Q. A particle of mass $m$ moving with a velocity $u$ makes an elastic one dimensional collision with a stationary particle of mass $m$ establishing a contact with it for extremely small time $T$. Their force of contact increases from zero to $F_{0}$ linearly in time $T/4$, remains constant for a further time $T/2$ and decreases linearly from $F_{0}$ to zero in further time $T/4$ as shown. The magnitude possessed by $F_{0}$ is
A) $\frac{mu}{T}$
B) $\frac{2mu}{T}$
C) $\frac{4mu}{3T}$
D) $\frac{3mu}{4T}$
Explanation
(c) Change in momentum = Impulse
= Area under force-time graph
$\therefore mv =$ Area of trapezium
$\Rightarrow mv = \frac{1}{2}\left( T + \frac{T}{2} \right)F_{0} \Rightarrow mv = \frac{3T}{4}F_{0} \Rightarrow F_{0} = \frac{4mu}{3T}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.