NCERT Grade 11-Law of Motion-Answers

NCERT Solutions for Class 11 Physics

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1.     (a) Constant speed ⇒ acceleration a=0

Hence, net force F = ma = 0

∴ No net force is acting on a drop of rain falling downwards with constant speed

(b) The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. Hence, no net force is acting on the floating cork.

(c) The kite is stationary in the sky, hence, as per Newton’s first law of motion, no net force is acting on the kite.

(d) Constant speed(30km/hr) ⇒ acceleration a = 0

Hence, net force F = ma = 0

Hence, no net force is acting on the car moving with a constant velocity.

(e) Since,the electron is far from all material objects & free from the influence of all fields. Hence, no net force is acting on the electron.

2.    0.5 N, in vertically downward direction, in all cases

Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as:

F = m × a

Where,

= Net force

= Mass of the pebble = 0.05 kg

a = g = 10 m/s2

∴ = ma = 0.05 × 10 = 0.5 N

The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction.

If the pebble is thrown at an angle of 45° with the horizontal, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble.

3.     (a) 1 N; vertically downward

Mass of the stone, = 0.1 kg

Acceleration of the stone, = g = 10 m/s2

As per Newton’s second law of motion, the net force acting on the stone,

F = ma = mg

= 0.1 × 10 = 1 N

This force acts vertically downwards.

(b) 1 N; acting vertically downward

The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction.

The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is 1 N.

(c) 1 N; acting vertically downward

It is given that the train is accelerating at the rate of 1 m/s2.

Therefore, the net force acting on the stone, F” = ma = 0.1 × 1 = 0.1 N

This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F,’ stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations.

Therefore, the net force acting on the stone is given only by acceleration due to gravity.

F = mg = 1 N

This force acts vertically downward.

(d) 1 N; in the direction of motion of the train

The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

Acceleration of the train, a = 0.1 m/s2

The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:

F = ma

= 0.1 × 1 = 0.1 N

4.    (i) The tension T on the string provides the necessary centriptetal force(fc) so that the particle can move in a circle.Thus,net force on the particle is T.Hence,the correct alternative is (i)

that is,

Where fc is the net force acting on the particle.

5.    Retarding force, F = -50N

Mass of the body, m = 20 kg

Initial velocity of the body, u = 15 m/s

Final velocity of the body, v = 0

Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:

Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:

i.e the body will stop within 6 seconds after the application of constant retarding force 50 N.

6.    0.18 N; in the direction of motion of the body

Mass of the body, m= 3 kg

Initial speed of the body, u= 2 m/s

Final speed of the body, v= 3.5 m/s

Time, t = 25 s

Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:

As per Newton’s second law of motion, force is given as:

F = ma

= 3 × 0.06 = 0.18 N

Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.

7.    Mass of the body, m= 5 kg

The given situation can be represented as follows:

The resultant of two forces is given as:

θ is the angle made by R with the force of 8 N

The negative sign indicates that is in the clockwise direction with respect to the force of magnitude 8 N.

As per Newton’s second law of motion, the acceleration (a) of the body is given as:

F = ma

2 m/s2, at an angle of 37° with a force of 8 N

8.    Initial speed of the three-wheeler, u = 36 km/h = 10 m/s

Final speed of the three-wheeler, v = 0 m/s

Time, t = 4 s

Mass of the three-wheeler, m = 400 kg

Mass of the driver, m’ = 65 kg

Total mass of the system, M = 400 + 65 = 465 kg

Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as:

The negative sign indicates that the velocity of the three-wheeler is decreasing with time.

Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as:

F = ma

= 465 × (-2.5) = -1162.5 N

The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

9.    Mass of the rocket, m = 20,000 kg

Initial acceleration, a = 5 m/s2

Acceleration due to gravity, g = 10 m/s2

Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation:

F – mg = ma

F = m (g + a)

= 2000 × (10 + 5)

= 2000 × 15 = 3 × 105 N

10.   Mass of the body, m= 0.40 kg

Initial speed of the body, u= 10 m/s due north

Force acting on the body, F = -8.0 N

Acceleration produced in the body, 

(i) At t = -5 s

Acceleration, a’ = 0 and u = 10 m/s

(ii) At t’ = 25 s

Acceleration, a” = -20 m/s2 and u = 10 m/s

(iii) At t = 100 s

For 0 ≤ t ≤ 30 s

a = – 20 m/s2

u = 10 m/s

As per the first equation of motion, for t = 30 s, final velocity is given as:

v = u + at

= 10 + (-20) × 30 = -590 m/s

For motion between 30 s to 100 s, i.e., in 70 s:

∴ Total distance is given as, s” = s1 + s2 = -8700 – 41300 = -50000m

11.   (a) 36 m/s, at an angle of 26.57° with the motion of the truck

(b) 10 m/s2

(a) Initial velocity of the truck, u = 0

Acceleration, a = 2 m/s2

Time, t = 10 s

As per the first equation of motion, final velocity is given as:

vy = ux + ayt

= 0 + 2 × 10 = 20 m/s

The final velocity of the truck and hence, of the stone is 20 m/s.

At t = 11 s, the horizontal component (vx) of velocity, in the absence of air resistance, remains unchanged, i.e.,

vx = 20 m/s

The vertical component (vy) of velocity of the stone is given by the first equation of motion as:

vy = uy + ayt

Where, t = 11 – 10 = 1 s and ay = g = 10 m/s2

∴ vy = 0 + 10 × 1 = 10 m/s

The resultant velocity (v) of the stone is given as:

Let θ be the angle made by the resultant velocity with the horizontal component of velocity, vx

(b) When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is 10 m/s2 acting downwards and stone covers a parabolic path.

12.   (a) At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically downwards due to force of gravity .

(b) At the mean position, the velocity of the bob is 1 m/s(maximum).If the string is cut at the mean position, then it will fall towards the Earth but under combined effect of gravity & horizontal velocity of the bob which is similar to the motion of projectile projected with some initial velocity. Hence, it will follows a parabolic path.

13.   (a) Mass of the man, m = 70 kg

Acceleration, a = 0

Using Newton’s second law of motion, we can write the equation of motion as:

R – mg = ma

Where, ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

∴ R = mg

= 70 × 10 = 700 N

∴ Reading on the weighing scale = 

(b) Mass of the man, m = 70 kg

Acceleration, a = downward

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma

R = m(g – a)

= 70(10 – 5) = 70 × 5

= 350 N

∴ Reading on the weighing scale = 

(c) Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 upward

Using Newton’s second law of motion, we can write the equation of motion as:

R – mg = ma

R = m(g + a)

= 70 (10 + 5) = 70 × 15

= 1050 N

∴ Reading on the weighing scale = 

(d) When the lift moves freely under gravity, acceleration a = g

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma

R = m(g – a)

= m (g – g) = 0

∴ Reading on the weighing scale =

The man will be in a state of weightlessness.

14.   (a) For t < 0

It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero.

For t > 4 s

It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 3 m from the origin. Hence, no force is acting on the particle.

For 0 < t < 4

It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.

(b) At t = 0

Impulse = Change in linear momentum

= mv – mu

Mass of the particle, m = 4 kg

Initial velocity of the particle, u = 0

At t = 4 s

Before t = 4 s, particle has a constant velocity 

After t = 4 s, particle is at rest i.e v = 0

∴ Impulse = m(v – u) = 4 (0 – 0.75) = -3kgms-1

15.   Horizontal force, F = 600 N

Mass of body A, m1 = 10 kg

Mass of body B, m2 = 20 kg

Total mass of the system m = m1 + m2 = 30 kg

Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as:

F = ma

When force F is applied on body A:

The equation of motion can be written as:

F – T = m1a

∴ T = F – m1a

= 600 – 10 × 20 = 400 N   …(i)

When force F is applied on body B:

The equation of motion can be written as:

F – T = m2a

T = F – m2a

∴ T = 600 – 20 × 20 = 200 N   ..(ii)

16.   The given system of two masses and a pulley can be represented as shown in the following figure:

Smaller mass, m1 = 8 kg

Larger mass, m2 = 12 kg

Tension in the string = T

Mass m2, owing to its weight, moves downward with acceleration a, and mass m1  moves upward.

Applying Newton’s second law of motion to the system of each mass:

For mass m1:

The equation of motion can be written as:

T – m1g = ma    …(i)

For mass m2:

The equation of motion can be written as:

m2g-T = m2a   ….(ii)

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

Therefore, the acceleration of the masses is 2 m/s2.

Substituting the value of a in equation (ii), we get:

Therefore, the tension in the string is 96 N.

17.   Let m, m1 and m2 be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.

Initial momentum of the system (parent nucleus) = 0

Let vand v2 be the respective velocities of the daughter nuclei having masses m1, and m2.

Total linear momentum of the system after disintegration = m1v1 + m2v2

According to the law of conservation of momentum:

Total initial momentum = Total final momentum

Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

18.   Mass of each ball = 0.05 kg

Initial velocity of each ball = 6 m/s

Magnitude of the initial momentum of each ball, pi = 0.3 kg m/s

After collision, the balls change their directions of motion without changing the magnitudes of their velocity.

Final momentum of each ball, p= -0.3 kg m/s

Impulse imparted to each ball = Change in the momentum of the system

= pf – pi

= -0.3 – 0.3 = -0.6 kg m/s

The negative sign indicates that the impulses imparted to the balls are opposite in direction.

19.   Mass of the gun, M= 100 kg

Mass of the shell, m= 0.020 kg

Muzzle speed of the shell, v= 80 m/s

Recoil speed of the gun = V

Both the gun and the shell are at rest initially.

Initial momentum of the system = 0

Final momentum of the system = mv + MV

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.

According to the law of conservation of momentum:

Final momentum = Initial momentum

mv + MV = 0

The recoil speed of the gun is 0.016 ms-1 or 1.6 cms-1

20.   The given situation can be represented as shown in the following figure.

Where,

AO = Incident path of the ball

OB = Path followed by the ball after deflection

∠AOB = Angle between the incident and deflected paths of the ball = 45°

∠AOP = ∠BOP = 22.5° = θ

Initial and final velocities of the ball = v

Horizontal component of the initial velocity = vcos θ along RO

Vertical component of the initial velocity = vsin θ along PO

Horizontal component of the final velocity = vcos θ acting along OS

Vertical component of the final velocity = vsin θ acting along OP

The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.

∴ Impulse imparted to the ball = Change in the linear momentum of the ball

= mv cosθ – (-mv cosθ)

= 2mv cos θ

Mass of the ball, m = 0.15 kg

Velocity of the ball, v = 54 km/h = 15 m/s

∴ Impulse = 2 × 0.15 × 15 cos 22.5° = 4.16 kg m/s

21.   Given mass of the stone, m = 0.25 kg

Radius of the circle, r = 1.5 m

Number of revolution per second, 

Angular velocity, 

The centripetal force acting on the stone is provided by the tension T, in the string, i.e.,

Now, given maximum tension in the string, Tmax = 200 N

Therefore we have,

Therefore, the maximum speed of the stone is 34.64 m/s.

22.   (b) According to the Newton’s first law of motion(law of inertia),a body will tend to maintain its linear path and momentum if there is no external force acting on it.At the instant when the string breaks, the stone will move in the direction of the velocity at that instant. Now for a particle in circular motion, the direction of velocity vector is tangential to the path of the particle at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

23.   (a) In order to pull a cart, a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward.

In empty space the mutual forces between the horse & the cart cancel out each other. Therefore, a horse cannot pull a cart and run in empty space.

(b) When a speeding bus stops suddenly, the lower portion of a passenger’s body, which is in contact with the seat, suddenly comes to res However, the upper portion tends to remain in motion (as per the first law of motion). As a result, the passenger’s upper body is thrown forward in the direction in which the bus was initially moving.

(c) While pulling a lawn mower, a force at an angle θ is applied on it, as shown in the following figure.

The vertical component of this applied force acts upward. This reduces the effective weight of the mower. W = mg – FSinθ

On the other hand, while pushing a lawn mower, a force at an angle θ is applied on it, as shown in the following figure.

In this case, the vertical component of the applied force acts in the direction of the weight of the mower. This increases the effective weight of the mower. W = mg – FSinθ

Since the effective weight of the lawn mower is lesser in the first case, pulling the lawn mower is easier than pushing it.

(d) According to Newton’s second law of motion, we have the equation of motion:

Where,

F = Stopping force experienced by the cricketer as he catches the ball

m = Mass of the ball

Δt = Time of impact of the ball within the hand

Now, It can be inferred from equation (i) that the impact force is inversely proportional to the impact time, i.e.,

Equation (ii) shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa.

While taking a catch, a cricketer moves his hand backward so as to increase the time of impact (Δt). This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.

24.   A ball rebounding between two walls located between at x= 0 and x= 2 cm; after every 2 s, the ball receives an impulse of magnitude 0.08 × 10-2 kg m/s from the walls

The given graph shows that a body changes its direction of motion after every 2 s. Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions x = 0 and x = 2 cm. Since the slope of the x-t graph reverses after every 2 s, the ball collides with a wall after every 2 s. Therefore, ball receives an impulse after every 2 s.

Mass of the ball, m= 0.04 kg

The slope of the graph gives the velocity of the ball. Using the graph, we can calculate initial velocity (u) as:

Velocity of the ball before collision, u = 10-2 m/s

Velocity of the ball after collision, v = -10-2 m/s

(Here, the negative sign arises as the ball reverses its direction of motion.)

Magnitude of impulse = Change in momentum

= |mv – mu| = |0.04 (-10-2 -10-2)| = |0.04 (-2 × 10-2)|

= 0.08 × 10-2 kg m/s

Hence, the required impulse is 0.08 × 10-2 kg m/s.

25.   Mass of the man, m= 65 kg

Acceleration of the belt, a = 1 m/s2

Coefficient of static friction, µ = 0.2

The net force F, acting on the man is given by Newton’s second law of motion as:

F’net = ma = 65 × 1 = 65 N

The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force F’s, exerted by the belt, i.e.,

F’net = fs

ma’ = µmg

∴ a’ = 0.2 × 10 = 2 m/s2

Therefore, the maximum acceleration of the belt up to which the man can stand still is 2 m/s2

26.   (a)The free body diagram of the stone at the lowest point is shown in the following figure:

According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,

Where, v1 = Velocity at the lowest point

The free body diagram of the stone at the highest point is shown in the following figure.

Using Newton’s second law of motion, we have:

Where, v2 = Velocity at the highest point

It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest points are respectively (mg-T1) and (mg+T2).

27.   (a) Mass of the helicopter, mh = 1000 kg

Mass of the crew and passengers, mp = 300 kg

Total mass of the system, m = mh + mp = 1300 kg

Acceleration of the helicopter, a = 15 m/s2

Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as:

R – mpg = mpa

i.e. R = mp(g + a)

= 300 (10 + 15) = 300 × 25

= 7500 N

Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward.

(b) Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as:

R’ – mg = ma

= m(g + a)

= 1300 (10 + 15) = 1300 × 25

= 32500 N

The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.

(c) The force on the helicopter due to the surrounding air is the reaction. As action & reaction are equal & opposite, therefore, force of 32500 N, directed upward. reaction, F’= 32500 N, vertically upwards.

28.   Speed of the water stream, v = 15 m/s

Cross-sectional area of the tube, A = 10-2 m2

Volume of water coming out from the pipe per second,

V = Av = 15 × 10-2 m3/s

Density of water, p = 103 kg/m3

Mass of water flowing out through the pipe per second = PXV= 150 kg/s

The water strikes the wall and does not rebound. Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as:

F = Rate of change of momentum

29.   (a) Force on the seventh coin is exerted by the weight of the three coins on its top.

Weight of one coin = mg

Weight of three coins = 3mg

Hence, the force exerted on the 7th coin by the three coins on its top is 3mg. This force acts vertically downward.

(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.

Weight of the eighth coin = mg

Weight of the ninth coin = mg

Weight of the tenth coin = mg

Total weight of these three coins = 3mg

Hence, the force exerted on the 7th coin by the eighth coin is 3mg. This force acts vertically downward.

(c) The 6th coin experiences a downward force because of the weight of the four coins (7th, 8th, 9th and 10th) on its top.

Therefore, the total downward force experienced by the 6th coin is 4 mg.

As per Newton’s third law of motion, the 6th coin will produce an equal reaction force on the 7th coin, but in the opposite direction. Hence, the reaction force of the 6th coin on the 7th coin is of magnitude 4 mg. This force acts in the upward direction.

30.   Speed of the aircraft, v = 720 km/h

Acceleration due to gravity, g = 10 m/s2

Angle of banking, θ = 15°

For radius r, of the loop, we have the relation:

31.   Radius of the circular track, r = 30 m

Speed of the train, v = 54 km/h = 15 m/s

Mass of the train, m = 106 kg

The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and tear of the rail

The angle of banking θ, is related to the radius (r) and speed (v) by the relation:

Therefore, the angle of banking is 36.87° approximately.

32.   750 N and 250 N in the respective cases; Method (b)

Mass of the block, m = 25 kg

Mass of the man, M = 50 kg

Acceleration due to gravity, g = 10 m/s2

Force applied on the block, F = 25 × 10 = 250 N

Weight of the man, W = 50 × 10 = 500 N

Case (a): When the man lifts the block directly

In this case, the man applies a force in the upward direction. This increases his apparent weight.

∴ Action by the man on the floor = 250 + 500 = 750 N

Case (b): When the man lifts the block using a pulley

In this case, the man applies a force in the downward direction. This decreases his apparent weight.

∴ Action by the man on the floor= 500 – 250 = 250 N

If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

33.   Case (a)

given Mass of the monkey, m = 40 kg

Acceleration due to gravity, g = 10 m/s

Maximum tension that the rope can bear, Tmax = 600 N

Acceleration of the monkey, a = 6 m/s2 upward

Using Newton’s second law of motion, we can write the equation of motion as:

T – mg = ma

∴ T = m(g + a)

= 40 (10 + 6)

= 640 N

Since T > Tmax the rope will break in this case.

Case (b)

Acceleration of the monkey is, a = 4m/s2 downwards

Using Newton’s second law of motion, we can write the equation of motion as:

mg – T = ma

∴ T = m (g – a)

= 40(10 – 4)

= 240 N

Since T < Tmax, the rope will not break in this case.

Case (c)

The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a = 0.

Using Newton’s second law of motion, we can write the equation of motion as:

T – mg = ma

T – mg = 0

∴ T = mg

= 400 N

Since T < Tmax, the rope will not break in this case.

Case (d)

When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g

Using Newton’s second law of motion, we can write the equation of motion as:

mg – T = mg

∴ T = m(g – g) = 0

Since T < Tmax, the rope will not break in this case.

34.   (a) Mass of body A, mA = 5 kg

Mass of body B, mB = 10 kg

Applied force, F = 200 N

Coefficient of friction, µs = 0.15

The force of friction is given by the relation:

fs = µ (mA + mB)g

= 0.15 (5 + 10) × 10

= 1.5 × 15 = 22.5 N leftward

Net force acting on the partition = 200 – 22.5 = 177.5 N rightward

As per Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.

Hence, the reaction of the partition will be 177.5 N, in the leftward direction.

(b) Force of friction on mass A:

fs = µmAg

= 0.15 × 5 × 10 = 7.5 N leftward

Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightward

As per Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i.e., 192.5 N acting leftward.

When the wall is removed, the two bodies will move in the direction of the applied force.

Net force acting on the moving system = 177.5 N

The equation of motion for the system of acceleration a, can be written as:

Net force causing mass A to move:

FA = mAa = 5 × 11.83 = 59.15 N

Net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N

This force will act in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3 N, acting opposite to the direction of motion.

35.   (a) Mass of the block, m = 15 kg

Coefficient of static friction, µ = 0.18

Acceleration of the trolley, a = 0.5 m/s2

As per Newton’s second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:

F = ma = 15 × 0.5 = 7.5 N

This force is acted in the direction of motion of the trolley. Force of static friction between the block and the trolley:

f = µmg

= 0.18 × 15 × 10 = 27 N

The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an stationary observer on the ground, the block will appear to be at rest.

When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.

(b) An observer, moving with the trolley, has some acceleration. This is the case of non- inertial frame of reference.The law of inertia is no longer valid.The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

36.   Mass of the box, m= 40 kg

Coefficient of friction,  = 0.15

Initial velocity, u = 0

Acceleration, a = 2 m/s2

Distance of the box from the end of the truck, s’ = 5 m

As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by:

F = ma

= 40 × 2 = 80 N

As per Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction f, acting between the box and the floor of the truck. This force is given by:

f = µmg

= 0.15 × 40 × 10 = 60 N

∴ Net force acting on the block:

Fnet = 80 – 60 = 20 N backward

The backward acceleration produced in the box is given by:

Using the second equation of motion, time t can be calculated as:

Hence, the box will fall from the truck after √20s from start.

The distance s, travelled by the truck in √20s is given by the relation:

37.   The coin revolves with the record in the case when the force of friction is enough to provide necessary centripetal force & if this force is not sufficient to provide necessary centripetal force then coin slips on the record

µ, ω are same for both the coins & we have different values of r for two coins.

To prevent slipping µmg ≥ mrω2

µmg ≥ mrω2

For 1st coin

Coin placed at 4 cm from the centre

Mass of each coin = m

Radius of the disc, r = 15 cm = 0.15 m

Frequency of revolution,

Coefficient of friction, µ = 0.15

In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.

Coin placed at 4 cm:

Radius of revolution, r’ = 4 cm = 0.04 m

Angular frequency, ω = 2πν

Frictional force, f = µg = 0.15 × 10 = 1.5N

Centripetal force on the coin:

Fcert = rω2

= 0.04 × (3.49)2

= 0.49 N

Since f > Fcert , the coin will revolve along with the record.

For 2nd coin

Coin placed at 14 cm:

Radius, r” = 14 cm = 0.14 m

Angular  frequency, ω = 3.49 s-1

Frictional force, f’ = 1.5 N

Centripetal force is given as:

Fcert = rω2

= 0.14 × (3.49)2

= 1.7 N

Since f < Fcert, the coin will slip from the surface of the record.

38.   In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.

The net force acting on the motorcyclist is the sum of the normal force (FN) and the force due to gravity (Fg = mg).

The equation of motion for the centripetal acceleration ca, can be written as:

Normal reaction is provided by the speed of the motorcyclist. At the minimum speed (νmin), FN = 0

39.   Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction, µ = 0.15

Frequency of rotation, v= 200 rev/min = 

The necessary centripetal force required for the rotation of the man is provided by the normal force (FN).

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force

(f = µFN)

Hence, the man will not fall until:

The minimum angular speed is given as:

40.   Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.

OP = R = Radius of the circle

N = Normal reaction

The respective vertical and horizontal equations of forces can be written as:

mg = Ncosθ …(i)

mlω2 = Nsinθ  ..(ii)

In ΔOPQ, we have:

l = R sinθ          … (iii)

Substituting equation (iii) in equation (ii), we get:

m(R sin θ)ω2 = N sin θ

mRω2 = N   ..(iv)

Substituting equation (iv) in equation (i), we get:

Since cosθ ≤ 1, the bead will remain at its lowermost point for

For 

On equating equations (v) and (vi), we get:

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