$A$ 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 $v = 0\, m s^{-1}$,Time $t = 0.01\, s$.
$(a)$ First,calculate the acceleration $a$ using the equation $v = u + at$:
$0 = 120 + a \times 0.01$
$a = -\frac{120}{0.01} = -12000\, m s^{-2}$ (The negative sign indicates retardation).
Now,calculate the distance of penetration $S$ using the equation $S = ut + \frac{1}{2}at^2$:
$S = (120 \times 0.01) + \frac{1}{2} \times (-12000) \times (0.01)^2$
$S = 1.2 - 0.6 = 0.6\, m$.
$(b)$ Calculate the average force $F$ using Newton's second law $F = ma$:
$F = (5 \times 10^{-3}\, kg) \times (12000\, m s^{-2})$
$F = 60\, N$.