Two identical balls `A` and `B` are released from the position shown in Fig. They collide elastically with each other on the horizontal portion. The ratio of heights attained by `A` and `B` after collision is (neglect friction)

Options

A. `1:4`
B. `2:1`
C. `4:13`
D. `2:11`

Answer

C

Explanation

After collision balls exchange their velocities
i.e., `v_A=sqrt(2gh) and v_B=sqrt(2g(4h))=2sqrt(2gh)`
Height attained by A will be `h_A=v_A^2/(2g)=h`

But path of B will be first straight line and then parabolic as shown in figure.
Using energy conservation for ball B.
`1/2mv_B^2=1/2mv^2+mgh " " or "" v=sqrt(6gh)`
`therefore h_B=h+(v^2sin^2 60^@)/(2g)=h+(9h)/4=(13h)/4`
Hence, `h_A/h_B=4/13`