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ed_saifa · « **on:** April 21, 2009, 08:10:09 pm »

Two objects $m_1$ and $m_2$ travelling with velocities $u_1$ and $u_2$ collide elastically. Then using the expressions from conservation of momentum and kinetic energy show that relative velocity of body 2 with respect to body 1 is equal and opposite before and after collision ie show that $u_1-u_2 = -(v_1-v_2)$
Thanks ^^

TrueTears · « **Reply #1 on:** April 21, 2009, 08:17:05 pm »

$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ ...[1]

$m_1u_1 + m_2(-u_2) = m_1(-v_1) + m_2(v_2)$ ...[2]

solve simultaneously $\implies$ this might help http://vcenotes.com/forum/index.php/topic,9668.msg134646.html#msg134646

Mao · « **Reply #2 on:** April 21, 2009, 08:29:49 pm »

here:

Kinetic energy: $\frac{m_1 u_1^2 + m_2 u_2^2}{2} = \frac{m_1 v_1^2 + m_2 v_2^2}{2}$

$\implies m_1 (u_1^2 - v_1^2) = m_2 (v_2^2 - u_2^2)$

$\implies m_1 (u_1 + v_1) (u_1 - v_1) = - m_2 (u_2 + v_2) (u_2 - v_2)$ ... [1]

Momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

$m_1 (u_1 - v_1) = - m_2 (u_2 - v_2)$ ... [2]

[2]/[1]: $u_1 + v_1 = u_2 + v_2, \; u_1 \neq v_1, \; u_2 \neq v_2$

$\implies u_1 - u_2 = - (v_1 - v_2)$

Shown as required. [The relative velocity of the particles is reversed after elastic collision.]

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