A bullet of mass 0.04 kg moving with a speed of 90 `mcdot 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?

Explanation

Let mass of the bullet, m = 0.04 kg
Initial velocity of the bullet, v = 90 m`cdot s^(-1)`
final velocity of the bullet, v = 0
Distance traversed of the bullet, s = 60 cm = 0.6 m
If a be the retardation of the bullet,
`v^(2) = u^(2) - 2as " " or, 0 = (90)^(2) - 2a xx 0.6`
`therefore" " a = ((90)^(2))/(2 xx 0.6) = 6750 m cdot s^(-2)`
So the actual resistive force, F = ma = 0.04 `xx 6750 = 270`N
The actual resistive force, and therefore, the retardation of the bullet may not be uniform . The answer therefore, only indicates the average resistive force.