WAEC 2013 · Paper 2 · Q18✱✱

(a)
A bullet of mass $0.084\text{ kg}$ is fired horizontally into a stationary block of mass $20\text{ kg}$ on a smooth horizontal floor. If they both move with a velocity of $0.24\text{ m s}^{-1}$ after impact, calculate, correct to two decimal places, the initial velocity of the bullet.
(b)
The distance, $s$ , travelled by a body at any time $t$ seconds is given by $s = \frac23t^3 - \frac72t^2 + 5t$ . Calculate, correct to three significant figures, the distance between the points when the body is momentarily at rest.

Worked solution (try it first)

Differentiate to get the velocity: $v = \dfrac{ds}{dt} = 2t^2 - 7t + 5$ .
The body is momentarily at rest when $v = 0$ : $(2t - 5)(t - 1) = 0$ , so $t = 1$ or $t = 2.5$ .
At $t = 1$ : $s = \frac23 - \frac72 + 5$
$= \frac{13}{6}\text{ m}$ .
At $t = 2.5$ : $s = \frac{125}{12} - \frac{175}{8} + \frac{25}{2}$
$= \frac{25}{24}\text{ m}$ .
Subtract: $\frac{13}{6} - \frac{25}{24} = \frac{27}{24}$
$= 1.125$ .
The distance between the two points is $1.13\text{ m}$ to 3 significant figures.