A body of mass 1 kg moving with velocity 1 m/s makes an elastic one dimensional collision with an identical stationary body. They are in contact for brief time 1 ms. Their force of interaction increas

Question

A body of mass 1 kg moving with velocity 1 m/s makes an elastic one dimensional collision with an identical stationary body. They are in contact for brief time 1 ms. Their force of interaction increases from zero to $F_{0}$ linearly in time 0.5 ms and decreases linearly to zero in further time 0.5 ms as shown in figure. The magnitude of force $F_{0}$ is

TestHub · Accepted Answer

The impulse equals the change in momentum. The ball and the identical stationary body exchange velocities in an elastic collision, so $Δ p = 1 \times 1 - 0 = 1 kg m/s$ in magnitude.

Impulse $=$ area under the $F$ - $t$ graph $= \frac{1}{2} \times (contact time) \times F_{0}$ . Taking the contact time as $1 ms = 1 0^{- 3} s$ : $\frac{1}{2} \times 1 0^{- 3} \times F_{0} = 1 \Rightarrow F_{0} = 2000 N = 2 kN$ .

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