Answer to Question #122331 in Physics for Samson

Question

A moving sphere has a head-on elastic collision with an initially stationary sphere. After collision, the kinetic energies of the two spheres are equal. Show that the mass ratio of the two spheres is 0.1716?

Answer 1

As far as the collision is elastic and there are no external forces on the system, the conservation laws of momentum and kineticenergy:

$$m_1 v_1 = m_1v_1' + m_2 v_2'\\ \dfrac{m_1v_1^2}{2} = \dfrac{m_1v_1'^2}{2} + \dfrac{m_2v_2'^2}{2}$$

where $v_1$ is the speed of the first sphere before the collision and $v_1'$ and $v_2'$ are the speed of the first and the second spheres respectively. As far, as the kinetic energies of the two spheres are equal after collision, we can write:

$$\dfrac{m_1v_1'^2}{2} = \dfrac{m_2v_2'^2}{2}\\$$

Substituting this into the second equation, get:

$$\dfrac{m_1v_1^2}{2} = \dfrac{m_1v_1'^2}{2} + \dfrac{m_2v_2'^2}{2} =\dfrac{m_1v_1'^2}{2} + \dfrac{m_1v_1'^2}{2} = m_1v_1'^2$$

Thus:

$$\dfrac{m_1v_1^2}{2} = m_1v_1'^2\\ v_1' = \dfrac{v_1}{\sqrt{2}}$$

Let's substitute this to the first equation (momentum conservation) and express $v_2'$:

$$m_1 v_1 = m_1\dfrac{v_1}{\sqrt{2}}+ m_2 v_2'\\$$

$$v_2' = \dfrac{m_1v_1(\sqrt{2}-1)}{\sqrt{2}m_2}$$

Again, from the eqation $\dfrac{m_1v_1'^2}{2} = \dfrac{m_2v_2'^2}{2}$ note, that:

$$\dfrac{m_2}{m_1} = \dfrac{v_1'^2}{v_2'^2}$$

Substituting here the expressions for $v_1'$ and $v_2'$ obtained before, get:

$$\dfrac{m_2}{m_1} = \dfrac{v_1^2}{2}\cdot \dfrac{2m_2^2 }{m_1^2v_1^2(\sqrt{2}-1)^2}$$

After expressing the ratio $m_2/m_1$, get:

$$\dfrac{m_2}{m_1} = (\sqrt{2}-1)^2 \approx 0.1716$$

QED.