A bullet of mass 50 g moving with a speed of 100 m s⁻¹ enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).

Given,

Mass of bullet, $m = 50\text{ g} = 0.05\text{ kg}$

Initial velocity, $u = 100\text{ ms}^{-1}$

Final velocity, $v = 0\text{ ms}^{-1}$

Distance travelled inside the block

$$s = 50\text{ cm} = 0.5\text{ m}$$

Using the equation

$$v^2 = u^2 + 2as$$

$$0^2 = (100)^2 + 2 \times a \times 0.5$$

$$0 = 10000 + a$$

$$a = -10000\text{ ms}^{-2}$$

Now, using Newton's second law,

$$F = ma$$

$$F = 0.05 \times (-10000)$$

$$F = -500\text{ N}$$

The negative sign shows that the force acts opposite to the direction of motion.

Therefore, the stopping force acting on the bullet is 500 N.