Question

Using second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sand-bag at a speed of $10^3$ m $s^{-1}$ and gets embedded after travelling 5 cm. Calculate:

(i) the resistive force exerted by the sand on the bullet.

(ii) the time taken by the bullet to come to rest.

• A. (i) $10^6$ N, (ii) $10^{-4}$ s
• B. (i) $10^5$ N, (ii) $10^{-4}$ s
• C. (i) $10^6$ N, (ii) $10^{-5}$ s
• D. (i) $10^5$ N, (ii) $10^{-5}$ s

Answer 1

Answer: (A)

(i) the resistive force exerted by the sand on the bullet is $10^6$ N. (ii) the time taken by the bullet to come to rest is $10^{-4}$ s.

Answer 2

1. Derivation of $\vec{F} = m\vec{a}$: Newton's second law: $\vec{F} = \frac{d\vec{p}}{dt}$. Momentum: $\vec{p} = m\vec{v}$. For constant mass: $\vec{F} = m \frac{d\vec{v}}{dt} = m\vec{a}$.

2. Given Data: Mass, $m = 10 \text{ g} = 0.01 \text{ kg}$ Initial velocity, $u = 10^3 \text{ m s}^{-1}$ Distance, $s = 5 \text{ cm} = 0.05 \text{ m}$ Final velocity, $v = 0 \text{ m s}^{-1}$

3. Calculate Acceleration ($a$): Using the kinematic equation $v^2 = u^2 + 2as$: $0^2 = (10^3)^2 + 2 \times a \times 0.05$ $0 = 10^6 + 0.1a$ $0.1a = -10^6$ $a = -10^7 \text{ m s}^{-2}$

4. Calculate Resistive Force ($F$): Using Newton's second law $F = ma$: $F = (0.01 \text{ kg}) \times (-10^7 \text{ m s}^{-2})$ $F = -10^6 \text{ N}$ The magnitude of the resistive force is $10^6 \text{ N}$.

5. Calculate Time ($t$): Using the kinematic equation $v = u + at$: $0 = 10^3 + (-10^7) \times t$ $10^7 t = 10^3$ $t = \frac{10^3}{10^7} = 10^{-4} \text{ s}$