Q154 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Physics
On a friction-less surface, a block of mass M moving at speed $v$ collides elastically with another block (of mass M) initially at rest. After collision, the first block moves at an angle $\theta$ to its initial direction and has speed $\dfrac{v}{3}$. The speed of the second block after collision is
- A$\dfrac{3v}{4}$
- B$\dfrac{\sqrt{3}\,v}{2}$
- Correct answerC$\dfrac{2\sqrt{2}\,v}{3}$
- D$\dfrac{3v}{\sqrt{2}}$
Solution
By the principle of conservation of kinetic energy for an elastic collision:
$\dfrac{1}{2} M v^2 = \dfrac{1}{2} M v_1^2 + \dfrac{1}{2} M v_2^2$
Substituting $v_1 = \dfrac{v}{3}$:
$\dfrac{1}{2} M v^2 = \dfrac{1}{2} M \left(\dfrac{v}{3}\right)^2 + \dfrac{1}{2} M v_2^2$
Canceling $\dfrac{1}{2} M$ from both sides:
$v^2 = \dfrac{v^2}{9} + v_2^2$
$v_2^2 = v^2 - \dfrac{v^2}{9}$
$v_2^2 = \dfrac{8v^2}{9}$
$v_2 = \dfrac{2\sqrt{2}v}{3}$
Answer: $\dfrac{2\sqrt{2}\,v}{3}$
$\dfrac{1}{2} M v^2 = \dfrac{1}{2} M v_1^2 + \dfrac{1}{2} M v_2^2$
Substituting $v_1 = \dfrac{v}{3}$:
$\dfrac{1}{2} M v^2 = \dfrac{1}{2} M \left(\dfrac{v}{3}\right)^2 + \dfrac{1}{2} M v_2^2$
Canceling $\dfrac{1}{2} M$ from both sides:
$v^2 = \dfrac{v^2}{9} + v_2^2$
$v_2^2 = v^2 - \dfrac{v^2}{9}$
$v_2^2 = \dfrac{8v^2}{9}$
$v_2 = \dfrac{2\sqrt{2}v}{3}$
Answer: $\dfrac{2\sqrt{2}\,v}{3}$
Hence, the correct answer is option: 'C'.