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![](https://qallery.app/diagrams/v2_work_seed_1/img-2.jpeg) A body of mass $m_1$ moving with velocity $u_1$ collides head-on with a body of mass $m_2$ at rest. After a perfectly elastic collision, the velocity $v_1$ of the first body is:

A$v_1 = \dfrac{m_1 - m_2}{m_1 + m_2} u_1$
B$v_1 = \dfrac{m_2 - m_1}{m_1 + m_2} u_1$
C$v_1 = u_1$
D$v_1 = -u_1$
Answer & Solution
Correct answer: A. $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2} u_1$
**Perfectly elastic collision** conserves both momentum and kinetic energy. With target at rest ($u_2 = 0$): $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2} u_1$. **Sanity checks:** - If $m_1 = m_2$: $v_1 = 0$, $v_2 = u_1$. The first body stops, the second moves with the original speed. (Classic Newton's cradle behaviour.) - If $m_2 \gg m_1$ (e.g. ball bouncing off a wall): $v_1 \approx -u_1$, the ball reverses. Option D is right *only* in this limit. - If $m_1 \gg m_2$: $v_1 \approx u_1$, the heavy body keeps going. Option C is right *only* in this limit.
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