In the figure shown, mass `2m` connected with a spring of force constant `k` is at rest and in equilibrium. A particle of mass `m` is released from height `4.5 mg//k` from `2m` . The particle stick to the block. Neglecting the duration of collision find time from the release of `m` to the moment when the spring has maximum compression.

Answer

5

Explanation

Velocity of the particle just before collision,
`u = sqrt(2gxx(4.5mg)/(K))`. Now just after collision, velocity `(V)` of the system of 'plate + particle ' is given by `mu = 3 mV`
`y = Asin (omegat +phi)`, where `A = (2mg)/(K)`,
`phi = (5pi)/(6), omega = sqrt((K)/(3m))`
The plate will be at rest again when `y =- A`
After simplification, `t = (2pi)/(3) sqrt((3m)/(K)) = pi//5s`.