Q. A mass of $100\ g$ strikes the wall with speed $5\ m/s$ at an angle as shown in figure and it rebounds with the same speed. If the contact time is $2 \times 10^{- 3}\ \sec$, what is the force applied on the mass by the wall
A) $250\sqrt{3}\ N$ to right
B) $250\ N$ to right
C) $250\sqrt{3}\ N$ to left
D) $250\ N$ to left
Explanation
(c) Force = Rate of change of momentum
Initial momentum ${\overrightarrow{P}}_{1} = mv\sin\theta\ \widehat{i} + mv\cos{\theta\ \widehat{j}}$
Final momentum ${\overrightarrow{P}}_{2} = - mv\sin\theta\ \widehat{i} + mv\cos{\theta\ \widehat{j}}$
$\therefore\overrightarrow{F} = \frac{\mathrm{\Delta}\overrightarrow{P}}{\mathrm{\Delta}t} = \frac{- 2mv\sin\theta}{2 \times 10^{- 3}}$
Substituting $m = 0.1\ kg,\ v = 5\ m/s,\ \theta = 60{^\circ}$
Force on the ball $\overrightarrow{F} = - 250\sqrt{3}N$
Negative sign indicates direction of the force
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.