An 11.0-g bullet traveling at 700 m/s strikes a block of wood fixed in place, penetrates, and travels a distance of 6.0 cm in the block before it stops. The average force exerted by the wooden block on the bullet was A) 6.4 * 10¹ N. B) 6.4 * 10² N. C) 4.5* 10⁴ N. D) 4.5 * 10⁵ N. E) 4.5 * 10⁶ N.

Question

Quizplus · Accepted Answer

The average force exerted by the wooden block on the bullet is equal to the change in momentum of the bullet divided by the time of impact. The change in momentum of the bullet is equal to its final momentum minus its initial momentum. The final momentum of the bullet is zero, since it stops inside the block. The initial momentum of the bullet is equal to its mass times its velocity. The time of impact is equal to the distance traveled by the bullet in the block divided by its velocity. Substituting these values into the equation for average force, we get:$$F = \frac{mv_f - mv_i}{t} = \frac{0 - (11.0 	imes 10^{-3} 	ext{ kg})(700 	ext{ m/s})}{6.0 	imes 10^{-2} 	ext{ m}} = 4.5 	imes 10^4 	ext{ N}$$

An 110-G Bullet Traveling at 700 M/s Strikes a Block of of Wood