Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only).

For the particle with the trajectory as shown in the graph, the force on the particle,

A. For t<0

We can observe from the graph that the line displaying the position of the particle is coinciding with the x axis, i.e. There is no displacement and hence no net force on the body before t=0.

For t >4 s

We can observe that the position-time graph attains a constant value after t=4, this means that the displacement is not changing, it has a constant value of 3 m and thus the particle is at rest. Thus the net force on the body is zero.

For 0<t<4 s

We can observe that the position-time graph has a constant slope at the given time. The velocity is thus constant and so the acceleration is zero. The net force acting on the particle is thus zero.

B. From the second Newton’s law of motion, Impulse, ‘I’ is defined as the change in the momentum of an object.

I = mv-mu = m(v-u)

Where,

‘m’ is the mass of the particle

‘v’ is the final velocity of the particle

‘u’ is the initial velocity of the particle.

At t=0

Given,

Mass of the particle, m = 4 kg

Initial velocity of the particle, u= 0

Final velocity of the particle can be obtained by calculating the slope of the position-time graph at t =4 s

⇒

Therefore,

⇒ I = 3 kg m/s

At t=4 s

Given,

Mass of the particle, m = 4 kg

Initial velocity of the particle,

Final velocity of the particle can be obtained by calculating the slope of the position-time graph after t =4, i.e. v = 0

Therefore,

⇒ I = -3 kg m/s