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#### 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 ?

6 viewed last edited 1 year ago An0nym0us
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Given that

Mass of the ball m = 0.04 kg

From the position time graph

Distance between the two walls \Delta s = 2 cm = 2 * 10^{-2} m

Time required to change ball direction \Delta t = 2 sec The graph displays a body changing its direction of motion every 2s. This condition can be seen physically as a ball rebounding between two stationary walls distanced by 2 cm. Because the slope of the x-t graph reverses every two seconds, the ball collides with a wall every two seconds. As a result, the ball receives an impulse every 2 seconds.

Step 1: Get an expression for the impulse received by the ball

Initial velocity of the ball = u m/s

The ball velocity after rebound = - v m/s

The graph's slope is identical in both cases (before and after bouncing)

u = - v

Velocity of the ball v = slope of the graph = \frac{\text{ change in distance}}{\text{ change in time }} = \frac{\Delta s}{\Delta t}

Initially momentum of the ball P_1= mu

Momentum of the ball after rebound P_2 = mv

We know that, Impulse is defined as the change in momentum

I = mv - mu

I = m(v - (-v))

I = 2mv = 2 m \frac{\Delta s}{\Delta t}

Step 2: Determining the magnitude of each ball's impulse

I = 2 * 0.04 kg * \frac{2 * 10^{-2} m}{2 sec}

I = 0.08 * 10^{-2} kg m/s

Hence, impulse received by the ball I = 0.08 * 10^{-2} kg m/s

Time between the two consecutive impulses = 2 seconds