# HC Verma Class XII Physics Chapter 5- Newton’s Laws of Motion

**Exercise : Solution of Questions on page Number : 76**

**Answer: 1**

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.

**Answer: 2**

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.

**Answer: 3**

(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.

**Answer: 4**

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.

**Answer: 5**

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

**Answer: 6**

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.

**Answer: 7**

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.

**Answer: 8**

(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.

**Exercise : Solution of Questions on page Number : 77**

**Answer: 9**

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.

**Answer: 10**

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

**Answer: 11**

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

**Answer: 12**

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.

**Answer: 13**

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.

**Answer: 14**

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

**Answer: 15**

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.

**Answer: 16**

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.

**Answer: 17**

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.

moving on the inclined plane, the tension is mg. So, by measuring the tension in the string we can differentiate between the two cases.

**objective II**

**Answer: 1**

(c) w1 + w2

From the free-body diagram,

**Answer: 2**

(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.

**Answer: 3**

(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.

**Answer: 4**

(a) Both the scales will read 10 kg.

From the free-body diagram,

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

**Answer: 5**

(c) mg cosθ

From the free-body diagram,

N = mg cosθ

Normal force exerted by the plane on the block is mg cosθ.

**Answer: 6**

(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

**Answer: 7**

(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.

**Answer: 8**

(a) zero

As the whole system is at rest, the resultant force on the charged particle at A is zero.

**Exercise : Solution of Questions on page Number : 78**

**Answer: 9**

(b) F1 may be equal to F2.

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

**Answer: 10**

(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, aA = F/mA.

Acceleration of particle B in downward direction due to air resistance, aB = F/mB.

Therefore, A will go higher than B.

**Answer: 11**

(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).

**Answer: 12**

(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.

**Answer: 13**

(a) t_{1}=t_{2}

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

So t_{1}=t_{2}

**Answer: 14**

(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.

**Objective II**

**Answer: 1**

(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.

**Answer: 2**

(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

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. Fnet ≠ 0.

**Answer: 3**

(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 S1 is inertial, then S2 will be inertial and if S1 is non-inertial, then S2 will be non-inertial.

**Answer: 4**

(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.

**Answer: 5**

**Exercise : Solution of Questions on page Number : 79**

**Objective II**

**Answer: 6**

(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:

aeff = g + a

N = maeff = mg + ma (N > mg)

(ii) Going down and speeding up:

aeff = g – a

N = mg – ma (N < mg)

(iii) Going down and slowing down:

aeff = g – (–a) = g + a

N = mg + ma (N > mg)

(iv) Going up and slowing down:

aeff = g – a

N = mg – ma (N < mg)

**Answer: 7**

(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.

**Answer: 8**

(d) F_{1} = 0, F_{2} = 0

a_{s2s1}=a …1

Acceleration of the particle w.r.t. to S_{1} = F1/m

Acceleration of the particle w.r.t. to S_{2} = F2/m

If we assume F_{1} = 0 and F_{2} = 0,

we can conclude that a_{s2s1}=0 …2

From equations (1) and (2), we can say that our assumption is wrong.

And F_{1} = 0, F_{2} = 0 is not possible.

**Answer: 9**

(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.

**Exercise**

**Answer: 1**

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.

**Answer: 2**

Given:

**Answer: 3**

Initial velocity of the electrons is negligible, i.e. u = 0.

**Answer: 4**

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.

**Answer: 5**

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:

**Answer: 6**

Given:

Mass of the particle, m = 50 g = 5 × 10−2 kg

Slope of the v-t graph gives acceleration.

At t = 2 s,

At t = 4 s,

Slope = 0

So, acceleration, a = 0

⇒ F = 0

At t = 6 sec,

**Answer: 7**

Let F’ = force exerted by the experimenter on block A and F be the force exerted by block A on block B.

**Answer: 8**

Given:

**Answer: 9**

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

**Answer: 10**

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

**Answer: 11**

Mass of block A, m = 5 kg

F = ma = 10 N

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

**Answer: 12**

(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

As the man moves up, θ increases, i.e. cosθ decreases. Thus, F increases.

(b) When the man is at depth h,

**Exercise : Solution of Questions on page Number : 80**

**Answer: 13**

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

N=m(g-a)⇒N=0.5(10-2)=4 N

So, the force exerted by the block A on the block B is 4 N.

**Answer: 14**

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

**Answer: 15**

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’=mg+m-a …(2)

This is minimum weight, N’ = 60 × 9.9 N

From equations (1) and (2), we have:

2 mg = 1306.8

So, the true mass of the man is 66 kg.

And true weight = 66 × 9.9 = 653.4 N

(b) Using equation (1) to find the acceleration, we get:

mg + ma = 72 × 9.9

**Answer: 16**

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,

Adding both the equations, we get:

Now, using equation (1), we get:

**Answer: 17**

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,

Suppose, further elongation, when the 1 kg block is added, is ∆x.

**Answer: 18**

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

When 1 kg body is added,

total mass = (2 + 1) kg = 3 kg

Let elongation be x’.

∴ kx’ = 3g + 3a = 3 × 9.8 + 6

So, further elongation = x’ − x

= 0.36 − 0.24 = 0.12 m.

**Answer: 19**

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

Let the mass of the balloon be M’ so that it can rise with a constant velocity v in the upward direction.

∴ Amount of mass that should be removed = M − M’.

**Answer: 20**

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,

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

**Answer: 21**

For the particle to move without being deflected and with constant velocity, the net force on the particle should be zero.

** Answer: 22**

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)

(a) t = 2 s, a = 3.266 ms−2, u = 0

So, the distance travelled by the body,

(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

**Answer: 23**

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.

After this time, the mass m1 also starts moving downward.

So, the string becomes tight again after 23 s.

**Answer: 24**

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,

So, contact force,

F = 20 + 1a

F = 20 + 4 = 24 N

**Exercise : Solution of Questions on page Number : 81**

**Answer: 25**

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

**Answer: 26**

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

From the free-body diagram of block of mass m1,

**Answer: 27**

Let the acceleration of the blocks be a.

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

**Answer: 28**

(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.

**Answer: 29**

**Answer: 30**

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

**Answer: 31**

Let the acceleration of mass M be a.

(c) Let T’ = resultant of tensions

**Answer: 32**

The free-body diagram of the system is shown below:

Let acceleration of the block of mass 2M be a.

So, acceleration of the block of mass M will be 2a.

M(2a) + Mgsinθ − T = 0

⇒ T = 2Ma + Mgsinθ …(i)

2T + 2Ma − 2Mg = 0

From equation (i),

2(2Ma + Mgsinθ) + 2Ma − 2Mg = 0

4Ma + 2Mgsinθ + 2Ma − Mg = 0

6Ma + 2Mgsin30° + 2Mg = 0

6Ma = Mg

**Exercise : Solution of Questions on page Number : 82**

**Answer: 33**

The free-body diagram of the system is shown below:

Block ‘m’ will have the same acceleration as that of M’, as it does not slip over M’.

From the free body diagrams,

T + Ma – Mg = 0 …(i)

T – M’a – Rsinθ = 0 …(ii)

Rsinθ – ma = 0

Rcosθ – mg = 0

Eliminating T, R and a from the above equations, we get:

**Answer: 34**

(a) 5a + T − 5g = 0

From free-body diagram (1),

T = 5g − 5a …..(i)

From free-body diagram (2),

T = 8g + 16a ……(ii)

From equations (i) and (ii), we get:

5g − 5a = 8g + 16a

(b) From free body diagram-3,

⇒ 8a − T = 0

⇒ T = 8a

Again, T + 5a − 5g = 0

From free body diagram-4,

8a + 5a − 5g = s0

(c) T + 1a − 1g = 0

From free body diagram-5

T = 1g − 1a …..(i)

Again, from free body diagram-6,

T2-2g-4a=0

⇒ T − 4g − 8a = 0 …..(ii)

From equation (i)

1g − 1a − 4g − 8a = 0

⇒a=g3 downward

**Answer: 35**

Given,

m1 = 100 g = 0.1 kg

m2 = 500 g = 0.5 kg

m3 = 50 g = 0.05 kg

The free-body diagram for the system is shown below:

From the free-body diagram of the 500 g block,

T + 0.5a − 0.5g = 0 …..(i)

From the free-body diagram of the 50 g block,

T1 + 0.05g − 0.05a = a ….(ii)

From the free-body diagram of the 100 g block,

T1 + 0.1a − T + 0.5g = 0 ….(iii)

From equation (ii),

T1 = 0.05g + 0.05a …..(iv)

From equation (i),

T1 = 0.5g − 0.5a …..(v)

Equation (iii) becomes

T1 + 0.1a − T + 0.05g = 0

From equations (iv) and (v), we get:

0.05g + 0.05a + 0.1a − 0.5g + 0.5a + 0.05g = 0

0.65a = 0.4 g

**Answer: 36**

Mass of the monkey, m = 15 kg,

The free-body diagram of the monkey is shown below:

From the free-body diagram,

T − [15g + 15(a)] = 0

T − [15g + 15(1)] = 0

⇒ T = 5 (10 + 1)

⇒ T = 15 × 11 = 165 N

The monkey should apply a force of 165 N to the rope.

Initial velocity, u = 0

s = 5 m

**Answer: 37**

Suppose the monkey accelerates upward with acceleration a and the block accelerates downward with acceleration a’.

Let force exerted by the monkey be F.

From the free-body diagram of the monkey, we get:

F− mg − ma = 0 …(i)

⇒ F = mg + ma

Again, from the free-body diagram of the block,

F + ma’ − mg = 0

mg + ma + ma’ − mg = 0 [From (i)]

⇒ ma = −ma’

⇒ a = −a’

If acceleration −a’ is in downward direction then the acceleration a’ will be in upward direction.

This implies that the block and the monkey move in the same direction with equal acceleration.

If initially they were at rest (no force is exerted by the monkey), then their separation will not change as time passes because both are moving same direction with equal acceleration.

**Answer: 38**

Let the acceleration of monkey A upwards be a, so that a maximum tension of 30 N is produced in its tail.

T − 5g − 30 − 5a = 0 …(i)

30 − 2g − 2a = 0 …(ii)

From equations (i) and (ii), we have:

T = 105 N (max.)

and a = 5 m/s2

So, A can apply a maximum force of 105 N on the rope to carry monkey B with it.

For minimum force, there is no acceleration of A and B.

Rewriting equation (i) for monkey A, we get:

T − 5g − 20 = 0

⇒ T = 70 N

∴ To carry monkey B with it, monkey A should apply a force of magnitude between 70 N and 105 N.

**Answer: 39**

(i) Given, mass of the man = 60 kg

Let W’ = apparent weight of the man in this case

From the free-body diagram of the man,

W’ + T − 60g = 0

⇒ T = 60g − W’ …(i)

From the free-body diagram of the box,

T − W’ = 30g = 0 …(ii)

From equation (i), we get:

60g − W’ − W’ − 30g = 0

⇒ W’ = 15g

Hence, the weight recorded on the machine is 15 kg.

(ii) To find his actual weight, suppose the force applied by the men on the rope is T because of which the box accelerates upward with an acceleration a’.Here we need to find T’.

Correct weight = W = 60g

From the free-body diagram of the man,

T’ + W − 60g − 60a = 0

⇒ T’ − 60a = 0

⇒ T’ = 60a …(i)

From the free-body diagram of the box,

T’ − W − 30g − 30a = 0

⇒ T’ − 60g − 30g − 30a = 0

⇒ T’ = 30a − 900 …(ii)

From equations (i) and (ii), we get:

T’ = 2T’ − 1800

T’ = 1800 N

So, the man should exert a force of 1800 N on the rope to record his correct weight on the machine.

**Exercise : Solution of Questions on page Number : 83**

**Answer: 40**

The force on the block which makes the body move down the plane is the component of its weight parallel to the inclined surface.

F = mg sinθ

Acceleration, g = sin θ

Initial velocity of block, u = 0

Distance to be covered

s = l

a = g sin θ

Using, s=ut+12at2, we get:

**Answer: 41**

Let the pendulum (formed by the ball and the string) make angle θ with the vertical.

(ii) Let the angle of the incline be θ.

From the diagram,

**Answer: 42**

The free-body diagram of the system is shown below: