Question

On a frictionless surface, a block of mass M moving at speed $\upsilon$ 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 $\frac{\upsilon }{3}$. The second block's speed after the collision is:

• A. $\frac{\sqrt{3}}{2}\upsilon$
• B. $\frac{2\sqrt{2}}{3}\upsilon$
• C. $\frac{3}{4}\upsilon$
• D. $\frac{3}{\sqrt{2}}\upsilon$

Answer 1

Answer: (B)

$\frac{2\sqrt{2}}{3}\upsilon$

In an elastic collision:$K_i=K_f$ where i and f refer to initial and final, K refers to kinetic energy of the particle.
$\therefore \frac{1}{2}m{v}^2+0=\frac{1}{2}m(\frac{v}{3}{)}^2+\frac{1}{2}m{v}^{\prime 2}$ where v' is the speed of the 2nd particle after collision.
$\therefore v^{\prime }=\frac{2\sqrt{2}}{3}v$