Two pendulum bobs of masses m and `2m` collide head on elastically at the lowest point in their motion. If both the balls are released from a height H above the lowest point, to what heights do they rise for the first time after collision?

Answer

B

Explanation

Given, `m_1=m`, `m_2=2m`, `v_1=-sqrt(2gH)` and `v_2=sqrt(2gH)`

Since, the collision is elastic. Using Eqs. (v) and (vi) discussed in the theory the velocities after collision are
`v_1^()'=((m-2m)/(m+2m))(-sqrt(2gH))+((4m)/(m+2m))sqrt(2gH)`
`=(sqrt(2gH))/(3)+(4sqrt(2gH))/(3)=5/3sqrt(2gH)`
and `v_2^()'=((2m-m)/(m+2m))(sqrt(2gH))+((2m)/(m+2m))(-sqrt(2gH))`
`=(sqrt(2gH))/(3)-(2sqrt(2gH))/(3)=-(sqrt(2gH))/(3)`

i.e. the velocities of the balls after the collision are as shown in figure.
Therefore, the heights to which the balls rise after the collision are:
`h_1=((v_1^()')^2)/(2g)` (using `v^2=u^2-2gh`)
or `h_1=((5/3sqrt(2gH))^2)/(2g)` or `h_1=25/9H`
and `h_2=((v_2^')^2)/(2g)`
or `h_2=(((sqrt(2gH))/(3))^2)/(2g)`
or `h_2=H/9`