Momentum: Question 1

Worked solution

Step 1: Assign signed velocities using the stated convention

Taking the direction of $P$‘s initial motion as positive: $u_P = 3\text{ m s}^{-1}, \qquad u_Q = -3\text{ m s}^{-1}$

($Q$ moves in the opposite direction to $P$, so its velocity is negative.)

Step 2: Find the total momentum before the collision

$p_{\text{before}} = m_P u_P + m_Q u_Q = (0.5)(3) + (0.5)(-3) = 1.5 - 1.5 = 0$

The two momenta are equal in size but opposite in sign, so they cancel exactly.

Step 3: Apply conservation of momentum to find the common velocity

Since $P$ and $Q$ coalesce, they share one common velocity $v$ afterwards, and the combined mass is $0.5+0.5=1\text{ kg}$: $m_P u_P + m_Q u_Q = (m_P+m_Q)v$ $0 = 1 \times v$ $v = 0\text{ m s}^{-1}$

The combined toy is momentarily brought to rest, because the two original momenta were equal and opposite.

Why the other options are wrong

• B ($3\text{ m s}^{-1}$): this comes from wrongly averaging the two speeds without giving $Q$‘s velocity a negative sign.
• C ($-3\text{ m s}^{-1}$): this simply reuses $Q$‘s original velocity, ignoring that momentum must be conserved for the whole system.
• D ($1.5\text{ m s}^{-1}$): this comes from adding the two (unsigned) speeds and dividing by $2$, rather than correctly signing $Q$‘s velocity before combining.

Final answer

• Common velocity $= \boxed{0}\text{ m s}^{-1}$, option A.