Question

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 $\frac{v}{3}$. The second block's speed after the collision is

• A. $\frac{3}{\sqrt 2}v$
• B. $\frac{\sqrt3}{ 2}v$
• C. $\frac{2\sqrt 2}{3}v$
• D. $\frac{3}{4}v$

Answer 1

The Correct Option is C

To solve this problem, we will use the principles of conservation of momentum and conservation of kinetic energy, due to the fact that the collision is elastic.

Conservation of Momentum

The initial momentum of the system is due to the moving block of mass \( M \) that is traveling at speed \( v \). The block at rest has no initial momentum. Therefore, the total initial momentum is given by:

p_{\text{initial}} = Mv

After the collision, both blocks will have some momentum. Let the velocity of the second block be \( v_2 \) and the angle of the first block with respect to its original path be \(\theta\).

p_{\text{final, x}} = M \times \frac{v}{3} \cos \theta + M \times v_2 \cos \phi = Mv

p_{\text{final, y}} = M \times \frac{v}{3} \sin \theta = M \times v_2 \sin \phi

Conservation of Kinetic Energy

The total kinetic energy of the system is conserved in elastic collisions:

\frac{1}{2}Mv^2 = \frac{1}{2}M\left( \frac{v}{3} \right)^2 + \frac{1}{2}Mv_2^2

v^2 = \left( \frac{v}{3} \right)^2 + v_2^2

Substitute \(\left( \frac{v}{3} \right)^2\):

v^2 = \frac{v^2}{9} + v_2^2

v_2^2 = v^2 - \frac{v^2}{9}

v_2^2 = \frac{9v^2 - v^2}{9}

v_2^2 = \frac{8v^2}{9}

v_2 = \sqrt{\frac{8v^2}{9}} = \frac{\sqrt{8}}{3}v = \frac{2\sqrt{2}}{3}v

Conclusion

The speed of the second block after the collision is \frac{2\sqrt{2}}{3}v, which corresponds to the correct option.