$A$ bullet of mass $5 \, g$ travelling at a speed of $120 \, m s^{-1}$ penetrates deeply into a fixed target and is brought to rest in $0.01 \, s$. Calculate: $(a)$ the distance of penetration in the target,$(b)$ the average force exerted on the bullet.

(N/A) Given: Mass $m = 5 \, g = 5 \times 10^{-3} \, kg$,initial velocity $u = 120 \, m s^{-1}$,final velocity $v = 0$,time $t = 0.01 \, s$.
$(a)$ To find the distance of penetration $(S)$,we first find the acceleration $(a)$ using the first equation of motion: $v = u + at$.
$0 = 120 + a \times 0.01$
$a = -120 / 0.01 = -12000 \, m s^{-2}$.
(The negative sign indicates retardation).
Now,using the second equation of motion: $S = ut + \frac{1}{2}at^2$.
$S = (120 \times 0.01) + \frac{1}{2} \times (-12000) \times (0.01)^2$
$S = 1.2 - 0.6 = 0.6 \, m$.
$(b)$ The average force $(F)$ exerted on the bullet is calculated using Newton's second law: $F = ma$.
$F = (5 \times 10^{-3} \, kg) \times (12000 \, m s^{-2})$
$F = 60 \, N$ (The force is retarding in nature).