An object of mass 4 kg is attached to a spring having spring constant `100(N)/(m)`. It performs simple harmonic motion on a smooth horizontal surface with an amplitude of 2 m. A 6 kg object is dropped vertically onto the 4 kg object when it crosses the mean position, and sticks to it. the change in amplitude of oscillation due to collision is

Options

A. `1m`
B. `zero`
C. `2[1-sqrt((2)/(5))]`
D. `2[1-(1)/(sqrt5)]`

Answer

C

Explanation

Time period of the system (object of mass 4 kg) before collision is `T_1=2pisqrt((4)/(100))`
After collision, time period of the combined mass is
`T_2=2pisqrt((10)/(100))`
We can apply momentum conservation for just before the collision and just after the collision in the horizontal direction.
`4A_1omega_1=10A_2omega_2impliesA(4xx2xxsqrt((100)/(4)))/(10xxsqrt((100)/(10)))=2sqrt((2)/(5))`m
So change in amplitude, `triangleA=A_1-A_2=2[1-sqrt((2)/(5))]`m