A bullet of mass 0.04 kg moving with a speed of 90 ms -1 enters a heavy wooden block and is stopped after a distance of 60 cm. The average resistive force exerted by the block on the bullet is

Question

A bullet of mass 0.04 kg moving with a speed of 90 ms -1 enters a heavy wooden block and is stopped after a distance of 60 cm. The average resistive force exerted by the block on the bullet is (A) 180 N (B) 220 N (C) 270 N (D) 320 N

Tardigrade Admin · Accepted Answer

a=(-u2/2 s)=(-902 ms -2/2(0.6))=-6750 ms -2 F=m a=(0.04 kg )(-6750 ms 2)=-270 N -ve sign indicates retarding force.

Q. A bullet of mass $0.04 k g$ moving with a speed of $90 m s^{- 1}$ enters a heavy wooden block and is stopped after a distance of $60 c m$ . The average resistive force exerted by the block on the bullet is

180 N

220 N

270 N

320 N

$a = \frac{- u ^{2}}{2 s} = \frac{- 9 0 ^{2} m s ^{- 2}}{2 ( 0.6 )} = - 6750 m s^{- 2}$
$F = ma = (0.04 k g) (- 6750 m s^{2}) = - 270 N$
$- v e$ sign indicates retarding force.

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