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 kg attached to it at a distance of 1 m, from the wall. A mass m = 0.5 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 with which the mass M will hit the wall when mass m released?

Answer 1

First we will use energy conservation

Loss in gravitational potential energy = gain in kinetic energy of the two blocks

$MgH - mgh = \frac{1}{2}Mv_1^2 + \frac{1}{2}mv_2^2$

$2*10*1 - 0.5*10*(\sqrt5 - 1) = \frac{1}{2}*2*v_1^2 + \frac{1}{2}*0.5*v_2^2$

$20 - 6.2 = v_1^2 + 0.25v_2^2$

also by constraint relation we know that

$v_1cos\theta = v_2$

where we know that

$cos\theta = \frac{2}{\sqrt5}$

now we have

$13.8 = v_1^2 + 0.25*0.8v_1^2$

$v_1^2 = \frac{13.8}{1.2}$

$v_1 = 3.4 m/s$

so the speed of the block M will be 3.4 m/s