Two blocks of masses 3 kg and 6kg rest on horizontal smooth surface. The 3 kg block is attached to a spring with a force constant `k=900Nm^(-1)` which is compressed 2m from beyond the equilibrium position. The 6 kg mass is at rest at 1m from mean position. 3kg mass strikes the 6 kg mass and the two stick together.

Options

A. velocity of the combined masses immediately after the collision is `10 ms^(-1)`
B. velocity of the combined masses imeediately after the collsion is `5 ms^(-1)`
C. amplitude of the resulting oscillation is `sqrt(2) m`
D. amplitude of the resulting oscillation is `sqrt(5)//2 m`

Answer

A, C

Explanation

Velocity of 3kg block just before collision
`= omegasqrt(a^(2) - x^(2)) = sqrt((k/m)(a^(2)-x^(2))`
`=sqrt((900/3)(2^(2) - 1^(2))) = 30 m//s`
Velocity of combined masses immediately after the collision
`= ((3)(30))/(3+6) = 10 m//s`
New angular frequency
`omega' = sqrt(k/m) = sqrt(900/9) = 10`
Therefore `v' = omega'sqrt(a'^(2) - x^(2))`
`rArr 10 = 10sqrt(a'^(2) - 1^(2)) rArr a' = sqrt(2m)`