A body of mass m_1 = 2 kg moving at v_1 = 3 m/s catches up…

A body of mass $m_1 = 2\ \mathrm{kg}$ moving at $v_1 = 3\ \mathrm{m/s}$ catches up with a body of mass $m_2 = 3\ \mathrm{kg}$ moving at $v_2 = 1\ \mathrm{m/s}$ along the same line in the same direction. Find their velocities after the collision if it is (a) perfectly inelastic, (b) elastic. What mass ratio $m_1/m_2$ would make the first body stop in an elastic collision?

Explanation of Method

Inelastic and elastic collision of two moving bodies.

Step 1: (a) Perfectly inelastic

\[ v = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{6 + 3}{5}. \]

\[v = 1.8\ \mathrm{m/s}\]

Step 2: (b) Elastic

\[ v_1' = \frac{(m_1 - m_2)v_1 + 2m_2v_2}{m_1 + m_2} = \frac{-3 + 6}{5}, \]

\[ v_2' = \frac{(m_2 - m_1)v_2 + 2m_1v_1}{m_1 + m_2}, \]

\[ v_2' = \frac{1 + 12}{5}. \]

\[v_1' = 0.6\ \mathrm{m/s}, \qquad v_2' = 2.6\ \mathrm{m/s}\]

Step 3: First body stopping

\[ (m_1 - m_2)v_1 + 2m_2v_2 = 0 \;\Longrightarrow\; \frac{m_1}{m_2} = 1 - \frac{2v_2}{v_1} = 1 - \frac23. \]

\[\frac{m_1}{m_2} = \frac13\]

Check

Energy (b): $ \tfrac12(2 \cdot 9 + 3 \cdot 1) = 10.5\ \mathrm{J} $, after $ \tfrac12(2 \cdot 0.36 + 3 \cdot 6.76) = 10.5\ \mathrm{J}\ \checkmark $

$\text{before}$ $3\ \mathrm{m/s}$ $1\ \mathrm{m/s}$ $\text{(a) inelastic}$ $1.8\ \mathrm{m/s}$ $\text{(b) elastic}$ $0.6\ \mathrm{m/s}$ $2.6\ \mathrm{m/s}$ $2\ \mathrm{kg}$ $3\ \mathrm{kg}$

Velocities to scale. In the inelastic case the bodies move off together; in the elastic case the faster, lighter body slows to 0.6 m/s and the heavier one speeds up to 2.6 m/s.

Result

\[\boxed{\,v = 1.8\ \mathrm{m/s}\,}\]

\[\boxed{\,v_1' = 0.6\ \mathrm{m/s}, \qquad v_2' = 2.6\ \mathrm{m/s}\,}\]

\[\boxed{\,\frac{m_1}{m_2} = \frac13\,}\]

Remark

Check. All the original numbers are correct, but its last algebraic step, $m_1/m_2 = v_2/v_1$, is wrong; the general result is $m_1/m_2 = 1 - 2v_2/v_1$. Both happen to give 1/3 here only because $v_1 = 3v_2$.