A particle of mass ' m ' , moving with a velocity ' v ' makes an elastic collision in one…

In an elastic head-on collision between two particles, the velocity of the particle gets exchanged, therefore net change of momentum of a particle is, $∆p=m∆v=m\left(v-0\right)=mv$

From newton's second law, the rate of change of momentum is force.

$$\begin{gathered} F=\frac{dp}{dt} \\ \Rightarrow Fdt=dp \\ \Rightarrow FT=∆p \end{gathered}$$

Here, $F$is force of contact during the collision for a small-time $T$that causes the $∆p$change in momentum of particle.

Thus, the area under the graph of $F-t$curve gives the change in momentum.

$$\begin{gathered} \text{Area}=\frac{1}{2}\times \left(\frac{T}{4}\right)\times \left(F_0\right)+\left(\frac{3T}{4}-\frac{T}{4}\right)\times \left(F_0\right)+\frac{1}{2}\times \left(T-\frac{3T}{4}\right)\times \left(F_0\right) \\ =\frac{3}{4}T{F}_0 \end{gathered}$$

Therefore, 

$$\begin{gathered} \text{Area}=∆p \\ \Rightarrow mv=\frac{3}{4}T{F}_0 \\ \Rightarrow F_0=\frac{4mv}{3T} \end{gathered}$$