HC Verma i Solutions for Class 12 Science Physics Chapter 5 Newton's Laws Of Motion are provided here with simple step-by-step explanations. These solutions for Newton's Laws Of Motion are extremely popular among class 12 Science students for Physics Newton's Laws Of Motion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma i Book of class 12 Science Physics Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s HC Verma i Solutions. All HC Verma i Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

Page No 76:

Question 1:

Answer:

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.

Page No 76:

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.

Answer:

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.

Page No 76:

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.

Answer:

(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.

Page No 76:

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.

Answer:

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.

Page No 76:

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.

Answer:

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

Page No 76:

Question 6:

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

Answer:

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.

 

Page No 76:

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.

 

Answer:

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.

Page No 76:

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.

Answer:

(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 m1 > m2.
Then tension in the string,
T=2m1m2gm1+m2



Page No 77:

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 m1 > m2.
Then tension in the string,
T=2m1m2gm1+m2

Answer:

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.

Page No 77:

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.

Answer:

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

Page No 77:

Question 11:

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

Answer:

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=dPdt , force in the first case will be more. That's why we are injured when we jump barefoot on a hard surface.

Page No 77:

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=dPdt , force in the first case will be more. That's why we are injured when we jump barefoot on a hard surface.

Answer:

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.

Page No 77:

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.

Answer:

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.

Page No 77:

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.

Answer:

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

Page No 77:

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.

Answer:

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.

Page No 77:

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.

Answer:

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.

Page No 77:

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.

Answer:

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 mg2+a2; 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.

Page No 77:

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 mg2+a2; 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.

Answer:

(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).

Page No 77:

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).

Answer:

(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.

Page No 77:

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.

Answer:

(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.

Page No 77:

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.

Answer:

(a) Both the scales will read 10 kg.





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

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

Page No 77:

Question 5:

(a) Both the scales will read 10 kg.





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

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

Answer:

(c) mg cosθ



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

Page No 77:

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.

Answer:

(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=mgcosθ+mgsinθcosθsinθN=mgcosθ

Page No 77:

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=mgcosθ+mgsinθcosθsinθN=mgcosθ

Answer:

(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.

Page No 77:

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.

Answer:

(a) zero

Using, Fnet= ma,
a = 0 Fnet = 0
As the whole system is at rest, the resultant force on the charged particle at A is zero.



Page No 78:

Question 9:

(a) zero

Using, Fnet= ma,
a = 0 Fnet = 0
As the whole system is at rest, the resultant force on the charged particle at A is zero.

Answer:

(b) F1 may be equal to F2.

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

Page No 78:

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. 

Answer:

(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.
mA > mB
aA < aB
S=ut+12at2
So, HA=ut-12(aA+g)t2
HB=ut-12(aB+g)t2
HA>HB

Therefore, A will go higher than B.

Page No 78:

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, aA = F/mA.
Acceleration of particle B in downward direction due to air resistance, aB = F/mB.
mA > mB
aA < aB
S=ut+12at2
So, HA=ut-12(aA+g)t2
HB=ut-12(aB+g)t2
HA>HB

Therefore, A will go higher than B.

Answer:

(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). 

Page No 78:

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). 

Answer:

(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, aeff=g-a.
t1=2Hg-a    ...(1)

In the downward direction of motion, aeff=g+a.
t2=2Hg+a    ...(2)
So, t1 > t2.

Page No 78:

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, aeff=g-a.
t1=2Hg-a    ...(1)

In the downward direction of motion, aeff=g+a.
t2=2Hg+a    ...(2)
So, t1 > t2.

Answer:

(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.

Page No 78:

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.

Answer:

(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.

Page No 78:

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.

Answer:

(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.

Page No 78:

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.

Answer:

(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 Fnet = ma = mdvdt = md(0)dt = m×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. Fnet ≠ 0.

Page No 78:

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 Fnet = ma = mdvdt = md(0)dt = m×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. Fnet ≠ 0.

Answer:

(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.

Page No 78:

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.

Answer:

(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.

Page No 78:

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.

Answer:

(a) F1F2 = F, for t < 0

At t < 0, the block is in equilibrium in the horizontal direction.
So, F1F2 = F
At t > 0, F2= 0 and F1= F.



Page No 79:

Question 6:

(a) F1F2 = F, for t < 0

At t < 0, the block is in equilibrium in the horizontal direction.
So, F1F2 = F
At t > 0, F2= 0 and F1= F.

Answer:

(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)

Page No 79:

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:
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:

(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.

Page No 79:

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.

Answer:

(d) F1 = 0, F2 = 0

as2s1=a    ...1   
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 as2s1=0    ...2
From equations (1) and (2), we can say that our assumption is wrong.
And F1 = 0, F2 = 0 is not possible.

Page No 79:

Question 9:

(d) F1 = 0, F2 = 0

as2s1=a    ...1   
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 as2s1=0    ...2
From equations (1) and (2), we can say that our assumption is wrong.
And F1 = 0, F2 = 0 is not possible.

Answer:

(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.

Page No 79:

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.

Answer:

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, S=ut+12at2, we get:
10=12a2210=2aa=5 m/s2
∴ Force, F = ma = 2 × 5 = 10 N

Page No 79:

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, S=ut+12at2, we get:
10=12a2210=2aa=5 m/s2
∴ Force, F = ma = 2 × 5 = 10 N

Answer:

Given:
Initial speed of the car, u = 40 km/hr =40003600=11.11 m/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=v2-u22s
a=02-11.1122×4=-123.438=-15.42 m/s2   
∴ Average force to be applied to stop the car, F = ma
⇒ F = 2000 × 15.42 ≈ 3.1 × 104 N

Page No 79:

Question 3:

Given:
Initial speed of the car, u = 40 km/hr =40003600=11.11 m/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=v2-u22s
a=02-11.1122×4=-123.438=-15.42 m/s2   
∴ Average force to be applied to stop the car, F = ma
⇒ F = 2000 × 15.42 ≈ 3.1 × 104 N

Answer:

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=v2-u22 S
  a=5×1052-02×1×10-3=25×10122×10-2a=12.5×1014 m/s2

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

Page No 79:

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=v2-u22 S
  a=5×1052-02×1×10-3=25×10122×10-2a=12.5×1014 m/s2

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

Answer:

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.

Page No 79:

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.

Answer:


 

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
T=F2      

Page No 79:

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
T=F2      

Answer:

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 = 153=5 m/s2
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 = -153=-5 m/s2
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.

Page No 79:

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 = 153=5 m/s2
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 = -153=-5 m/s2
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.

Answer:

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'-F=mAa    ...(1)
For block B,
F = mBa    ...(2)               
Dividing equation (1) by (2), we get:
F'F-1=mAmB
F'=F1+mAmB     
∴ Force exerted by the experimenter on block A is F1+mAmB.

Page No 79:

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'-F=mAa    ...(1)
For block B,
F = mBa    ...(2)               
Dividing equation (1) by (2), we get:
F'F-1=mAmB
F'=F1+mAmB     
∴ Force exerted by the experimenter on block A is F1+mAmB.

Answer:

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=v2-u22s, we get:
a=-3022×10-3=-4.5×105 m/s2   
 Force, F = ma
⇒ F =
4 × 10−6 × 4.5 × 105
       = 1.8 N

Page No 79:

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=v2-u22s, we get:
a=-3022×10-3=-4.5×105 m/s2   
 Force, F = ma
⇒ F =
4 × 10−6 × 4.5 × 105
       = 1.8 N

Answer:

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=Fm
a=kxm=150.20.3=30.3=10 m/s2  
So, the initial acceleration when the particle is released from a point x = 20 cm is 10 m/s2.

Page No 79:

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=Fm
a=kxm=150.20.3=30.3=10 m/s2  
So, the initial acceleration when the particle is released from a point x = 20 cm is 10 m/s2.

Answer:

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

     

F1 = -k1x   (compressed)
    F2 = -k2x   (expanded)
ma=F1+F2
a=-xk1+k2m        

i.e. k1+k2xm opposite the displacement or towards the mean position.

Page No 79:

Question 11:

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

     

F1 = -k1x   (compressed)
    F2 = -k2x   (expanded)
ma=F1+F2
a=-xk1+k2m        

i.e. k1+k2xm opposite the displacement or towards the mean position.

Answer:

Mass of block A, m = 5 kg
 F = ma = 10 N
a=105=2 m/s2
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+12at2, we get:
0.2=0+12×2t2
t2 = 0.2
t = 0.44 s ≈ 0.45 s

Page No 79:

Question 12:

Mass of block A, m = 5 kg
 F = ma = 10 N
a=105=2 m/s2
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+12at2, we get:
0.2=0+12×2t2
t2 = 0.2
t = 0.44 s ≈ 0.45 s

Answer:

(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
F=mg2cosθ
As the man moves up, θ increases, i.e. cosθ decreases. Thus, F increases.

(b) When the man is at depth h,
cosθ=hd/22+h2

F=mg2cosθ =mg2h/d22+h2 =mg4hd2+4h2



Page No 80:

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
F=mg2cosθ
As the man moves up, θ increases, i.e. cosθ decreases. Thus, F increases.

(b) When the man is at depth h,
cosθ=hd/22+h2

F=mg2cosθ =mg2h/d22+h2 =mg4hd2+4h2

Answer:

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=mg-aN=0.510-2=4 N
So, the force exerted by the block A on the block B is 4 N.

Page No 80:

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
N=mg-aN=0.510-2=4 N
So, the force exerted by the block A on the block B is 4 N.

Answer:

(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-a=mg-a
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 :

       

T+ma=mgT=mg-a
T = 0.05 (9.8 − 1.2) = 0.43 N

(e) Goes down with deceleration 1.2 m/s2 :
       
T+m-a=mgT=mg+a
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

Page No 80:

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-a=mg-a
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 :

       

T+ma=mgT=mg-a
T = 0.05 (9.8 − 1.2) = 0.43 N

(e) Goes down with deceleration 1.2 m/s2 :
       
T+m-a=mgT=mg+a
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

Answer:


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
m=1306.82×9.9=66 kg   
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
a=72×9.9-66×9.966=9.9×666=9.911a=0.9 m/s2

Page No 80:

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'=mg+m-a    ...(2)
This is minimum weight, N' = 60 × 9.9 N

From equations (1) and (2), we have:
2 mg = 1306.8
m=1306.82×9.9=66 kg   
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
a=72×9.9-66×9.966=9.9×666=9.911a=0.9 m/s2

Answer:

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,
mAa=T-mAg-mAg10    ...1
mBa=mBg+mBg10-T    ...2

Adding both the equations, we get:
amA+mB=mB-mAg+mB-mAg10

Putting value of the masses,we get:
9a=33g10ag=1130    ...3
Now, using equation (1), we get:
T=mAa+g+g10
The reading of the spring balance = 2Tg=2gmAa+g+g10
2×1.5ag+1+110=31130+1+110= 4.4 kg

Page No 80:

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,
mAa=T-mAg-mAg10    ...1
mBa=mBg+mBg10-T    ...2

Adding both the equations, we get:
amA+mB=mB-mAg+mB-mAg10

Putting value of the masses,we get:
9a=33g10ag=1130    ...3
Now, using equation (1), we get:
T=mAa+g+g10
The reading of the spring balance = 2Tg=2gmAa+g+g10
2×1.5ag+1+110=31130+1+110= 4.4 kg

Answer:

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
x=mgk=2×9.8100 =19.6100=0.1960.2 m

Suppose, further elongation, when the 1 kg block is added, is x.
Then,kx+x=m'g
kx= 3g − 2g = g
x=gk=9.8100=0.0980.1 m

Page No 80:

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
x=mgk=2×9.8100 =19.6100=0.1960.2 m

Suppose, further elongation, when the 1 kg block is added, is x.
Then,kx+x=m'g
kx= 3g − 2g = g
x=gk=9.8100=0.0980.1 m

Answer:

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
x=23.6100=0.2360.24 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
x'=35.4100       =0.3540.36 m
So, further elongation = x' − x
                                  = 0.36 − 0.24 = 0.12 m.

Page No 80:

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
x=23.6100=0.2360.24 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
x'=35.4100       =0.3540.36 m
So, further elongation = x' − x
                                  = 0.36 − 0.24 = 0.12 m.

Answer:

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)
M=B+kvg
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
M'=B+kvg
∴ Amount of mass that should be removed = M − M'.
M=B+kvg-B-kvg      =B+kv-B+kvg      =2kvg=2Mg-Bg      =2M-Bg

Page No 80:

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)
M=B+kvg
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
M'=B+kvg
∴ Amount of mass that should be removed = M − M'.
M=B+kvg-B-kvg      =B+kv-B+kvg      =2kvg=2Mg-Bg      =2M-Bg

Answer:

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,
U-mg=mg6U=7 mg6

      

m=6U7g                ....i

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

Mg-U=Mg6Mg-Mg6=UM=6U5g                ....ii

Mass added=6U5g-6U7g
                   =6U7-535 g=6U·235 g
From equation (i), m=6U7g
∴ Mass added=25m .

Page No 80:

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,
U-mg=mg6U=7 mg6

      

m=6U7g                ....i

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

Mg-U=Mg6Mg-Mg6=UM=6U5g                ....ii

Mass added=6U5g-6U7g
                   =6U7-535 g=6U·235 g
From equation (i), m=6U7g
∴ Mass added=25m .

Answer:

For the particle to move without being deflected and with constant velocity, the net force on the particle should be zero.
F+mg=0
v×A+mg=0 v×A=-mg

 vAsinθ=mg
v=mgAsinθ
v will be minimum when sinθ = 1.
⇒ θ = 90°
vmin=mgA

Page No 80:

Question 22:

For the particle to move without being deflected and with constant velocity, the net force on the particle should be zero.
F+mg=0
v×A+mg=0 v×A=-mg

 vAsinθ=mg
v=mgAsinθ
v will be minimum when sinθ = 1.
⇒ θ = 90°
vmin=mgA

Answer:

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,
 Tm1g = m1a    ...(i)

For mass m2,
m2gT= m2a    ...(ii)

Adding equations (i) and (ii), we get:
g(m2m1) = a(m1 + m2)
a=gm2-m1m1+m2     =9.8×0.6-0.30.6+0.3     =3.266 m/s2

(a) t = 2 s, a = 3.266 ms−2, u = 0
So, the distance travelled by the body,
S=ut+12at2  =0+123.26622=6.5 m

(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

Page No 80:

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,
 Tm1g = m1a    ...(i)

For mass m2,
m2gT= m2a    ...(ii)

Adding equations (i) and (ii), we get:
g(m2m1) = a(m1 + m2)
a=gm2-m1m1+m2     =9.8×0.6-0.30.6+0.3     =3.266 m/s2

(a) t = 2 s, a = 3.266 ms−2, u = 0
So, the distance travelled by the body,
S=ut+12at2  =0+123.26622=6.5 m

(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:

a = 3.26 m/s2, T = 3.9 N
After 2 s, velocity of mass m1,
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
t=6.529.823 sec
After this time, the mass m1 also starts moving downward.
So, the string becomes tight again after 23 s.

Page No 80:

Question 24:

a = 3.26 m/s2, T = 3.9 N
After 2 s, velocity of mass m1,
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
t=6.529.823 sec
After this time, the mass m1 also starts moving downward.
So, the string becomes tight again after 23 s.

Answer:

Mass per unit length=330 kg/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,
m1a=F-20    ...im2a=32-F    ...ii Adding both the equations, we get:a=12m1+m2=123=4 m/s2         
So, contact force,
F = 20 + 1a
F = 20 + 4 = 24 N



Page No 81:

Question 25:

Mass per unit length=330 kg/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,
m1a=F-20    ...im2a=32-F    ...ii Adding both the equations, we get:a=12m1+m2=123=4 m/s2         
So, contact force,
F = 20 + 1a
F = 20 + 4 = 24 N

Answer:

Mass of each block is 1 kg, sin θ1=45, sin θ2=35.
The free-body diagrams for both the boxes are shown below:


 
mgsinθ1T = ma    ...(i)
 T − mgsinθ2 = ma    ...(ii)

Adding equations (i) and (ii),we get:
mg(sinθ1sinθ2) = 2ma
⇒ 2a = g (sinθ1 − sinθ2)
a=g5×12      =g10

Page No 81:

Question 26:

Mass of each block is 1 kg, sin θ1=45, sin θ2=35.
The free-body diagrams for both the boxes are shown below:


 
mgsinθ1T = ma    ...(i)
 T − mgsinθ2 = ma    ...(ii)

Adding equations (i) and (ii),we get:
mg(sinθ1sinθ2) = 2ma
⇒ 2a = g (sinθ1 − sinθ2)
a=g5×12      =g10

Answer:

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:
am1+m2=m2g-m2g2          because F=m2g2a=m2g2m1+m2
So, the acceleration of mass m1,
a=m2g2m1+m2, towards the right.

Page No 81:

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:
am1+m2=m2g-m2g2          because F=m2g2a=m2g2m1+m2
So, the acceleration of mass m1,
a=m2g2m1+m2, towards the right.

Answer:

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=gm1-m2m1+m2
a=3g7=29.47      =4.2 m/s2     
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
 a=5g+15     =g+0.2 m/s2

Page No 81:

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=gm1-m2m1+m2
a=3g7=29.47      =4.2 m/s2     
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
 a=5g+15     =g+0.2 m/s2

Answer:

 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, (a1a2) and (a1 + a2), as shown.
From figure 2, T − 1g − 1a1 = 0    ...(i)
From figure 3, T2-2g-2a1-a2=0    ...ii
From figure 4, T2-3g-3a1+a2=0    ...iii

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(a1a2) = 3g − 3 (a1a2)
5a1 + a2 = g    ...(v)

Solving equations (iv) and (v), we get:
a1=2g29and a2=g-5a1a2=g-10g29=19g29So, a1-a2=2g29-19g29=-17g29and a1+a2=2g29+19g29=21g29
So, accelerations of m1, m2 and m3 are 19g29up, 17g29 down and 21g29down , respectively.
Now, u = 0, s = 20 cm = 0.2 m
a2=19g29           s=ut+12at20.2=12×1929gt2t=0.25 s

Page No 81:

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, (a1a2) and (a1 + a2), as shown.
From figure 2, T − 1g − 1a1 = 0    ...(i)
From figure 3, T2-2g-2a1-a2=0    ...ii
From figure 4, T2-3g-3a1+a2=0    ...iii

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(a1a2) = 3g − 3 (a1a2)
5a1 + a2 = g    ...(v)

Solving equations (iv) and (v), we get:
a1=2g29and a2=g-5a1a2=g-10g29=19g29So, a1-a2=2g29-19g29=-17g29and a1+a2=2g29+19g29=21g29
So, accelerations of m1, m2 and m3 are 19g29up, 17g29 down and 21g29down , respectively.
Now, u = 0, s = 20 cm = 0.2 m
a2=19g29           s=ut+12at20.2=12×1929gt2t=0.25 s

Answer:


For m1 to be at rest, a1= 0.
T m1g = 0
T = m1g    ...(i)
For mass m2,
T/2 − 2g = 2a
T = 4a + 4g    ...(ii)
For mass m3,
3g T/2= 2a
T = 6g − 6a    ...(ii)

From equations (ii) and (iii), we get:
3T – 12g = 12g – 2T 
T = 24g/5= 4.08g
Putting the value of T in equation (i), we get:
m1= 4.8kg

Page No 81:

Question 30:


For m1 to be at rest, a1= 0.
T m1g = 0
T = m1g    ...(i)
For mass m2,
T/2 − 2g = 2a
T = 4a + 4g    ...(ii)
For mass m3,
3g T/2= 2a
T = 6g − 6a    ...(ii)

From equations (ii) and (iii), we get:
3T – 12g = 12g – 2T 
T = 24g/5= 4.08g
Putting the value of T in equation (i), we get:
m1= 4.8kg

Answer:

The free-body diagrams for both the bodies are shown below:  
     
T + ma =mg    ...(i)
and T = ma    ...(ii)

From equations (i) and (ii), we get:
ma + ma = mg
⇒ 2ma = g
a=g2=105=5 m/s2
From equation (ii),
T = ma = 5 N

Page No 81:

Question 31:

The free-body diagrams for both the bodies are shown below:  
     
T + ma =mg    ...(i)
and T = ma    ...(ii)

From equations (i) and (ii), we get:
ma + ma = mg
⇒ 2ma = g
a=g2=105=5 m/s2
From equation (ii),
T = ma = 5 N

Answer:

Let the acceleration of mass M be a.
So, the acceleration of mass 2M will be a2.

(a) 2M(a/2) − 2T = 0
Ma = 2T
T + Ma − Mg = 0
Ma2+Ma=Mg            3Ma=2Mga=2g3

(b) Tension, T=Ma2=M2×2g3=Mg3


(c) Let T' = resultant of tensions
               
T'=T2+T2=2TT'=2T=2Mg3Again, tanθ=TT=1θ=45°
So, the force exerted by the clamp on the pulley is 2 Mg3 at an angle of 45° with the horizontal.

Page No 81:

Question 32:

Let the acceleration of mass M be a.
So, the acceleration of mass 2M will be a2.

(a) 2M(a/2) − 2T = 0
Ma = 2T
T + Ma − Mg = 0
Ma2+Ma=Mg            3Ma=2Mga=2g3

(b) Tension, T=Ma2=M2×2g3=Mg3


(c) Let T' = resultant of tensions
               
T'=T2+T2=2TT'=2T=2Mg3Again, tanθ=TT=1θ=45°
So, the force exerted by the clamp on the pulley is 2 Mg3 at an angle of 45° with the horizontal.

Answer:


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
a=g6
Hence, the acceleration of mass M=2a=2×g6=g3 up the plane.



Page No 82:

Question 33:


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
a=g6
Hence, the acceleration of mass M=2a=2×g6=g3 up the plane.

Answer:

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:
M=M'+mcot θ-1

Page No 82:

Question 34:

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:
M=M'+mcot θ-1

Answer:

(a) 5a + T − 5g = 0
From free-body diagram (1),
T = 5g − 5a             .....(i)
Again, 12T-4g-8a=0
T − 8g − 16a = 0

      
          (1)

From free-body diagram (2),
T = 8g + 16a                    ......(ii)

From equations (i) and (ii), we get:
5g − 5a = 8g + 16a
21a=-3g-a=-97
So, the acceleration of the 5 kg mass is 97  m/s2 upward and that of the 4 kg mass is 2a=2g7 downward.

(b) From free body diagram-3,
4a-T2=0


                  (2)


⇒ 8aT = 0
T = 8a
Again, T + 5a − 5g = 0
From free body diagram-4,

8a + 5a − 5g = s0
⇒ 13a − 5g = 0
a=5g13           downward

Acceleration of mass 2 kg is 2a=1013 g and 5 kg is 5g13.

(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
Acceleration of mass 1 kg is g3 upward,
Acceleration of mass 2 kg is 2g3  downward.
         
                  (3)

Page No 82:

Question 35:

(a) 5a + T − 5g = 0
From free-body diagram (1),
T = 5g − 5a             .....(i)
Again, 12T-4g-8a=0
T − 8g − 16a = 0

      
          (1)

From free-body diagram (2),
T = 8g + 16a                    ......(ii)

From equations (i) and (ii), we get:
5g − 5a = 8g + 16a
21a=-3g-a=-97
So, the acceleration of the 5 kg mass is 97  m/s2 upward and that of the 4 kg mass is 2a=2g7 downward.

(b) From free body diagram-3,
4a-T2=0


                  (2)


⇒ 8aT = 0
T = 8a
Again, T + 5a − 5g = 0
From free body diagram-4,

8a + 5a − 5g = s0
⇒ 13a − 5g = 0
a=5g13           downward

Acceleration of mass 2 kg is 2a=1013 g and 5 kg is 5g13.

(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
Acceleration of mass 1 kg is g3 upward,
Acceleration of mass 2 kg is 2g3  downward.
         
                  (3)

Answer:

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+ 0.1aT + 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 + 0.1aT + 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
a=0.40.65g      =4065g=813g     downward
So, the acceleration of the 500 gm block is 8g13downward.

Page No 82:

Question 36:

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+ 0.1aT + 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 + 0.1aT + 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
a=0.40.65g      =4065g=813g     downward
So, the acceleration of the 500 gm block is 8g13downward.

Answer:

Mass of the monkey, m = 15 kg,
Acceleration of the monkey in the upward direction, a = 1 m/s2

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
Using, s=ut+12at2, we get:
5=0+12×1×t2t2=5×2t=10 s
Hence, the time required to reach the ceiling is 10 s.

Page No 82:

Question 37:

Mass of the monkey, m = 15 kg,
Acceleration of the monkey in the upward direction, a = 1 m/s2

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
Using, s=ut+12at2, we get:
5=0+12×1×t2t2=5×2t=10 s
Hence, the time required to reach the ceiling is 10 s.

Answer:

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:
Fmgma = 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.

Page No 82:

Question 38:

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:
Fmgma = 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:

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

         (i)


       (ii)

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.
T1 = weight of monkey B
T1 = 20 N
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.

Page No 82:

Question 39:

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

         (i)


       (ii)

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.
T1 = weight of monkey B
T1 = 20 N
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:

(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 = 60gW'    ...(i)

From the free-body diagram of the box,
TW' = 30g = 0    ...(ii)

From equation (i), we get:
60gW' − 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.



Page No 83:

Question 40:

(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 = 60gW'    ...(i)

From the free-body diagram of the box,
TW' = 30g = 0    ...(ii)

From equation (i), we get:
60gW' − 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.

Answer:



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:
l=0+12gsinθt2t2=2lgsinθTime taken, t=2lgsinθ

Page No 83:

Question 41:



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:
l=0+12gsinθt2t2=2lgsinθTime taken, t=2lgsinθ

Answer:

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


From the free-body diagram,
Tcosθ − mg = 0
Tcosθ = mg
T=mgcos θ             ...i
And, maT sin θ = 0
⇒ ma = T sin θ
T=masin θ                 ....iitan θ=agθ=tan-1 ag
So, the angle formed by the ball with the vertical is tan-1 ag .

(ii) Let the angle of the incline be θ.
From the diagram,


ma cos θ = mg sin θ
sin θcos θ=agtan θ=agθ=tan-1 ag
So, the angle of incline is tan-1 ag.

Page No 83:

Question 42:

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


From the free-body diagram,
Tcosθ − mg = 0
Tcosθ = mg
T=mgcos θ             ...i
And, maT sin θ = 0
⇒ ma = T sin θ
T=masin θ                 ....iitan θ=agθ=tan-1 ag
So, the angle formed by the ball with the vertical is tan-1 ag .

(ii) Let the angle of the incline be θ.
From the diagram,


ma cos θ = mg sin θ
sin θcos θ=agtan θ=agθ=tan-1 ag
So, the angle of incline is tan-1 ag.

Answer:

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

    

The two bodies are separated because the elevator is moving downward with an acceleration of 12 m/s2 (>g) and the body moves with acceleration, g = 10 m/s2                [Freely falling body]

Now, for the block:
g = 10 m/s2, u = 0, t = 0.2 s
So, the distance travelled by the block is given by
s=ut+12at2 =0+1210×0.22=5×0.04 =0.2 m=20 cm
The displacement of the body is 20 cm during the first 0.2 s.



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