Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 5.16:

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Answer:

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:

 … (iii)

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.

Question 5.17:

A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Answer:

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 v₁ and v₂ be the respective velocities of the daughter nuclei having masses m₁ and m₂.

Total linear momentum of the system after disintegration = 

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.

Question 5.18:

Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s^–1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Answer:

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, pf = –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.

Question 5.19:

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s^–1, what is the recoil speed of the gun?

Answer:

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

Question 5.20:

A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

Answer:

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 along OS

Vertical component of the final velocity = vsin θ 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

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

Question 5.21:

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Answer:

Mass of the stone, m = 0.25 kg

Radius of the circle, r = 1.5 m

Number of revolution per second, 

Angular velocity, ω = 

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

Maximum tension in the string, T_max = 200 N

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

Question 5.22:

If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:

(a) the stone moves radially outwards,

(b) the stone flies off tangentially from the instant the string breaks,

(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?

Answer:

Answer: (b)

When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

Question 5.23:

Explain why

(a) a horse cannot pull a cart and run in empty space,

(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,

(c) it is easier to pull a lawn mower than to push it,

(d) a cricketer moves his hands backwards while holding a catch.

Answer:

(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.

An empty space is devoid of any such reaction force. 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 rest. 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 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.

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.

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 with the hand

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.

Question 5.24:

Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

Figure 5.17

Answer:

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

= 0.08 × 10^–2 kg m/s

Question 5.25:

Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s^–2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)

Figure 5.18

Answer:

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:

 = 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 fs, exerted by the belt, i.e.,

∴a‘ = 0.2 × 10 = 2 m/s²

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