WAEC 2022 · Paper 2 · Q15

(a)
Two particles, $P$ and $Q$ , of masses $3\text{ kg}$ and $1.5\text{ kg}$ respectively moved in opposite directions. $P$ moved with a velocity of $5\text{ m s}^{-1}$ while $Q$ moved with a velocity of $7\text{ m s}^{-1}$ . The particles collided head-on and moved in the same direction after collision. The difference in the velocities after collision is $\frac34\text{ m s}^{-1}$ , where the final velocity of $P$ ( $V_P$ ) is greater than the final velocity of $Q$ ( $V_Q$ ). Find the velocities of $P$ and $Q$ after collision.
Separate values with commas, e.g. 3, −2
(b)
A ball is thrown vertically upwards. The height, $h$ metres, after a time $t$ seconds, is given by $h = 5 + 30t - 5t^2$ . Find the: (i) velocity of the ball after 2 seconds; (ii) maximum height the ball reached.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Take $P$ 's direction as positive.
Momentum before $= 3(5) + 1.5(-7) = 15 - 10.5 = 4.5$ .
Momentum after $= 3V_P + 1.5V_Q$ .
Momentum is conserved: $3V_P + 1.5V_Q = 4.5$ , so $2V_P + V_Q = 3$ .
The velocities differ by $\frac34$ : $V_P - V_Q = \frac34$ .
Add the equations: $3V_P = \frac{15}{4}$ , so $V_P = \frac54\text{ m s}^{-1}$ and $V_Q = \frac12\text{ m s}^{-1}$ .