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 \(\theta\) to its initial direction and has a speed v/3 . The second blocks speed after the collision is :
Answer & explanation
Correct answer: option 2
\(\frac{1}{2}Mv^2 = \frac{1}{2}M(\frac{v}{3})^2 + \frac{1}{2}Mv'^2\)
v' = \(\frac{2\sqrt{2}}{3}v\)