Particle 1 experiences a perfectly elastic collision with a stationary particle 2. Determine their mass ratio, if
(a) after a head-on collision the particles fly apart in the opposite directions with equal velocities,
(b) the particles fly apart symmetrically relative to the initial motion direction of particle I with the angle of divergence `theta=60^@`.

Explanation

(a) When the particles fly apart in opposite direction with equal velocities (say v), then from conservation of momentum,
`m_1u+0=(m_2-m_1)v` (1)
and from conservation of kinetic energy,
`1/2m_1u^2=1/2m_1v^2+1/2m_2v^2`
or, `m_1u^2=(m_1+m_2)v^2`
From Eq. (1) and (2)
`m_1u^2=(m_1+m_2)(m_1^2u^2)/((m_2-m_1)^2)`
or, `m_2^2-3m_1m_2=0`
Hence `m_1/m_2=1/3` as `m_2=0`

(b) When they fly apart symmetrically relative to the initial motion direction with the angle of divergence `theta=60^@`,
From conservation of momentum, along horizontal and vertical direction,
`m_1u_1=m_1v_1cos(theta//2)+m_2v_2cos(theta//2)` (1)
and `m_1v_1sin (theta//2)=m_2v_2sin (theta//2)`
or, `m_1v_1=m_2v_2` (2)
Now, from conservation of kinetic energy,
`1/2m_1v_1^2+0=1/2m_1v_1^2+1/2m_2v_2^2` (3)
From (1) and (2),
`m_1u_1=cos(theta//2)(m_1v_1+(m_1v_1)/(m_2)m_2)=2m_1v_1cos(theta//2)`
So, `u_1=2v_1cos(theta//2)`
From (2), (3), and (4)
`4m_1cos^2(theta//2)v_1^2=m_1v_1^2+(m_2m_1^2v_1^2)/(m_2^2)`
or, `4cos^2(theta//2)=1+m_1/m_2`
or, `m_1/m_2=4cos^2theta/2-1`
and putting the value of `theta`, we get, `m_1/m_2=2`