The left block in figure collides inelastically with the right block and sticks to it. Find the amplitude of the resulting simple harmonic motion.

Explanation

Assuming the collision to last for a small interval only, we can apply the principle of conservation of momentum.
Momentum before collision = momentum after collision
`mv + 0 = (m + m) v. ," "v. = (v)/(2)`
The kinetic energy of the blocks is `= (1)/(2) (2m) xx ((v)/(2))^(2) = (1)/(4)mv^(2)`
This is also the total energy of vibration as the spring is unstretched at this moment. It the amplitude is .A. the total energy can also be written as `(1)/(2) KA^(2)`
Thus, `(1)/(2)KA^(2) = (1)/(4)mv^(2) rArr A^(2) = (mv^(2))/(2k) rArr A = sqrt((m)/(2k))v`