Momentum: Question 10

Worked solution

Step 1: Total momentum before the collision

$Q$ starts at rest, so $u_Q=0$. Total momentum before the collision: $p_{\text{before}} = m_P u_P + m_Q u_Q = (0.8)(5) + m(0) = 4\text{ kg m s}^{-1}$

Step 2: Total momentum after the collision

$P$ and $Q$ couple together, so they share one common velocity of $2\text{ m s}^{-1}$ afterwards, and the combined mass is $(0.8+m)$: $p_{\text{after}} = (m_P+m_Q)v = (0.8+m)(2) = 1.6 + 2m$

Step 3: Apply conservation of momentum and solve for $m$

$p_{\text{before}} = p_{\text{after}}$ $4 = 1.6 + 2m$ $2m = 2.4$ $m = 1.2$

Why the other options are wrong

• B ($0.8\text{ kg}$): this wrongly assumes $Q$‘s mass must equal $P$‘s mass, without using the momentum equation at all.
• C ($2.0\text{ kg}$): this comes from solving $4=2m$, which omits $P$‘s own momentum contribution ($1.6\text{ kg m s}^{-1}$) from the combined momentum after the collision.
• D ($3.2\text{ kg}$): this comes from solving $4=0.8+m$, forgetting to multiply the combined mass by the common velocity of $2\text{ m s}^{-1}$.

Final answer

• Mass of $Q$ $= \boxed{1.2}\text{ kg}$, option A.