Momentum: Question 2

Worked solution

Setting up the sign convention

Take the common direction in which both spheres are travelling as positive. So: $u_A = 8\text{ m s}^{-1}, \qquad u_B = 2\text{ m s}^{-1}$

Part (a): Total momentum before the collision

Momentum is mass times velocity, $p=mv$, so the total momentum of the system is the sum of each sphere’s momentum: $p_{\text{before}} = m_A u_A + m_B u_B = (3)(8) + (5)(2) = 24 + 10 = 34\text{ kg m s}^{-1}$

Part (b): Velocity of $B$ after the collision

Since no external horizontal forces act during the impact, momentum is conserved: the total momentum after the collision must still equal $34\text{ kg m s}^{-1}$.

Let $v_B$ be $B$‘s velocity immediately after the collision. Using $A$‘s given velocity of $1\text{ m s}^{-1}$: $m_A v_A + m_B v_B = p_{\text{before}}$ $(3)(1) + (5)v_B = 34$ $3 + 5v_B = 34$ $5v_B = 31$ $v_B = 6.2\text{ m s}^{-1}$

Since $v_B$ is positive, $B$ continues to move in the same direction as before the collision, it speeds up from $2\text{ m s}^{-1}$ to $6.2\text{ m s}^{-1}$, while the lighter sphere $A$ slows down from $8\text{ m s}^{-1}$ to $1\text{ m s}^{-1}$. This is consistent, since $B$ ($6.2\text{ m s}^{-1}$) now moves faster than $A$ ($1\text{ m s}^{-1}$), so $A$ does not catch up with $B$ again.

Final answers

• (a) Total momentum before the collision $= \boxed{34}\text{ kg m s}^{-1}$
• (b) $v_B = \boxed{6.2}\text{ m s}^{-1}$, in the same direction as the spheres’ original motion