Physics, asked by riya48086, 1 year ago

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?

Answers

Answered by AbhijithPrakash
78

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.

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