Question

A bullet of mass 0.04 kg moving with a speed of 90 m s⁻¹ enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?

Answer 1

The retardation 'a' of the bullet (assume constant) is given by

$a = \frac{-u^2}{2s} = \frac{-90 \times 90}{2 \times 0.6} \text{ m s}^{-2} = -6750\text{ m s}^{-2}$

The retarding force, by the Second Law of motion, is

     $= 0.04 \text{ kg} \times 6750 \text{ m s}^{-2} = 270 \text{ N}$

The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force.