Momentum: Question 3

Worked solution

Part (a): Total momentum before the collision

Take the direction of $P$‘s initial motion as positive. Truck $Q$ moves directly towards $P$, i.e. in the opposite direction, so its velocity is negative: $u_P = 5\text{ m s}^{-1}, \qquad u_Q = -2\text{ m s}^{-1}$

The total momentum before the collision is: $p_{\text{before}} = m_P u_P + m_Q u_Q = (800)(5) + (1200)(-2) = 4000 - 2400 = 1600\text{ kg m s}^{-1}$

Part (b): Common velocity after coupling

Since $P$ and $Q$ couple together, they move afterwards with one common velocity $v$, and the combined mass is: $m_P + m_Q = 800 + 1200 = 2000\text{ kg}$

By conservation of momentum, the total momentum is unchanged by the collision: $m_P u_P + m_Q u_Q = (m_P+m_Q)v$ $1600 = 2000v$ $v = \frac{1600}{2000} = 0.8\text{ m s}^{-1}$

Since $v$ is positive, the coupled trucks move at $0.8\text{ m s}^{-1}$ in the direction of $P$‘s original motion.

Part (c): Effect of reversing $Q$‘s direction

If $Q$ had instead moved at $2\text{ m s}^{-1}$ in the same direction as $P$, its momentum would be $+2400\text{ kg m s}^{-1}$ rather than $-2400\text{ kg m s}^{-1}$. The total momentum before collision would then be: $4000 + 2400 = 6400\text{ kg m s}^{-1}$

instead of the original $1600\text{ kg m s}^{-1}$. Because the two trucks’ momenta would now add together rather than partly cancelling each other out, the total (and hence the common velocity after coupling, $6400 \div 2000 = 3.2\text{ m s}^{-1}$) would be larger. So the coupled trucks would move faster than in the original scenario.

Final answers

• (a) $u_P = \boxed{5}\text{ m s}^{-1}$, $u_Q = \boxed{-2}\text{ m s}^{-1}$; total momentum before $= \boxed{1600}\text{ kg m s}^{-1}$
• (b) Common velocity $= \boxed{0.8}\text{ m s}^{-1}$, in the direction of $P$‘s original motion
• (c) Faster. The momenta would add instead of partly cancelling, giving a larger total momentum