Question

A string with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of 2 m from the wall, has a point mass $M=2\text{kg}$ attached to it at a distance of 1 m from the wall. A mass $m=0.5\text{kg}$ attached at the free end is held at rest so that the string is horizontal between the wall and the pulley and vertical beyond the pulley. What will be the speed in m/s upto 1 decimal place with which the mass $M$ will hit the wall when the mass m is released and $g=9.8{\text{ms}}^{-2}$?

Answer 1

When mass $m$ is released from rest, the heavier mass $M$ loses potential energy and it strikes the wall with velocity $v$ and the string remains stretched. It increases the kinetic energy of mass M and increases the kinetic and potential energies of mass m.
Decrease in potential energy of mass M
$=Mg\times 1=2.0\times g\times 1=2g$ where $g=9.8{\text{ms}}^{-2}$
Increase in kinetic energy of mass $M=\frac{1}{2}M{v}^2=\frac{1}{2}\times 2.0\times v^2=v^2$
Increase in potential energy of mass $m=mg(\sqrt{5}-1)=\frac{g}{2}(\sqrt{5}-1)$
$(\because$ Distance through which mass m is raised)
$=AB+BC-AC=1+\sqrt{5}-2=(\sqrt{5}-1)\text{m}$
Increase in kinetic energy of mass $m=\frac{1}{2}m{v}^2=\frac{1}{2}\times 0.5\times v^2=\frac{v^2}{4}$
As the total energy must be conserved, the decrease in the potential energy of the mass M equals the increase in the potential and kinetic energies of both the masses.
$\therefore Mg\times 1=\frac{1}{2}M{v}^2+mg(\sqrt{5}-1)+\frac{1}{2}m{v}^2$
$\therefore 2g=v^2+\frac{1}{2}g(\sqrt{5}-1)+\frac{1}{4}{v}^2\Rightarrow g[2-\frac{\sqrt{5}}{2}+\frac{1}{2}]=\frac{5}{4}{v}^2=g\left(1.38\right)$
or $v^2=1.38\times \frac{9.8\times 4}{5}\therefore v=3.28\text{m/s}$