Question

A spring is connected with plank and other end of spring is connected with a block of mass m. Initially spring is stretched by a distance x${}_0$ and block is connected with a thread which is connected to other end of the plank as shown. If thread is cut, what will be maximum speed of the plank.

Answer 1

Take system: spring + plank + block
Observation for the system: No external force acts on the system in the horizontal direction.
Conclusion:
Both linear momentum and mechanical energy will be conserved for the system.
Analysis of the problem:
The speed of the plank will be maximum when all the potential energy of spring converted into kinetic energy of the system.
The initial linear momentum of the system is zero and no external forces are acting on the system, the linear momentum of the system will be zero always.
The motion of the block and plank will be such that they will always move in opposite direction or both will be at rest to justify conservation of linear momentum.
Let speed of block and plank (at maximum case) be v and V respectively. Using conservation of linear momentum at this situation
$mv=MV$ (i)
Now applying conservation of mechanical energy
$\Delta K=\Delta U=0$
$[(\frac{1}{2}M{V}^2+\frac{1}{2}m{v}^2)-0]+[0-\frac{1}{2}k{x}_0^2]=0$ (ii)
From equation (i), $v=\frac{MV}{m}$
Substituting this value in Equation (ii), we get
$[\frac{1}{2}M{V}^2+\frac{1}{2}m{\left(\frac{MV}{m}\right)}^2]=\frac{1}{2}k{x}_0^2$
$\Rightarrow \frac{1}{2}[M+\frac{M^2}{m}]v^2=\frac{1}{2}k{x}_0^2$
$V^2=\frac{km{x}_0^2}{(M^2+Mm)}\Rightarrow V=\sqrt{\frac{km}{M(M+m)}}\cdot x_0$