Two shopping trolleys, X and Y, roll in the same straight line across a smooth, flat, horizontal car park. Trolley X has mass 15 kg and moves at 2 m s^-1. Trolley Y has mass 10 kg and moves at 0.5 m s^-1 in the same direction, some distance ahead of X. X catches up with Y and th…

A Level Maths 9709: Two Shopping Trolleys Colliding and Locking Together

Setting up the sign convention

Take the common direction of motion as positive. So: $u_X = 2\text{ m s}^{-1}, \qquad u_Y = 0.5\text{ m s}^{-1}$

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 trolley’s momentum: $p_{\text{before}} = m_X u_X + m_Y u_Y = (15)(2) + (10)(0.5) = 30 + 5 = 35\text{ kg m s}^{-1}$

Part (b): Common velocity after the collision

Since $X$ and $Y$ lock together, they share one common velocity $v$ afterwards, and the combined mass is: $m_X + m_Y = 15 + 10 = 25\text{ kg}$

By conservation of momentum, the total momentum is unchanged by the collision: $m_X u_X + m_Y u_Y = (m_X+m_Y)v$ $35 = 25v$ $v = \frac{35}{25} = 1.4\text{ m s}^{-1}$

Since $v$ is positive, the combined trolley moves at $1.4\text{ m s}^{-1}$ in the direction of $X$ ‘s original motion.

Final answers

(a) Total momentum before $= \boxed{35}\text{ kg m s}^{-1}$
(b) Common velocity $= \boxed{1.4}\text{ m s}^{-1}$ , in the direction of the trolleys’ original motion