Question

A particle of mass $m$ moving with velocity u makes an elastic one dimensional collision with a stationary particle of mass m. They are in contact for a short time $T$. Their force of interaction increases from zero to ${F_0}$​​ linearly in time $T/2$​ and decreases linearly to zero in further time $T/2$​ .The magnitude of ${F_0}$​ is:

Answer 1

Hint:The question can only be solved using the impulse concepts and the concept of change in momentum. Also notice that the question says “elastic one dimensional collision”.

Complete step by step answer:
Given that the mass of the particle is \[m\] and it is moving with a velocity $u$ and it makes a “elastic one dimensional collision” with another stationary particle of the same mass. So we are to find the magnitude of ${{F}_{0}}$ when the force of interaction increases from zero to ${{F}_{0}}$​ linearly in time $T/2$​ and decreases linearly to zero in further time $T/2$​ after the elastic collision takes place.

We know that in an elastic one-dimensional collision, a particle rebounds with the same speed in the opposite dimension. That is, the first ball will stop and the stationary ball will start to move with the velocity $u$, therefore, Impulse will be equal to the change in momentum. And the change in momentum will be $0-mu=-mu$ so, the magnitude will be $mu$. Now, we know that
$F=\dfrac{dp}{dt} \\
\Rightarrow Fdt=dp \\ $
Now, we integrate the left hand side from time $T=0$ to $T=T$. And on the right hand side, we integrate from initial momentum to final momentum.
$\int\limits_{0}^{T}{Fdt}=\int\limits_{{{p}_{i}}}^{{{p}_{f}}}{dp}$
Also the Left hand side is the area under the graph of Force and time. Therefore the equation becomes
$\dfrac{1}{2}{{T}_{0}}{{F}_{0}}=mu \\
\therefore {{F}_{0}}=\dfrac{2mu}{{{T}_{0}}}$

Note:The collision here is elastic and so the solution was simple, however, if the collision was inelastic, the solution would take a different turn as we need to consider the velocity before the collision and the velocity after the collision, similar case for the momentum.