A problem on collision of two elastic spheres

Question

Two elastic spheres, each of mass $m$ collide directly. Show that the energy lost during the impact is $m(u^2-v^2)/4$, where $u$ and $v$ are their relative velocities before and after impact. If the velocity of one of the spheres be exactly reversed by the impact, then show that the energy lost is four times that of the sphere whose velocity is reversed. ($e=1/2$).

I have no idea to solve it. Please help me solve it in full. Thanking you in advance.

Answer 1

The laws of physics take the same form in all inertial reference frames.

So attach a frame of reference to the center of mass of the two spheres. This is inertial because of the conservation of momentum. Then the velocities of the two balls are $\pm \frac{u}{2}$ before collision, and $\pm \frac{v}{2}$ after. Hence, for the system of two spheres

$E_k = \frac{1}{2}m(\frac{1}{2}u)^2 + \frac{1}{2}m(\frac{1}{2}u)^2 = \frac{1}{4}mu^2$ before, and similarly $E_k = \frac{1}{4}mv^2$ after. So the energy lost is $\frac{1}{4}m(u^2-v^2)$.

For the rest you will need to use an inertial frame relative fixed to the ground as the individual kinetic energies depend on the choice of frame.

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Introduce the following symbols:

• $u_1,u_2: $ velocity of spheres $1,2$ prior to impact
• $v_1,v_2: $ velocity of spheres $1,2$ after impact

Case 1

Given $v_1 = -u_1$. Then by conservation of momentum, and given the coefficient of restitution $e=\frac{1}{2}$ (i.e. relative velocity $v = \frac{1}{2}u$), we have:

$\begin{align} v_2 &+ v_1 &= u_1 &+ u_2 \\ v_2 &- v_1 &= \frac{1}{2}(u_1 &- u_2) \\ \text{reducing to} \\ v_2 &- u_1 &= u_1 &+ u_2 \\ v_2 &+ u_1 &= \frac{1}{2}(u_1 &- u_2) \end{align}$

Solve to get $v_2 = \frac{1}{3}u_1 \tag{1}$, $u_2 = -\frac{5}{3}u_1 \tag{2}$

Then for the loss of kinetic energy

$\Delta E_k = \frac{m}{2}({u_1}^2 - {v_1}^2) + \frac{m}{2}({u_2}^2 - {v_2}^2) = 0 + \frac{m}{2}(\frac{25}{9}-\frac{1}{9}){u_1}^2 = \frac{m}{2}(\frac{8}{3}{u_1}^2) = \frac{8}{3} \times \frac{1}{2}mu^2$

So the loss of energy is $\frac{8}{3}$ the energy of the reversed ball [this is not the 4x stipulated in the exercise - I have checked it against conservation of momentum, coefficient of restitution and the energy loss relationship from part 1, and believe this number to be true].

Case 2

Given $v_2 = -u_2$. You will get a permutation of the results for Case 1, and the same total energy loss.