MHT CET · Physics · Center of Mass Momentum and Collision
A ball of mass \(m\) moving with speed \(v\) collides elastically with an identical stationary ball which is initially at rest. After collision the first ball moves at an angle \(\pi\) to its initial direction and has speed \(\left(\frac{v}{3}\right)\). The second ball moves in a straight line after the collision. Then speed of the second ball after collision is
- A \(\frac{2}{\sqrt{3}} v\)
- B \(\frac{2 \sqrt{2}}{3} v\)
- C \(\frac{4}{3} v\)
- D \(\frac{3}{\sqrt{2}} v\)
Answer & Solution
Correct Answer
(B) \(\frac{2 \sqrt{2}}{3} v\)
Step-by-step Solution
Detailed explanation
For elastic collision:
Using energy conservation, \(\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2=\frac{1}{2} m_1 v_1^{\prime 2}+\frac{1}{2} m_2 v_2^{\prime 2}\)
Here, \(\frac{1}{2} m v^2+0=\frac{1}{2} m(v / 3)^2+\frac{1}{2} m v_2^{\prime 2}\)
Or \(v_2^{\prime 2}=\frac{8}{9} v^2\)
Or \(v_2^{\prime}=\frac{2 \sqrt{2}}{3} v\)
Using energy conservation, \(\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2=\frac{1}{2} m_1 v_1^{\prime 2}+\frac{1}{2} m_2 v_2^{\prime 2}\)
Here, \(\frac{1}{2} m v^2+0=\frac{1}{2} m(v / 3)^2+\frac{1}{2} m v_2^{\prime 2}\)
Or \(v_2^{\prime 2}=\frac{8}{9} v^2\)
Or \(v_2^{\prime}=\frac{2 \sqrt{2}}{3} v\)
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