Source: High school physics (Chinese)
Problem Sets:
Problem
An object with mass $m_1=2.0$ kg undergoes a one-dimensional elastic collision with a stationary object of mass $m_2$. After the collision, the first object continues to move in its original direction with a speed equal to one-fourth of its initial speed.
Let the initial velocity of $m_1$ be $v_1$ and its final velocity be $v_1' = v_1/4$. The initial velocity of $m_2$ is $v_2=0$. For a 1D elastic collision, both momentum and kinetic energy are conserved. Momentum conservation: $m_1 v_1 = m_1 v_1' + m_2 v_2'$. For elastic collisions, the relative speed of approach equals the relative speed of separation: $v_1 - v_2 = v_2' - v_1'$. Using $v_2=0$ and $v_1' = v_1/4$:
$$v_1 = v_2' - v_1/4 \implies v_2' = \frac{5}{4}v_1$$Substitute $v_1'$ and $v_2'$ into the momentum equation:
$$m_1 v_1 = m_1 (\frac{1}{4}v_1) + m_2 (\frac{5}{4}v_1)$$ $$m_1 = \frac{1}{4}m_1 + \frac{5}{4}m_2 \implies \frac{3}{4}m_1 = \frac{5}{4}m_2 \implies m_2 = \frac{3}{5}m_1$$