A body of mass `1 kg` moving with velocity `1 m//s` makes an elastic one dimesional collision with an identical stationary body. They are in contact for brief time `1 sec`. Their force of interaction increases form zero to `F_(0)` linearly in time `0.5s` and decreases linearly to zero in further time `0.5 sec` as shown in figure. Find the magnitude of force `F_(0)` in newton.

Answer

2

Explanation

In the one dimensional elastic collision with one body at rest, the body moving initially comes to rest & the on which was at rest earlier starts moving the velocity that first body had before collision. So, if `m` & `V_(0)` be the mass & velocity of body,
the change in momentum `= mV_(0) rArr int Fdt = mV_(0) rArr int Fdt = mV_(0) rArr F = (2mV_(0))/(Deltat) = 2 N`