A `2kg ` mass is attached to a spring of force constant `600 N//m` and rests on a smooth horizontal surface. A second mass of `1kg` slides along the surface toward the first at `6m//s`.
(a) Find the amplitude of oscillation if the masses make a perfectly inelastic collision and remain together on the spring. what is the period of oscillation ?
(b) Find the amplitude and period of oscillation if the collision is perfectly elastic.
( c) For each case , write down the position `x` as a function of time `t` for the mass attached to the spring, assuming that the collision occurs at time `t = 0`. What is the impulse given to the `2kg mass in each case ?

Explanation

a. From conservation of liner momentum.
`1 xx 6 = (1 + 2)nu`
`:. Nu = 2m//s`
Now from conservation of mechanical energy.
`(1)/(2) mv^(2) = (1)/(2) kA^(2)`
`A = (sqrt((m)/(k))) nu = (sqrt((3)/(600))) xx 2`
or `A = 0.141 m = 14.1 cm`
`T = 2 pi sqrt((m)/(k)) = 2 pi sqrt((3)/(600)) = 0.44 s`

b. For perfectly elastic collision:
`nu_(2)' = ((m_(2) - m_(1))/(m_(1) + m_(2))) nu_(2) + ((2 m_(1))/(m_(1) + m_(2))) nu_(1)`
or `nu_(2)' = ((2 xx 1)/(1 + 2)) xx 6 = 4m//s`
`A = (sqrt((m)/(k)) nu_(2) = ((2)/(600))^(1//2) xx 4`
or `A = 0.23 m = 23 cm`
`T = 2 pi sqrt((m)/(k)) = 2 pi sqrt ((2)/(600)) = 0.36 s`
c. In the first case, `omega = (2 pi)/(T) = (2 pi)/(0.44) = 14.28 rad//s`
and amplitude `A = 14.1 cm`
`x = A sin omega t = (14.1 cm) sin (14.28 t)`
In the second case, `omega = (2 pi)/(T) = (2 pi)/(0.36) = 17.45 rad//s`
and amplitude `A = 23 cm`
`x = 23 sin (17.45 t) cm`
Impulase, `J = Delta P = 2 xx 2 = 4 N-s` in the first case and `4xx 2 = 8N-s` in the second case.