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

Answer

A

Explanation

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