Momentum: Question 7

Worked solution

Part (a): Setting up the momentum equation

Take the common direction of motion as positive. Before the collision: $u_A = 6\text{ m s}^{-1}, \qquad u_B = 4\text{ m s}^{-1}$

Total momentum before the collision: $p_{\text{before}} = m_A u_A + m u_B = (180)(6) + m(4) = 1080 + 4m$

After the collision, $A$ has velocity $3\text{ m s}^{-1}$ and $B$ has velocity $10\text{ m s}^{-1}$, so: $p_{\text{after}} = m_A v_A + m v_B = (180)(3) + m(10) = 540 + 10m$

By conservation of momentum, $p_{\text{before}} = p_{\text{after}}$: $1080 + 4m = 540 + 10m$

Collecting the $m$ terms: $1080 - 540 = 10m - 4m$ $540 = 6m$ $m = \frac{540}{6} = 90$

Check: momentum before $= (180)(6)+(90)(4)=1080+360=1440\text{ kg m s}^{-1}$; momentum after $= (180)(3)+(90)(10)=540+900=1440\text{ kg m s}^{-1}$. The two totals agree, confirming $m=90\text{ kg}$.

Part (b): Can $A$ collide with $B$ again?

Immediately after the collision, $A$ moves at $3\text{ m s}^{-1}$ and $B$ moves at $10\text{ m s}^{-1}$, both in the same (positive) direction. Since $B$ is now moving faster than $A$ in the same direction, the distance between them only increases from this point onward, so $A$ cannot collide with $B$ again.

Final answers

• (a) $m = \boxed{90}\text{ kg}$
• (b) No. $B$ moves away from $A$ at a greater speed ($10\text{ m s}^{-1}$ compared with $A$‘s $3\text{ m s}^{-1}$) in the same direction, so they cannot meet again