One end of an ideal spring is fixed to a wall at origin `O` and axis of spring is parallel to x-axis. A block of mass `m = 1kg` is attached to free end of the spring and it is performing SHM. Equation of position of the block in co-ordinate system shown in figure is `x = 10 + 3 sin (10t)`. Here, t is in second and `x` in `cm`. Another block of mass `M = 3kg`, moving towards the origin with velocity `30 cm//s` collides with the block performing SHM at `t = 0` and gets stuck to it. Calculate

(a) new amplitude of oscillations,
(b) neweqution for position of the combined body,
( c) loss of energy during collision. Neglect friction.

Options

A. 3cm
B. 20cm
C. 10cm
D. 100cm

Answer

A

Explanation

Conserving linear momentum.
`(1+3)v=1xx0.3+3(-0.3)`
`v=-0.15m//s `
Negative sign indicates that combined body starts to move leftward. But at the instant of collision, spring is its natural length or combined body is in equilibrium position.
Hence, at t=0, phase of combined body becomes equal to`pi`
Therefore, new amplitude of oscillation is `a'=(|v|)/(omega)=(0.15)/(5)=0.03m=3cm`