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

Options

A. (a) `(mu)/(t_0)`
B. (b) `(2mu)/(t_0)`
C. (c) `(mu)/(2t_0)`
D. (d) None of these

Answer

B

Explanation

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)`