Momentum: Question 4

Worked solution

Part (a): Total momentum before pushing off

Both skaters start at rest, so each has zero velocity, and hence zero momentum: $p_{\text{before}} = m_A(0) + m_B(0) = 0\text{ kg m s}^{-1}$

Part (b): Velocity of $B$ after pushing off

Take the direction in which $A$ moves off as positive. By conservation of momentum, the total momentum immediately after pushing off must still equal the total momentum before, which was $0$: $m_A v_A + m_B v_B = p_{\text{before}}$ $(55)(3.5) + (70)v_B = 0$ $192.5 + 70v_B = 0$ $v_B = -\frac{192.5}{70} = -2.75\text{ m s}^{-1}$

The negative sign shows that $B$ moves in the direction opposite to $A$. So $B$ moves off at $2.75\text{ m s}^{-1}$, away from $A$ in the opposite direction to the one $A$ takes.

Part (c): Why they must move in opposite directions

Before pushing off, the total momentum of the system is $0$ (part (a)). Since no external horizontal forces act on the skaters (the ice is frictionless), momentum is conserved, so the total momentum immediately afterwards must still be $0$: $m_A v_A + m_B v_B = 0 \quad\Longrightarrow\quad m_B v_B = -m_A v_A$

Since $m_A$ and $m_B$ are both positive, $v_B$ must have the opposite sign to $v_A$. This means that however hard, or in whichever direction, the skaters push, they must always end up moving apart in opposite directions along the line. One of them cannot simply set off without the other moving the other way, or total momentum would no longer be zero.

Final answers

• (a) Total momentum before $= \boxed{0}\text{ kg m s}^{-1}$
• (b) $v_B = \boxed{-2.75}\text{ m s}^{-1}$, i.e. $2.75\text{ m s}^{-1}$ in the direction opposite to $A$
• (c) Total momentum stays at $0$, so $m_Bv_B = -m_Av_A$ forces $v_A$ and $v_B$ to have opposite signs