) g h}{(M + m)}]$

The distance travelled by the bigger block during this time is,
$S = v_{1} T$

Answer:

It is given that h is much lesser than the radius of the earth.

Mass of the earth, M_e = 6 × 10²⁴ kg
Mass of the block, M_b = 3 × 10²⁴ kg
Let V_e be the velocity of earth and V_b be the velocity of the block.

Let the earth and the block be attracted by gravitational force.

Thus, according to the conservation law of energy, the change in the gravitational potential energy will be the K.E. of block.
$G M_{e} M_{b} (\frac{1}{R + (\frac{h}{2})} - \frac{1}{R + h}) = (\frac{1}{2}) M_{e} \times V_{e}^{2} + (\frac{1}{2}) M_{b} \times V_{b}^{2}$ ...(1)

As only the internal force acts in this system, the momentum is conserved.
M_eV_e = M_bV_b
$\Rightarrow V_{e} = \frac{M_{b} V_{b}}{M_{e}} . . . (i i)$

Question 63:

It is given that h is much lesser than the radius of the earth.

Mass of the earth, M_e = 6 × 10²⁴ kg
Mass of the block, M_b = 3 × 10²⁴ kg
Let V_e be the velocity of earth and V_b be the velocity of the block.

Let the earth and the block be attracted by gravitational force.

Thus, according to the conservation law of energy, the change in the gravitational potential energy will be the K.E. of block.
$G M_{e} M_{b} (\frac{1}{R + (\frac{h}{2})} - \frac{1}{R + h}) = (\frac{1}{2}) M_{e} \times V_{e}^{2} + (\frac{1}{2}) M_{b} \times V_{b}^{2}$ ...(1)

As only the internal force acts in this system, the momentum is conserved.
M_eV_e = M_bV_b
$\Rightarrow V_{e} = \frac{M_{b} V_{b}}{M_{e}} . . . (i i)$

Answer:

According to the question, the collision of the two bodies of mass m is not head-on. Thus, the two bodies move in different directions.

Let the velocity vectors of the two bodies after collision be v₁ and v₂.

As the collision in the question is elastic, momentum is conserved.

On applying the law of conservation of momentum in X-direction, we get:

$m u_{1} + m \times 0 = m v_{1} \cos α + m v_{2} \cos β . . . (i)$

On applying the law of conservation of momentum in Y-direction, we get:

$0 = m v_{1} \sin α - m v_{2} \sin β . . . (i i)$

Question 64:

According to the question, the collision of the two bodies of mass m is not head-on. Thus, the two bodies move in different directions.

Let the velocity vectors of the two bodies after collision be v₁ and v₂.

As the collision in the question is elastic, momentum is conserved.

On applying the law of conservation of momentum in X-direction, we get:

$m u_{1} + m \times 0 = m v_{1} \cos α + m v_{2} \cos β . . . (i)$

On applying the law of conservation of momentum in Y-direction, we get:

$0 = m v_{1} \sin α - m v_{2} \sin β . . . (i i)$

Answer:

It is given that the mass of both the bodies (small particle and spherical body) is same.
Let the velocity of the particle be v.

We break the particle velocity into two components: v cosα (normal to the sphere) and v sinα (tangential to the sphere).

From figure,

cosα $= \sqrt{r^{2} - ρ^{2}}$ and sinα $= \frac{ρ}{r}$

Here, the collision occurs due to the component v cosα.

After an elastic collision, bodies of same mass exchange their respective velocities.
Thus, the spherical body will have a velocity v cosα, while the particle will not have any component of velocity in this direction.

However, the tangential velocity of the particle v sinα will remain unaffected.

Thus, we have:
Velocity of the sphere = v cosα $= \frac{v}{r} \sqrt{r^{2} - ρ^{2}}$
And,
Velocity of the particle = v sinα $= \frac{v ρ}{r}$

Centre of Mass Linear Momentum Collision

Class 11th Concepts Of Physics Part 1 HC Verma Solution

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Class 11^th Concepts Of Physics Part 1 HC Verma Solution