Laws Of Motion

As we know, uniform translatory motion is defined as if all the particles of body moves in same straight line i.e. direction will be constant and with the same velocity.

So from above definition we can say that option (a) is totally wrong, option (b) is also wrong because option talking about circular path & option (d) is obviously wrong.

Therefore, option (a) is correct.

question, a metre scale is moving with uniform velocity that means velocity of meter scale is constant also we know change in constant quantity is zero so rate of change of velocity (i.e. acceleration) will also be zero.

Now from Newton’s 2^nd law i.e. _net but we have a(acceleration) is zero i.e. , so . _net

∴ Net force on metre scale must be zero so from our options (c) & (d) are absolutely wrong.

Now torque is the cross product of position vector and force vector also we have force vector in this case is zero i.e. _net

And torque is so putting force vector value

hence from here it is obvious that options (a) is wrong.

Therefore, option (b) is correct.

We know that change in momentum =final momentum – initial momentum 0r now we have given mass 150g i.e. m₁=m₂=m=0.15kg also and so

So

Therefore, option (c) is correct.

We have derived from above question that

Also we know that magnitude of any vector can be obtained by so

Therefore, option (c) is correct.

As we know newton 2^nd law state that rate of change of momentum is equal to external force applied on anybody or mathematically _ext now conservation of momentum mean _ext or so

Hence we can understand conservation of momentum by 2^nd law so options (a) & (b) are wrongs.

Now let’s look to 3^rd law which state that every action has equal and opposite reaction or in another word action force is equal to reaction force in magnitude but opposite in direction.

i.e. we can also write this as

From here the law of conservation of momentum also proved.

Therefore, option (d) is correct.

As we know rate of change of momentum is force . So in this case direction of force acting on player will be the same as the direction of change in momentum. Now to understand this in better let’s look the image below

& shows player’s northward and westward directions respectively. Now shows south direction so is towards south-west which means force on player will act on this direction.

Therefore, option (c) is correct.

We have given in question x(t) = pt + qt² + rt³ so to find the force we will differentiate above position equation two times i.e.

We have where

So

Now first derivative of above equation

_t=2

Hence

Therefore, option (a) is correct.

We have to find the time at which the final velocity along the y axis which means x-component will be zero. We have

_x now X-component

Therefore, option (b) is correct

Car start from rest mean initial velocity u=0, also and t=2s mass is given as m.

As we know so

Now force exerted on the car i.e. here this force is due to engine of car so whatever this force is actually internal force in another words due to friction force the car moves in eastward direction.

Therefore, option (b) is correct.

We have given mass=m, for t<0

Now let’s look for time interval

Also we know,

1^st derivative of x(t) w.r.t. time will give velocity

⇒

2^nd derivative of x(t) w.r.t. time will give velocity

⇒

Then,

Here, this force is obviously time dependent so force is not constant hence option (d) is correct.

Now let’s look to option (a) i.e. At t then

⇒

So from this we can say option (a) also correct.

Now for option (b) i.e. between t = 0 s and t = (1/4) s, as we know impulse is change in momentum or so at t = (1/4) s impulse will be

here we have to keep in mind that F(t) varies from t=0 to maximum 1/8 s, so F (1/4) will be replaced by F (1/8) also from above we have found

⇒ so option(b) is also correct.

Option (c) is absolutely wrong as we have above found that force is acting.

Option (e) is also wrong because we have already calculated the impulse.

We have given Let acceleration in body A and B is ‘a’.

Body A will move along with body B by force F till the force of friction between surface of A and B is larger or equal to zero.

Now taking system A+B then acceleration will be

⇒

⇒

⇒

So force on A

⇒ F_AB

⇒F_AB

⇒F_AB

If F_ABis equal or smaller than f₂ then body A will move along with body B.

So f₂=F_AB or

…(i)

N=Reaction force by B on A

[N₂=Normal reaction on B along with A by surface]

…. (ii)

From (i)

so adding this with equation (ii) we can say F …(iii)

F = 0.45mg Newton is the maximum force on B. so that A and B can move together. So option (e) is correct.

Both bodies can move together if F is less than or equal to 0.45mg Newton.

So options (a) and (b) are also correct and rejects the option (c) as 0.5mg>0.45mg.

For option (d): Minimum force which can move A and B together

⇒F_min

⇒F_min

⇒F_min

Given force in option (d) 0.1mg Newton <0.25mg Newton. So body A and B will not move i.e.

Bodies A and B will remain in rest hence option (d) is also correct.

Let’s consider a case in which normal reaction i.e.

N from figure also we know friction

Now from figure taking whole as a system then

when m₁ will up and m₂ will down.

Putting f in this equation

⇒

⇒ or in simply

⇒ from this option (a) is totally wrong while option (b) is correct.

Now if body m₁ moves down and m₂ moves up then, direction of friction force (f) becomes upward (opp. to motion).

⇒

⇒

⇒

Hence option (d) is correct but option (c) is wrong.

In question it is mention that plane below block A has μ co-efficient of friction also block B lying on frictionless surface.

Let’s consider two cases

Case 1: When A just start i.e.

Body A moves up and B down the plane

mg will cancel out

Or so option (b) is correct.

Case 2: When A moves upward and B downward

⇒

⇒

⇒

from this we can say option (a) is wrong and option (c) is correct.

Now if B moves upward and A downward then

⇒

⇒

⇒

⇒

So from here we can say that as θ₁ increases also increases but decreases also sum of θ₁ and θ₂ is 90° so θ₁> θ₂ and also right.

Hence from here body B can move up that means option (d) is wrong.

Given that speed of each ball v=5 m/s and mass of each ball m=0.005 kg.

So initial momentum of each ball

⇒

The direction of the velocity of each ball is reversed after the collision so final momentum of each ball

⇒

So from above result it is obvious that option (d) is correct.

As we know change in momentum of each ball is equal to impulse imparted to each ball

⇒

⇒

⇒

So from here we can say options (a) and (b) are wrong and option (d) is correct.

We have given mass m=10 kg, F₂=8N, F₁=6N

Let’s look this figure to understand better here R is resultant force

Now we know F …(1)

So …(2)

…(3)

Hence from equations (1) & (2) option (a) is correct and from equations (1) & (3) option (c) also our acceleration is different hence options (b) & (d) are wrong.

Let m₁ and m₂ be the mass of girl with bicycle and stone respectively i.e. m₁=50kg & m₂=0.5kg

Also we have given u₁=5m/s, u₂=5m/s and v₂=15 m/s and we have to find v₁

As we know by law of conservation of momentum

initial momentum (girl, bicycle, stone) =final momentum (cycle and girl) + stone

i.e.

Therefore, the speed of cycle and girl decreased by

Let the acceleration be ‘a’ when the lift is descending, then the apparent weight decreases on weighing scale

So

Apparent weight due to reaction force by the lift on weighing scale will be

Reading of weighing scale

As we know impulse is change in momentum which can be written in terms of force

We have given mass of body m=2 kg, initial velocity (v₁) at t=0 is zero. Now in the interval x-t graph is straight line that mean velocity of body will remain constant.

So v₂

And when the slope of graph is zero so velocity will also be zero i.e. v₃=0

Now or Impulse=change in momentum

Impulse at

Impulse at

So from above observation impulse at increased by and at it is decreased by .

On seeing a child on the road when a person applies breaks suddenly, the lower part of person slows rapidly with the car, but the upper part of driver continues to move with same speed in the same direction due to the inertia of motion his head car hit with steering.

we have mass m=2kg and

V at 2 sec,

So momentum at 2 sec p (2) =mv (2)

Now

So force then

Hence momentum and force at 2 sec will be and respectively.

Let f (frictional force) at y-axis and F at x-axis

Now when a small force F₁ is applied on a heavier box, this box does not move, at this state, force of friction f₁ is equal to F₁. Increasing force on box does not move till F = f_s (the maximum static frictional or limiting force).

After force f_s, the frictional force decrease i.e. less kinetic force F_k<f_s is applied on body and it starts to move with less friction

In diagram A is equivalent to limiting frictional force and a B is equivalent to kinetic frictional force.

Packing of fragile materials in paper, straw or bubble wrap is a common practice to avoid damage via transportation. During transportation, a vehicle can experience sudden turns, bumps or halts. As the body continues to be in its state of motion as by the law of inertia, such incidents can cause the objects to move and fall. Large forces offered by the point of contact such as walls or floors can easily break them. If we wrap the objects in paper or straw, the materials can effectively absorb impact and the object can move within the straw for a distance before coming to a complete stop. The force is thereby reduced and the damage can be avoided.

Tough cement floor is robust and tough whereas a muddy ground is soft and cushiony. When a child falls on a cement floor, the force acts on the child instantly and thus the impact is more. To reduce the impact of force, we increase the time of action of the force or we could increase the distance travelled by the object. This is what happens in a muddy ground. As the momentum is gradually changed over a due course of time in a muddy ground, the impact is less and the child suffers less pain.

(a)To solve this question, we look into the theory of Impulse. Impulse is defined as the change in momentum and is written as:

where, I is the impulse, F is the force applied on the object, Δt is the time of application of force, m is the mass of the object and v is the velocity of the object.

Here, m=500gm=0.5kg

v_initial = 0 m s^-1

v_final = 25 m s^-1

Therefore,

I= mΔv = m(v_final - v_initial)= 0.5 ( 25 - 0) = 12.5 N s. (ANS)

(b) Here, m=500gm=0.5kg

v_initial = -25 m s^-1

v_final = 12.5 m s^-1

There, change in momentum is:

Δp = mΔv = 0.5 [ 12.5 - (-25)]

= 18.75 kg m s^-1 (ANS)

The mountain roads are generally constructed in a winding fashion so as to increase friction and thereby reduce skidding of vehicles. This comes from the definition of friction for an object placed at a slope of angle θ .

Here, f is the frictional force, μ is the coefficient of friction, m is the mass of the object, g is the acceleration due to gravity and θ is the angle made by the object with the surface.

Now, winding the road means decreasing the θ with respect to ground. This will increase friction as cosine will increase. Hence the frictional force increases. Going straight up means going at a larger angle so the friction will decrease.

Here, θ’ is the angle made by the slope of the mountain with the ground whereas, θ is the angle made by each turn with the ground. As it is clear that θ ’> θ.

.

To solve such questions, we first draw the free body diagram to see what forces are acting on the system. Here, we can see that the weight of the body is acting downwards and given by “mg”. The corresponding tension on the string AB is given by “T”. In addition to these forces, we apply a third force via string CD in the downward direction given by “F”. So,

T = F + mg

As the tension is the greatest force, given by the sum of the weight and the additional force applied, string AB will break.

In case of a jerk, we are applying a large amount of force for a really small duration of time. This instantaneous force is not transmitted through the block to the string AB. So, in this situation, string CD will break.

In such questions, we first draw the free body diagrams. From the above diagram, we write the Newton’s equation for each mass.
Here, M = 5 kg
m = 3 kg

a = 2 m s^-2

Therefore, the equations are:

Substituting the values of m, M, a and g and solving the above equations gives us:

T₁ = 94.4 N

T₂ = 35.4 N

For system to be in equilibrium, the forces should balance out.

For block B, the equation is:

—————(1)

For block A the equation is:

—————(2)

Substituting value of F from

So the weight of the block should be 50N.

Here, the force by the finger is given by “F”, the normal force is “N”, the frictional force is “f” and the weight of the body is acting downwards given by “Mg”.

We first balance the horizontal forces:

Now we balance the horizontal forces. Here, we have to stop the block from falling. So the acceleration should be zero.
—————(1)

The frictional force is given by:

—————(2)

Substituting (2) in (1)

Given, the height of the cliff is,

Height of the cliff, h=500 m

Mass of the ball, m= 1 kg

Mass of the gun, M=100 kg

Range of the ball, R= 400 m

Acceleration due to gravity, g= 10 m s^-2

Let the final velocity of the ball be u and the of the gun be v.

Now we know, time of flight is

And range is given by:

So, u=400/10 = 40 m s^-2

Therefore, the velocity is:

The recoil velocity of the gun is 0.4 m s^-1.

The force acting on any particle is mass(m) times acceleration(a).

Here, we need to find the acceleration from the above graphs.

From the x-t graph, we can see the relation that

x=t

And from the y-t curve,

y=t²

so,

ax = x’’ = 0 m s^-2

ay = y’’ = 2 ms^-2

So, the force is

Fx = 0

Fy = m x 2 = 0.5 x 2 = 1 N

Hence the force is 1N and is along y-axis.

Here, the initial velocity of the coin is, u = 20 m s¹

And the acceleration of the elevator is, a = 2 m s^-2

Here, let’s think this problem from the elevator reference frame. For the person inside the elevator going up with acceleration 2 m s^-2 will experience a net acceleration of (g+a) which is 12 m s^-2.

And as the coin return back to its original position (because we are in the elevator reference frame), the net displacement (s) is zero.

So we can use the laws of motion to solve the problem.

Here, a’=a+g = 12 m s^-2

So the time of flight of the coin is 3.33 seconds.

(a)We know force is a vector quantity. From the theory of planes, we know that two intersecting lines form a plane.

Let these lines be F₁ and F₂. This means that the vector sum, F₁ + F₂ is also on this plane.

As the body is moving with uniform speed, the acceleration is zero. So the net of all forces is zero.

F₁ + F₂ + F₃ =0

So, F₃ = - (F₁ + F₂)

This means even F₃ lies on the same plane as F₁ and F₂.

Thus the forces are coplanar.

(b) As the sum of all forces are zero, the net torque in any direction is zero. For example, torque about “O” is

Τ = OP x (F₁ + F₂ + F₃)

Τ = 0 as (F₁ + F₂ + F₃)=0

Hence, torque acting about any point is also zero.

Here, we have two cases, one with friction and one without friction. The distance travelled by the block in both cases is same.

Let the distance travelled by the block be “s” the mass of the block be “m”, the coefficient of friction be “μ” and the acceleration due to gravity be “g”. The angle made by the incline with the ground is θ .

Case:1 No friction

The equation of motion is

Here, initial velocity, u = 0

Acceleration in the incline frame of reference is, a=g(sinθ)

Case:2 With Friction

Here, we first calculate the acceleration.

From the free body diagram,

Using this acceleration in the equation of motion. The time taken is given by t’.

We know t’=pt

Substituting the values of t’, and t

Squaring both sides and simplifying,

Given, θ=45°

Here, as the velocity shows different behavior at different times, we have to divide time into intervals to specify velocity.

Now, we can differentiate velocity to calculate Forces(F).

Hence, the total force is

To find the total time, we find the time of each segments.

Segment:ABC

Here, the frictional force is

And the centripetal force is

Therefore,

Segment:DEF

Again, the centripetal force is equal to friction.

Now, the time taken is distance/speed.

For ABC, the distance is

And for DEF we have

Therefore,

Hence the time taken is 86.3 seconds.

(a) Here, the x and y coordinates are

x = A cos ωt

y = B sin ωt

The equation of an ellipse is:

Substituting x and y, we find that LHS = RHS.

So the trajectory is an ellipse.

(b) To find force, we need to find acceleration.

This can be found by differentiating r two times with respect to t.

And

We know,

In case (a), we have only horizontal velocity. From the conservation of energy at the point when ball hits the ground,

K.E = P.E

So, the total velocity is

At the point of contact, v_h= 0

So, from the conservation of energy,

Hence, both techniques will yield the same speed.

In this problem, we have to resolve forces horizontally and vertically.

Case: 1 Vertically

Case: 2 Horizontally

(a)For this condition to be true, the forces must balance each other.
The frictional force(f) is given by:

Therefore,

where, μ is the coefficient of friction, N is the normal to the surface, g is the acceleration due to gravity and θ is the angle of the incline.

So for θ equal to tan-1(μ), the box just starts to slide.

(b) From the figure, we can resolve the forces for 𝜶 > 𝜽:

(c)In this scenario, as we trying to move the box up the plane, the direction of friction changes and is now acting down the plane.

The equation of force is:

(d)Here, the acceleration is in the direction of the force. Thus resolving the forces, we have,

(a) From the diagram, the force on the floor(F) is given by:

Where m is the mass of the crew and a is the upwards acceleration.

(b) Here, we have to consider all the weight if the system. The force due to rotor F_r is:

(c) This force is just the reaction force of Fr. So the force due to surrounding air is 6.25 x 10⁴ N.