Newton's Laws of Motion

Question 1:

Question 2:

No, the accelerating elevator will affect the weight of both sides of the beam balance. So, the net effect of the accelerating elevator cancels out, and we get the actual mass.

Question 3:

During free fall:
Acceleration of the boy = Acceleration of mass M = g
Acceleration of mass M w.r.t. boy, a = 0
So, the force exerted by the box on the boy's head = M × a = 0

Yes, the force greatly increases during the period he balances himself after striking the ground because of the weight of the box.

Question 4:

(a) In the car, the path of the coin will be vertically downward because the only force acting on the coin is gravity in the downward direction.

(b) In a free falling elevator, the coin as well as the person will be in a condition of weightlessness. So, the coin will remain stationary w.r.t. the person.

Question 5:

If no force acts on the particle it cannot change its direction. So, it is not possible for a particle to describe a curved path if no force acts on it.

Yes, the answer depends on the frame of reference chosen to view the particle if the frame of reference describes a curved path.

Question 6:

We are pushed forward because of the inertia of motion, as our body opposes the sudden change.

Question 7:

We can't find a body whose acceleration is zero with respect to all other bodies in the universe because every body in the universe is moving with respect to other bodies.As we live on earth which itself is accelerates due to its revolution around the sun and spinning about its own axis, so whatever observations and measurements ,we make , are w.r.t to earth which itself is not an inertial frame.Similarly all other planets are also in motion around the sun so tdeally no inertial frame is possible.

 

Question 8:

Yes, if the force on the object is zero, its acceleration w.r.t. all the other objects will we zero. So, the frame will necessarily be an inertial frame.

Question 9:

(a)


The reading of the balance = Tension in the string
And tension in the string = 20g
So, the reading of the balance = 20g = 200 N

(b) If the balance is heavy, the reading will not change because the weight of spring balance does not affect the tension in the string.

(c) If the blocks have unequal masses, the spring balance will accelerate towards the heavy block with an acceleration a. Then the reading will be equal to the tension in the string.

Suppose m_{1 >} m₂.
Then tension in the string,
$T=\frac{2m_1{m}_{2}g}{m_1+m_2}$

Question 10:

No. The acceleration of the particle can also be zero if the vector sum of all the forces is zero, i.e. no net force acts on the particle.

Question 11:

The force of friction acting between my feet and ground is responsible for my deceleration.

Question 12:

In both the cases, change in momentum is same but the time interval during which momentum changes to zero is less in the first case. So, by $F=\frac{dP}{dt}$ , force in the first case will be more. That's why we are injured when we jump barefoot on a hard surface.

Question 13:

The forces on the rope must be equal and opposite, according to Newton's third law. But not all the forces acting on each team are equal. The friction between one team and the ground does not depend on the other team and can be larger on one side than on the other. In addition, the grips on the rope need not be equal and opposite. Thus, the net force acting on each team from all sources need not be equal.

Question 14:

Air applies a velocity-dependent force on the parachute in upward direction when the parachute opens. This force opposes the gravitational force acting on the spy. Hence, the net force in the downward direction decreases and the spy decelerates.

Question 15:

Yes, this is an example of Newton's third law of motion, which sates that every action has an equal and opposite reaction.

Question 16:

Yes, the forces due to the spring on the two blocks are equal and opposite.
But it's not an example of Newton's third law because there are three objects (2 blocks + 1 spring). Spring force on one block and force by the same block on the spring is an action-reaction pair.

Question 17:

No, w.r.t. the ground frame, the person's head is not really pushed backward.
As the train moves, the  lower  portion of the passenger's body starts moving with the train, but the upper portion tries to be in rest according to Newton's first law and hence, the passenger seems to be pushed backward.

Question 1:

No, a person sitting inside the compartment can't tell just by looking at the plumb line whether the train is accelerating on a horizontal straight track or moving on an incline.
When the train is accelerating along the horizontal, the tension in the string is $m\sqrt{g^2+a^2}$; when it is moving on the inclined plane, the tension is mg. So, by measuring the tension in the string we can differentiate between the two cases.

Question 2:

(c) w₁ + w₂



From the free-body diagram,
(w₁ + w₂) – N = 0
N = w₁ + w₂
The ceiling pulls the chain by a force (w₁ + w₂).

Question 3:

(b) the ground on the horse

The horse pushes the ground in the backward direction and, in turn, the ground pushes the horse in the forward direction, according to Newton's third law of motion.

Question 4:

(d) the road

The car pushes the ground in the backward direction and according to the third law of motion, reaction force of the ground in the forward direction acts on the car.

Question 5:

(a) Both the scales will read 10 kg.





From the free-body diagram,
K₁x₁ = mg = 10$\times$9.8 = 98 N
K₂x₂ = K₁x₁  
So,  K₁x₁ = K₂x₂ = 98 N

Therefore, both the spring balances will read the same mass, i.e. 10 kg.

Question 6:

(c) mg cosθ



From the free-body diagram,
N = mg cosθ
Normal force exerted by the plane on the block is mg cosθ.

Question 7:

(b) mg/cosθ

 
Free-body Diagram of the Small Block of Mass 'm'

The block is at equilibrium w.r.t. to wedge. Therefore,
mg sinθ = ma cosθ
⇒ a = gtanθ  

Normal reaction on the block is
N = mg cosθ + ma sinθ

Putting the value of a, we get:
N = mg cosθ + mg tanθsinθ

$N=mg\cos \theta +mg\frac{\sin \theta }{\cos \theta }\sin \theta N=\frac{mg}{\cos \theta }$

Question 8:

(d) remain standing

If the earth suddenly stops attracting objects placed near its surface, the net force on the person will become zero and according to the first law of motion, the person will remain standing.

Question 9:

(a) zero

Using, F_net= ma,
a = 0 ⇒ F_{net = 0}
As the whole system is at rest, the resultant force on the charged particle at A is zero.

Question 10:

(b) F₁ may be equal to F₂.

Any force applied in the direction opposite the motion of the particle decelerates it to rest. 

Question 11:

(b) A will go higher than B.

Let the air exert a constant resistance force = F (in downward direction).
Acceleration of particle A in downward direction due to air resistance, a_A = F/m_A.
Acceleration of particle B in downward direction due to air resistance, a_B = F/m_B.
m_A > m_B
a_A < a_B
$S=ut+\frac{1}{2}a{t}^2$
$\mathrm{So}, H_A=ut-\frac{1}{2}(a_A+g)t^2$
$H_B=ut-\frac{1}{2}(a_B+g)t^2$
$H_A>H_B$

Therefore, A will go higher than B.

Question 12:

(c) remain at the top of the wedge

Downward gravitational force will be balanced by the upward pseudo force (because of the motion of the wedge in downward direction). The block will remain at its position, as both the box and the inclined plane are falling with the same acceleration (g). 

Question 13:

(b) t₁ > t₂

Let acceleration due to air resistance force be a.
Let H be maximum height attained by the particle.
Direction of air resistance force is in the direction of motion.

In the upward direction of motion, $a_{\mathrm{eff}}=\left|g-a\right|$.
$t_1=\sqrt{\frac{2H}{\left|g-a\right|}} ...\left(1\right)$

In the downward direction of motion, $a_{\mathrm{eff}}=g+a$.
$t_2=\sqrt{\frac{2H}{g+a}} ...\left(2\right)$
So, t₁ > t₂.

Question 14:

(a) t₁ = t₂

After the coin is dropped, the only force acting on it is gravity, which is same for both the cases.
So t₁ = t₂.

Question 1:

(c) x

The moving train does not put any extra force on the alpha particle and the recoiling nucleus. So, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger, will be same as before, i.e. x.

Question 2:

(b) cannot be an inertial frame because the earth is revolving around the sun
(d) cannot be an inertial frame because the earth is rotating about its axis

A reference frame attached to the earth cannot be an inertial frame because the earth is revolving around the sun and also rotating about its axis.

Question 3:

(c) the frame may be inertial but the resultant force on the particle is zero
(d) the frame may be non-inertial but there is a non-zero resultant force

According to Newton's second law which says that net force acting on the particle is equal to rate of change of momentum ( or mathematically F = ma),  so if a particle is at rest then F_net = ma = m$\frac{dv}{dt} = m\frac{d\left(0\right)}{dt} = m\times 0 = 0$.
Now, if the frame is inertial, then the resultant force on the particle is zero.

If the frame is non-inertial,
vector sum of all the forces plus a pseudo force is zero.
i.e. F_net ≠ 0.

Question 4:

(a) Both the frames are inertial.
(b) Both the frames are non-inertial.

 S₁ is moving with constant velocity w.r.t frame S₂. So, if S₁is inertial, then S₂will be inertial and if S₁is non-inertial, then S₂will be non-inertial.

Question 5:

(a) AB
(c) CD

Slope of the x-t graph gives velocity. In the regions AB and CD, slope or velocity is constant, i.e. acceleration is zero. Hence, from the second law, force is zero in these regions.

Question 6:

(a) F₁= F₂ = F, for t < 0

At t < 0, the block is in equilibrium in the horizontal direction.
So, F₁= F₂ = F
At t > 0, F₂= 0 and F₁= F.

Question 7:

(b) going up and speeding up
(c) going down and slowing down

It means normal force exerted by the floor of the elevator on the person is greater that the weight of the person.
i.e. N > mg

(i) Going up and speeding up:
a_eff = g + a
N = ma_eff = mg + ma (N > mg)

(ii) Going down and speeding up:
a_eff= g – a
N = mg – ma (N < mg)

(iii) Going down and slowing down:
a_eff = g – (–a) = g + a
N = mg + ma (N > mg)

 (iv) Going up and slowing down:
 a_eff = g – a
  N = mg – ma (N < mg)

Question 8:

(c) going up with uniform speed
(d) going down with uniform speed

Tension in the cable = Weight of the elevator
Or, total upward force = total downward force
That is, there's no acceleration or uniform velocity.
So, the elevator is going up/down with uniform speed.

Question 9:

(d) F₁ = 0, F₂ = 0

$a_{s_2{s}_1}=a ...\left(1\right)$   
Acceleration of the particle w.r.t. to S₁ = F₁/m
Acceleration of the particle w.r.t. to S₂ = F₂/m
If we assume  F₁ = 0 and F₂ = 0,
we can conclude that $a_{s_2{s}_1}=0 ...\left(2\right)$
From equations (1) and (2), we can say that our assumption is wrong.
And F₁ = 0, F₂ = 0 is not possible.

Question 1:

(d) He might have used a non-inertial frame

If no force is acting on a particle and yet, its acceleration is non-zero, it means the observer is in a non-inertial frame.

Question 2:

Given:
Mass of the block, m = 2 kg,
Distance covered, S = 10 m and initial velocity, u = 0
Let a be the acceleration of the block.
Using, $\mathrm{S}=ut+\frac{1}{2}a{t}^2$, we get:
$$\begin{gathered} 10=\frac{1}{2}a\left(2^2\right) \\ \Rightarrow 10=2a \\ \Rightarrow a=5 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
∴ Force, F = ma = 2 × 5 = 10 N

Question 3:

Given:
Initial speed of the car, u = 40 km/hr $=\frac{4000}{3600}=11.11 \mathrm{m}/\mathrm{s}$
Final speed of the car, v = 0
Mass of the car, m = 2000 kg
Distance to be travelled by the car before coming to rest, s = 4m
Acceleration, $a=\frac{v^2-u^2}{2\mathrm{s}}$
$\Rightarrow a=\frac{0^2-{\left(11.11\right)}^2}{2\times 4}=\frac{-123.43}{8}=-15.42 \mathrm{m}/{\mathrm{s}}^2$
∴ Average force to be applied to stop the car, F = ma
⇒ F = 2000 × 15.42 ≈ 3.1 × 10⁴ N

Question 4:

Initial velocity of the electrons is negligible, i.e. u = 0.
Final velocity of the electrons, v = 5 × 10⁶ m/s
Distance travelled by the electrons,
s = 1 cm = 1 × 10⁻² m
∴ Acceleration, $a=\frac{v^2-u^2}{2 \mathrm{S}}$
  $$\begin{gathered} \Rightarrow a=\frac{{\left(5\times {10}^5\right)}^2-0}{2\times 1\times {10}^{-3}}=\frac{25\times {10}^{12}}{2\times {10}^{-2}} \\ \Rightarrow a=12.5\times {10}^{14} \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

So, force on the electrons, F = ma
⇒ F = 9.1 × 10⁻³¹ × 12.5 × 10⁻¹⁴
      = 1.1 × 10⁻¹⁵ N

Question 5:

The free-body diagrams for both the blocks are shown below:
        


From the free-body diagram of the 0.3 kg block,
T = 0.3g
⇒ T= 0.3 × 10 = 3 N

Now, from the free-body diagram of the 0.2 kg block,
T₁= 0.2g + T
⇒ T₁= 0.2 × 10 + 3 = 5 N
∴ The tensions in the two strings are 5 N and 3 N, respectively.

Question 6:

 

Let a be the common acceleration of the blocks.
For block 1,
$F-T=ma$    ...(1)
For block 2,
T = ma    ...(2)

Subtracting equation (2) from (1), we get:
$F-2T=0$
$\Rightarrow T=\frac{F}{2}$      

Question 7:

Given:
Mass of the particle, m = 50 g = 5 × 10⁻² kg
Slope of the v-t graph gives acceleration.
At t = 2 s,
Slope = $\frac{15}{3}=5 \mathrm{m}/{\mathrm{s}}^2$
So, acceleration, a = 5 m/s²
F = ma = 5 × 10⁻² × 5
⇒ F = 0.25 N along the motion.

At t = 4 s,
Slope = 0
So, acceleration, a = 0
⇒ F = 0

At t = 6 sec,
Slope = $\frac{-15}{3}=-5 \mathrm{m}/{\mathrm{s}}^2$
So, acceleration, a = − 5 m/s²
F = ma = − 5 × 10⁻² × 5
⇒ F = − 0.25 N along the motion
or, F = 0.25 N opposite the motion.

Question 8:

Let F' = force exerted by the experimenter on block A and F be the force exerted by block A on block B.
      
Let a be the acceleration produced in the system.

For block A,
$F\text{'}-F=m_{A}a$    ...(1)
For block B,
F = m_Ba    ...(2)               
Dividing equation (1) by (2), we get:
$\frac{F\text{'}}{F}-1=\frac{m_A}{m_B}$
$\Rightarrow F\text{'}=F\left(1+\frac{m_{\mathrm{A}}}{m_{\mathrm{B}}}\right)$
∴ Force exerted by the experimenter on block A is $F\left(1+\frac{m_{\mathrm{A}}}{m_{\mathrm{B}}}\right)$.

Question 9:

Given:
Radius of a raindrop, r = 1 mm = 10⁻³ m
Mass of a raindrop, m = 4 mg = 4 × 10⁻⁶ kg
Distance coved by the drop on the head, s = 10⁻³ m
Initial speed of the drop, v = 0
Final speed of the drop, u = 30 m/s
Using $a=\frac{v^2-u^2}{2s}$, we get:
$a=\frac{-{\left(30\right)}^2}{2\times {10}^{-3}}=-4.5\times {10}^5 m/s^2$
 Force, F = ma
⇒ F = 4 × 10⁻⁶ × 4.5 × 10⁵
       = 1.8 N

Question 10:

Displacement of the particle from the mean position, x = 20 cm = 0.2 m
k = 15 N/m
Mass of the particle, m = 0.3 kg
Acceleration, $a=\frac{\left|\mathit{F}\right|}{m}$
$\Rightarrow a=\frac{kx}{m}=\frac{15\left(0.2\right)}{0.3}=\frac{3}{0.3}=10 \mathrm{m}/{\mathrm{s}}^2$
So, the initial acceleration when the particle is released from a point x = 20 cm is 10 m/s².

Question 11:

Let the block m be displaced towards left by displacement x.

     

∴ F₁ = $-$k₁x   (compressed)
    F₂ = $-$k₂x   (expanded)
$ma=F_1+F_2$
$\Rightarrow a=\frac{-x\left(k_1+k_2\right)}{m}$

i.e. $\left(k_1+k_2\right)\frac{x}{m}$ opposite the displacement or towards the mean position.

Question 12:

Mass of block A, m = 5 kg
 F = ma = 10 N
$\Rightarrow a=\frac{10}{5}=2 \mathrm{m}/{\mathrm{s}}^2$
As there is no friction between A and B, when block A moves, block B remains at rest in its position.

Initial velocity of A, u = 0
Distance covered by A to separate out,
s = 0.2 m
Using $s=ut+\frac{1}{2}a{t}^2$, we get:
$0.2=0+\frac{1}{2}\times 2t^2$
⇒ t² = 0.2
⇒ t = 0.44 s ≈ 0.45 s

Question 13:

(a) At any depth, let the ropes makes an angle θ with the vertical.



From the free-body diagram,
Fcosθ + Fcosθ − mg = 0
2Fcosθ = mg
$\Rightarrow F=\frac{mg}{2cos\theta }$
As the man moves up, θ increases, i.e. cosθ decreases. Thus, F increases.

(b) When the man is at depth h,
$\cos \theta =\frac{h}{\sqrt{{\left(d/2\right)}^2+h^2}}$

$$\begin{gathered} F=\frac{mg}{2\cos \theta } \\ =\frac{mg}{2\left[h/\sqrt{{\left({\displaystyle \frac{d}{2}}\right)}^2+h^2}\right]} \\ =\frac{mg}{4h}\sqrt{d^2+4h^2} \end{gathered}$$

Question 14:

When the elevator is descending, a pseudo-force acts on it in the upward direction, as shown in the figure.  

 

From the free-body diagram of block A,
$mg-N=ma$
$$\begin{gathered} N=m\left(g-a\right) \\ \Rightarrow N=0.5\left(10-2\right)=4 \mathrm{N} \end{gathered}$$
So, the force exerted by the block A on the block B is 4 N.

Question 15:

(a) When the elevator goes up with acceleration 1.2 m/s²:

        

$T=mg+ma$
⇒ T = 0.05 (9.8 + 1.2) = 0.55 N

(b) Goes up with deceleration 1.2 m/s² :

        
$T=mg+m\left(-a\right)=m\left(g-a\right)$
⇒ T  = 0.05 (9.8 − 1.2) = 0.43 N

(c) Goes up with uniform velocity:

        
$T=mg$
⇒ T = 0.05 × 9.8 = 0.49 N

(d) Goes down with acceleration 1.2 m/s² :

       

$$\begin{gathered} T+ma=mg \\ \Rightarrow T=m\left(g-a\right) \end{gathered}$$
⇒ T = 0.05 (9.8 − 1.2) = 0.43 N

(e) Goes down with deceleration 1.2 m/s² :
       
$$\begin{gathered} T+m\left(-a\right)=mg \\ \Rightarrow T=m\left(g+a\right) \end{gathered}$$
⇒ T = 0.05 (9.8 + 1.2) = 0.55 N

(f) Goes down with uniform velocity:

       

$T=mg$
⇒ T = 0.05 × 9.8 = 0.49 N

Question 16:

Maximum weight will be recorded when the elevator accelerates upwards.
Let N be the normal reaction on the person by the weighing machine.
So, from the free-body diagram of the person,

                   
$N=mg+ma$    ...(1)
This is maximum weight, N = 72 × 9.9 N

When decelerating upwards, minimum weight will be recorded.
$N\text{'}=mg+m\left(-a\right)$    ...(2)
This is minimum weight, N' = 60 × 9.9 N

From equations (1) and (2), we have:
2 mg = 1306.8
$\Rightarrow m=\frac{1306.8}{2\times 9.9}=66 \mathrm{kg}$   
So, the true mass of the man is 66 kg.
And true weight = 66 $\times$ 9.9 = 653.4 N

(b) Using equation (1) to find the acceleration, we get:
mg + ma = 72 × 9.9
$$\begin{gathered} \Rightarrow a=\frac{72\times 9.9-66\times 9.9}{66}=\frac{9.9\times 6}{66}=\frac{9.9}{11} \\ \Rightarrow a=0.9 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Question 17:

Let the left and right blocks be A and B, respectively.
And let the acceleration of the 3 kg mass relative to the elevator be 'a' in the downward direction.



From the free-body diagram,
$m_{A}a=T-m_{A}g-\frac{m_{A}g}{10} ...\left(1\right)$
$m_{B}a=m_{B}g+\frac{m_{B}g}{10}-T ...\left(2\right)$

Adding both the equations, we get:
$a\left(m_A+m_B\right)=\left(m_B-m_A\right)g+\left(m_B-m_A\right)\frac{g}{10}$

Putting value of the masses,we get:
$$\begin{gathered} 9a=\frac{33g}{10} \\ \Rightarrow \frac{a}{g}=\frac{11}{30} ...\left(3\right) \end{gathered}$$
Now, using equation (1), we get:
$T=m_A\left(a+g+\frac{g}{10}\right)$
The reading of the spring balance = $\frac{2T}{g}=\frac{2}{g}{m}_A\left(a+g+\frac{g}{10}\right)$
$$\begin{gathered} \Rightarrow 2\times 1.5\left(\frac{a}{g}+1+\frac{1}{10}\right)=3\left(\frac{11}{30}+1+\frac{1}{10}\right) \\ = 4.4 \mathrm{kg} \end{gathered}$$

Question 18:

Given,
mass of the first block, m = 2 kg
k = 100 N/m
Let elongation in the spring be x.
             
From the free-body diagram,
kx = mg
$$\begin{gathered} x=\frac{mg}{k}=\frac{2\times 9.8}{100} \\ =\frac{19.6}{100}=0.196\approx 0.2 \mathrm{m} \end{gathered}$$

Suppose, further elongation, when the 1 kg block is added, is $∆x$.
Then,$k\left(x+∆x\right)=m\text{'}g$
⇒ k$∆x$= 3g − 2g = g
$\Rightarrow ∆x=\frac{g}{k}=\frac{9.8}{100}=0.098\approx 0.1 \mathrm{m}$

Question 19:

When the ceiling of the elevator is going up with an acceleration 'a', then a pseudo-force acts on the block in the downward direction.
      
a = 2 m/s²

From the free-body diagram of the block,
kx = mg + ma
⇒ kx = 2g + 2a
        = 2 × 9.8 + 2 × 2
        = 19.6 + 4
$\Rightarrow x=\frac{23.6}{100}=0.236\approx 0.24 \mathrm{m}$

When 1 kg body is added,
total mass = (2 + 1) kg = 3 kg
Let elongation be x'.
∴ kx' = 3g + 3a = 3 × 9.8 + 6
$$\begin{gathered} \Rightarrow x\text{'}=\frac{35.4}{100} \\ =0.354\approx 0.36 \mathrm{m} \end{gathered}$$
So, further elongation = x' − x
                                  = 0.36 − 0.24 = 0.12 m.

Question 20:

Let M be mass of the balloon.
Let the air resistance force on balloon be F .
Given that F ∝ v.
⇒ F= kv,
where k = proportionality constant.

     
When the balloon is moving downward with constant velocity,
B + kv = Mg    ...(i)
$\Rightarrow M=\frac{B+kv}{g}$
Let the mass of the balloon be M' so that it can rise  with a constant velocity v in the upward direction.
 B = Mg + kv
$\Rightarrow M\text{'}=\frac{B+kv}{g}$
∴ Amount of mass that should be removed = M − M'.
$$\begin{gathered} ∆M=\frac{B+kv}{g}-\frac{B-kv}{g} \\ =\frac{B+kv-B+kv}{g} \\ =\frac{2kv}{g}=\frac{2\left(Mg-B\right)}{g} \\ =2\left\{M-\frac{B}{g}\right\} \end{gathered}$$

Question 21:

Let U be the upward force of water acting on the plastic box.
Let m be the initial mass of the plastic box.
When the empty plastic box is accelerating upward,
$$\begin{gathered} U-mg=\frac{mg}{6} \\ \Rightarrow U=\frac{7 mg}{6} \end{gathered}$$

      

$\Rightarrow m=\frac{6U}{7g} ....\left(i\right)$

Let M be the final mass of the box after putting some sand in it.

$$\begin{gathered} Mg-U=\frac{Mg}{6} \\ \Rightarrow Mg-\frac{Mg}{6}=U \\ \Rightarrow M=\frac{6U}{5g} ....\left(ii\right) \end{gathered}$$

Mass added$=\frac{6U}{5g}-\frac{6U}{7g}$
                   $$\begin{gathered} =\frac{6U\left(7-5\right)}{35 g} \\ =\frac{6U·2}{35 g} \end{gathered}$$
From equation (i), $m=\frac{6U}{7g}$
∴ Mass added$=\frac{2}{5}m$ .

Question 22:

For the particle to move without being deflected and with constant velocity, the net force on the particle should be zero.
$\overrightarrow{F}+m\overrightarrow{g}=0$
$$\begin{gathered} \Rightarrow \left(\overrightarrow{v}\times \overrightarrow{A}\right)+\overrightarrow{mg}=0 \\ \Rightarrow \left(\overrightarrow{v}\times \overrightarrow{A}\right)=-\overrightarrow{mg} \end{gathered}$$

 $\left|vA\sin \theta \right|=\left|mg\right|$
$\therefore v=\frac{mg}{A\sin \theta }$
v will be minimum when sinθ = 1.
⇒ θ = 90°
$\therefore v_{min}=\frac{mg}{A}$

Question 23:

The masses of the blocks are m₁ = 0.3 kg and m₂ = 0.6 kg

The free-body diagrams of both the masses are shown below:

        
For mass m₁,
 T − m₁g = m₁a    ...(i)

For mass m₂,
m₂g − T= m₂a    ...(ii)

Adding equations (i) and (ii), we get:
g(m₂ − m₁) = a(m₁ + m₂)
$$\begin{gathered} \Rightarrow a=g\left(\frac{m_2-m_1}{m_1+m_2}\right) \\ =9.8\times \frac{0.6-0.3}{0.6+0.3} \\ =3.266 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

(a) t = 2 s, a = 3.266 ms⁻², u = 0
So, the distance travelled by the body,
$$\begin{gathered} \mathrm{S}=ut+\frac{1}{2}a{t}^2 \\ =0+\frac{1}{2}\left(3.266\right)2^2=6.5 \mathrm{m} \end{gathered}$$

(b) From equation (i),
T = m₁ (g + a)
   = 0.3 (3.8 + 3.26) = 3.9 N

(c)The force exerted by the clamp on the pulley,
F = 2T = 2 × 3.9 = 7.8 N

Question 24:

a = 3.26 m/s², T = 3.9 N
After 2 s, velocity of mass m_1,
v = u + at = 0 + 3.26 × 2
   = 6.52 m/s upward

At this time, m₂ is moving 6.52 m/s downward.

At time 2 s, m₂ stops for a moment. But m₁ is moving upward with velocity 6.52 m/s. It will continue to move till final velocity (at highest point) becomes zero.

Here, v = 0, u = 6.52 m/s
a = −g = − 9.8 m/s²
v = u + at = 6.52 + (−9.8)t
$\Rightarrow t=\frac{6.52}{9.8}\approx \frac{2}{3} \sec$
After this time, the mass m₁ also starts moving downward.
So, the string becomes tight again after $\frac{2}{3} \mathrm{s}$.

Question 25:

Mass per unit length$=\frac{3}{30} \mathrm{kg}/\mathrm{cm}$ = 0.10 kg/cm
Mass of the 10 cm part, m₁ = 1 kg
Mass of the 20 cm part, m₂ = 2 kg

Let F = contact force between them.



From the free-body diagram,
$$\begin{gathered} m_{1}a=F-20 ...\left(\mathrm{i}\right) \\ {m}_{2}a=32-F ...\left(\mathrm{ii}\right) \\ \mathrm{Adding} \mathrm{both} \mathrm{the} \mathrm{equation}\text{s}, \mathrm{we} \mathrm{get}: \\ a=\frac{12}{m_1+m_2}=\frac{12}{3}=4 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$         
So, contact force,
F = 20 + 1a
F = 20 + 4 = 24 N

Question 26:

Mass of each block is 1 kg, $\sin {\mathrm{\theta }}_1=\frac{4}{5}$, $\sin {\mathrm{\theta }}_2=\frac{3}{5}$.
The free-body diagrams for both the boxes are shown below:


 
mgsinθ₁ − T = ma    ...(i)
 T − mgsinθ₂ = ma    ...(ii)

Adding equations (i) and (ii),we get:
mg(sinθ₁− sinθ₂) = 2ma
⇒ 2a = g (sinθ₁ − sinθ₂)
$$\begin{gathered} \Rightarrow a=\frac{g}{5}\times \frac{1}{2} \\ =\frac{g}{10} \end{gathered}$$

Question 27:

The free-body diagrams for both the blocks are shown below:
       

From the free-body diagram of block of mass m₁,
m₁a = T − F    ...(i)

From the free-body diagram of block of mass m₂,
m₂a = m₂g − T    ...(ii)

Adding both the equations, we get:
$$\begin{gathered} a\left(m_1+m_2\right)=m_{2}g-\frac{m_{2}g}{2} \left[\mathrm{because} F=\frac{m_{2}g}{2}\right] \\ \Rightarrow a=\frac{m_{2}g}{2\left(m_1+m_2\right)} \end{gathered}$$
So, the acceleration of mass m₁,
$a=\frac{m_{2}g}{2\left(m_1+m_2\right)}, \mathrm{towards} \text{the} \mathrm{right}.$

Question 28:

Let the acceleration of the blocks be a.
The free-body diagrams for both the blocks are shown below:

    

From the free-body diagram,
m₁a = m₁g + F − T    ...(i)

Again, from the free-body diagram,
m₂a = T − m₂g − F    ...(ii)

Adding equations (i) and (ii), we have:
$a=g\frac{m_1-m_2}{m_1+m_2}$
$$\begin{gathered} \Rightarrow a=\frac{3g}{7}=\frac{29.4}{7} \\ =4.2 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
Hence, acceleration of the block is 4.2 m/s².

After the string breaks, m₁ moves downward with force F acting downward. Then,

       
m₁a = F + m₁g
5a = 1 + 5g
 $$\begin{gathered} \Rightarrow a=\frac{5g+1}{5} \\ =g+0.2 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Question 29:

 The free-body diagram for mass m₁ is shown below:
      
          (Figure 1)
The free-body diagram for mass m₂ is shown below:

     
         (Figure 2)
The free-body diagram for mass m₃ is shown below:

            
        (Figure 3)

Suppose the block m₁ moves upward with acceleration a₁ and the blocks m₂ and m₃ have relative acceleration a₂ due to the difference of weight between them.

So, the actual acceleration of the blocks m₁, m₂ and m₃ will be a₁, (a₁ − a₂) and (a₁ + a₂), as shown.
From figure 2, T − 1g − 1a₁ = 0    ...(i)
From figure 3, $\frac{T}{2}-2g-2\left(a_1-a_2\right)=0 ...\left(\mathrm{ii}\right)$
From figure 4, $\frac{T}{2}-3g-3\left(a_1+a_2\right)=0 ...\left(\mathrm{iii}\right)$

From equations (i) and (ii), eliminating T, we get:
1g + 1a₂ = 4g + 4 (a₁ + a₂)
5a₂ − 4a₁ = 3g    ...(iv)

From equations (ii) and (iii), we get:
2g + 2(a₁ − a₂) = 3g − 3 (a₁ − a₂)
5a₁ + a₂ = g    ...(v)

Solving equations (iv) and (v), we get:
$$\begin{gathered} a_1=\frac{2g}{29} \\ \mathrm{and} a_2=g-5a_1 \\ \Rightarrow a_2=g-\frac{10g}{29}=\frac{19g}{29} \\ \mathrm{So}, a_1-a_2=\frac{2g}{29}-\frac{19g}{29}=-\frac{17g}{29} \\ \mathrm{and} a_1+a_2=\frac{2g}{29}+\frac{19g}{29}=\frac{21g}{29} \end{gathered}$$
So, accelerations of m₁, m₂ and m₃ are $\frac{19g}{29}\left(\mathrm{up}\right), \frac{17g}{29} \left(\mathrm{down}\right) \mathrm{and} \frac{21g}{29}\left(\mathrm{down}\right)$ , respectively.
Now, u = 0, s = 20 cm = 0.2 m
$$\begin{gathered} a_2=\frac{19g}{29} \\ \therefore s=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 0.2=\frac{1}{2}\times \frac{19}{29}g{t}^2 \\ \Rightarrow t=0.25 \mathrm{s} \end{gathered}$$