A particle of mass `m_1` collides elastically with a stationary particle of mass `m_2(m_1gtm_2)`. Find the maximum angle through which the striking particle may deviate as a result of the collision.

Explanation

From conservation of momentum
`vecp_1=overset(rarr')p_1+overset(rarr')p_2`
so `(vecp_1-overset(rarr')p_1)^2=p_1^2-2p_1p_1^(')cos theta_1+p_1^('^2)=p_2^('^2)`
From conservation of energy
`(p_1^2)/(2m_1)=(p_1^'2)/(2m_1)+(p_2^'2)/(2m_2)`
Eliminating `p_2'` we get
`0=p_1^('2)(1+m_2/m_1)-2p_1^'p_1costheta_1+p_1^2(1-m_2/m_1)`
This quardratic equation for `p_1'` has a real solution in terms of `p_1` and `costheta_1` only if
`4cos^2theta_1 ge 4(1-m_2^2/m_1^2)`
or `sin^2theta le m_2^2/m_1^2`
or `sin theta_1 le +m_2/m_1` or `sin theta_1 ge -m_2/m_1`
This clearly implies (since only + sign makes sense) that
`sin theta_(1max)=m_2/m_1`.