Question

Block of mass `m_(2)` is in equilibrium and at rest. The mass `m_(1)` moving with velocity `u` vertically downwards collids with `m_(2)` and sticks to it. Find the energy of oscillation.

Answer 1

Let the velocity of `m_(1)` & `m_(2)` be `v` after collision
By conservation of momentum
Hence , `KE = (1)/(2)(m_(1) + m_(2))v^(2) = (1)/(2)(m_(1)^(2))/(m_(1) + m_(2))u^(2)`

`m_(2)g = Kx_(1)`
`(m_(1) + m_(2))g = Kx_(2)`
`PE = (1)/(2)K(x_(2) - x_(1))^(2) = (1)/(2)K((m_(1)g)/(K))^(2) rArr PE = (1)/(2)(m_(1)^(2)g^(2))/(K)`
Therefore energy of oscillion is -
`E = KE + PE`
`E = (1)/(2)(m^(2)u^(2))/(m_(1) + m_(2)) + (m_(1)^(2)g^(2))/(2K) rArr = (1)/(2)[(m_(1)^(2)u^(2))/(m_(1) + m_(2)) + (m_(1)^(2)g^(2))/(K)]` Ans.