Q.Two ice skaters of masses $50\ \text{kg}$ and $70\ \text{kg}$, standing together at rest on frictionless ice, push each other apart. If the $50\ \text{kg}$ skater moves off with a speed of $3\ \text{m/s}$, find the speed with which the $70\ \text{kg}$ skater moves off.

Given: $m_1 = 50\ \text{kg}$ moving off at $v_1 = 3\ \text{m/s}$; $m_2 = 70\ \text{kg}$ moving off at unknown speed $v_2$, in the opposite direction. Before pushing apart, both skaters are at rest together, so the total momentum of this isolated (frictionless-ice) system is zero, and it must remain zero immediately afterward: $m_1 v_1 + m_2 (-v_2) = 0$ (taking the first skater's direction as positive, so the second skater's velocity is $-v_2$). Solving for $v_2$: $$v_2 = \frac{m_1 v_1}{m_2} = \fra …

Set the total momentum after the push equal to the total momentum before (zero, since both start at rest), and …

• Forgetting that the two skaters must move in opposite directions for their momenta to cancel back to zero. …