Question

Two objects moving in opposite directions with the same speed v undergo a totally inelastic collision, and half the initial kinetic energy is lost. Find the ratio of their masses.

Answer 1

Answer:

$\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875$

Explanation:

We name M the larger mas, and m the smaller mass, so base on this we write the following conservation of momentum equation for the collision:

$P_i=P_f\\M\,v_i -m\,v_i=(M+m)\,v_f\\(M-m)\,v_i = (M+m)\,v_f\\\\\frac{v_i}{v_f} = \frac{(M+m)}{(M-m)}$

We can write this in terms of what we are looking for (the quotient of masses $\frac{m}{M}$:

$\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}$

We use now the information about  Kinetic Energy of the system being reduced in half after the collision:

$K_i=2\,K_f\\\frac{1}{2} (M+m)\,v_i^2= 2*\frac{1}{2} (M+m)v_f^2\\v_i^2=2\,v_f^2\\\frac{v_i^2}{v_f^2} =2$

We can combine this last equation with the previous one to obtain:

$\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}\\(\frac{v_i}{v_f})^2 = \frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}\\2=\frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}$

where solving for the quotient m/M gives:

$\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875$