Question

Two blocks with masses $m_{1}=1 \, kg \, and \, m_{2}=2 \, kg$ are connected by a spring constant $k=24 \, N \, m^{- 1}$ and placed on a frictionless horizontal surface. The block $m_{1}$ is imparted an initial velocity $v_{0}=12 \, cm \, s^{ \, - 1}$ to the right, the value of amplitude of oscillation (in $cm$ ) of the block with respect to other is

Answer 1

The amplitude of oscillations will be the maximum compression in the spring. At the time of maximum compression velocities of both the blocks are equal say $v$ , then using law of conservation of momentum,
$m_{1}v_{0}=\left(m_{1} + m_{2}\right)v$
or $1\times 12=\left(1 + 2\right)v \, \, or \, \, v=4 \, cm \, s^{- 1}$
Using law of conservation of energy, we have
$\frac{1}{2}m_{1}v_{0}^{2}=\frac{1}{2}kx^{2}+\frac{1}{2}\left(m_{1} + m_{2}\right)v^{2}$
Putting the value and solving we get $x=2$ cm