NCERT Class 11 Physics Solutions: Chapter 5 – Laws of Motion Part 9

Doorsteptutor material for NSO is prepared by world's top subject experts: fully solved questions with step-by-step explanation- practice your way to success.

Question 5.24:

Figure 5.24 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 shows the position-time graph of a body of mass.

Figure 5.24

Loading Image


A ball rebounding between two walls located between at; after every 2 s, the ball receives an impulse of magnitude 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 between two stationary walls situated between positions. Since the slope of the x-t graph reverses after every , the ball collides with a wall after every 2 s. Therefore, ball receives an impulse after every 2 s.

Mass of the ball,

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,

Velocity of the ball after collision,

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

Magnitude of impulse = Change in momentum

Question 5.25:

Figure 5.25 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 showes the question condition.

Figure 5.25

Loading Image


Mass of the man,

Acceleration of the belt,

Coefficient of static friction,

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

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.,

Therefore, the maximum acceleration of the belt up to which the man can stand stationary is

Question 5.26:

A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

Lowest and highest point

Lowest Point

Highest Point

( a )

( b )

( c )

( d )

and denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.


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

The free body diagram of the stone at the lowest point

Figure 5.26(A)

Loading Image

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.,


= Velocity at the lowest point

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

The free body diagram of the stone at the highest point

Figure 5.26(B)

Loading Image

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


= 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 and (T + mg).