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A particle of mass m moving with velocity u makes an elastic one-dimensional collision with a stationary particle of mass `m`. They come in contact for a very small time `t_0`. Their force of interaction increases from zero to `F_0` linearly in time `0.5t_0`, and decreases linearly to zero in further time `0.5t_0` as shown in figure. The magnitude of `F_0` is

A

`(mu)/(t_0)`

B

`(2mu)/(t_0)`

C

`(mu)/(2t_0)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B

In elastic collision between equal masses velocities are interchanged.
Therefore change in momentum in any one particle is `mu`.
Now, `|Deltap|=|Impu lse|`
`="area under" F-t graph`
`:. mu=1/2xxt_0xxF_0`
`:. F_0=(2mu)/(t_0)`
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