Q. On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed $\frac{v}{3}$. The second block's speed after the collision is :
A) $\frac{\sqrt{3}}{2}v$
B) $\frac{2\sqrt{2}}{3}v$
C) $\frac{3}{4}v$
D) $\frac{3}{\sqrt{2}}v$
Explanation
[Solution] Kinetic energy is conserved: $v^2 = \frac{v^2}{9} + v'^2 \Rightarrow v' = \frac{2\sqrt{2}}{3}v$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.