Q. A particle of mass $m$, initially at rest, is acted upon by a variable force $F$ for a brief interval of time $T$. It begins to move with a velocity $u$ after the force stops acting. $F$ is shown in the graph as a function of time. The curve is semicircle
A) $u = \frac{\pi F_{0}^{2}}{2m}$
B) $u = \frac{\pi T^{2}}{8m}$
C) $u = \frac{\pi F_{0}T}{4m}$
D) $u = \frac{F_{0}T}{2m}$
Explanation
(c) Initially particle was at rest. By the application of force its momentum increases
Final momentum of the particle = Area of $F$-$t$ graph
$\Rightarrow mu =$ Area of semi circle
$mu = \frac{\pi r^{2}}{2} = \frac{\pi r_{1}r_{2}}{2} = \frac{\pi(F_{0})(T/2)}{2} \Rightarrow u = \frac{\pi F_{0}T}{4m}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.