A particle of mass ' m ' , moving with a velocity ' v ' makes an...

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.

$F = \dfrac{d p}{d t}$
$⇒ F d t = d p$
$⇒ FT = ∆ p$

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.

$\text{Area} = \dfrac{1}{2} \times \left(\dfrac{T}{4}\right) \times \left({F}_{0}\right) + \left(\dfrac{3T}{4} - \dfrac{T}{4}\right) \times \left({F}_{0}\right) + \dfrac{1}{2} \times \left(T - \dfrac{3T}{4}\right) \times \left({F}_{0}\right)$
$= \dfrac{3}{4}T{F}_{0}$

Therefore,

$\text{Area} = ∆ p$
$⇒ mv = \dfrac{3}{4}T{F}_{0}$
$⇒ {F}_{0} = \dfrac{4mv}{3T}$