Elastic Collision

Conservation of Momentum

In any collision, total momentum is conserved:

$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$

An elastic collision also conserves kinetic energy. Solving both conservation equations simultaneously gives the final velocities:

$v_1' = \frac{(m_1 - m_2)v_1 + 2m_2 v_2}{m_1 + m_2}$

$v_2' = \frac{(m_2 - m_1)v_2 + 2m_1 v_1}{m_1 + m_2}$

Scenario	Result
Equal masses ( $v_2=0$ )	$v_1'=0$ , $v_2'=v_1$ (complete transfer)
Heavy hits light ( $m_1 \gg m_2$ )	$v_1' \approx v_1$ , $v_2' \approx 2v_1$ (light bounces fast)
Light hits heavy ( $m_1 \ll m_2$ )	$v_1' \approx -v_1$ , $v_2' \approx 0$ (light bounces back)

$m_1$	$v_1$	$m_2$	$v_2$	$v_1'$
1	10	1	0	0.0000 (stops)
2	10	1	0	3.3333
1	0	2	5	6.6667

Implement elasticV1(m1, v1, m2, v2) returning the post-collision velocity of mass 1.

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