Question

Two blocks of masses `3kg` block is attached to a spring with a force constant, `k k = 900N//ma` which is compressed `2m` initially from its equilibrium position. When `3kg` mass is released, it strikes the `6kg` mass and the two stick togther in an inelastic collision.

The common velocity of the blocks after collision is
A. `10 m//s`
B. `30m//s`
C. `15m//s`
D. `2m//s`

Answer 1

Correct Answer - A
Angular frequency of oscillation is given by `omega = sqrt((k)/(m)) = 10 sqrt(3)`. Velocity of `3kg` block at the instant it hits the `6kg` block is given by `v = omega sqrt(A^(2)-x^(2)) = 10 sqrt(3) sqrt(2^(2) -1^(2)) rArr 30 m//s`
During collision moment is conserved thus
`m_(1)v_(1) = (m_(1)+m_(2)) v_(C ) rArr 3 xx 30 = 9 xx v_(C)`
or hence, common velocity of combined mass `=(v_(C )) = 10m//s`