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 ?

Answer

A, B, C, D

Explanation

(a) `omega = sqrt ((k)/(m)) = sqrt((600)/(2 + 1)) = 10sqrt(2) rad//s`
From conservation of liner momentum (at mean position) velocity of combined mass will be `2 m//s`.
This is the maximum velocity of combined mass.
`:. v = omega A`
or `A = (v)/(omega) = (2)/(10 sqrt (2))m`
`= 0.141 m = 14.1cm`
`T = (2pi)/(omega) = (2pi)/(10 sqrt (2)) = 0.44 s`
(b) If the collision is elastic,
`omega' = sqrt((600)/(2)) = 10 sqrt (3) rad//s`
In elastic collision,
`v'_(2) = ((m_(2) - m_(1))/(m_(2) + m_(1)))v_(2) + ((2m_(1))/(m_(1) + m_(2)))v_(1)`
`= ((2 xx 1)/(1 + 2))(6) = 4m//s` (As `v_(2) = 0`)
This is maximum velocity.
`:. v'_(2) = omega' A'`
or `A' = (v'_(2))/(omega') = (4)/(10 sqrt (3))`
`= 0.23m = 23cm`
`T' = (2pi)/(omega') = (2pi)/(10 sqrt (3)) = 0.36 s`
( c) In both cases journey is started from mean position
`:. x = +- Asin omega t`
will be the displacement - time eqution. For impulse we can apply the equation. Impulse = change in liner momentum.