A 10.0-g bullet traveling horizontally at 755 m/s strikes a stationary target and stops after penetrating 14.5 cm into the target. What is the average force of the target on the bullet? A)1.97 × 10⁴ N B)2.07 × 10⁵ N C)6.26 × 10³ N D)3.13 × 10⁴ N E)3.93 × 10⁴ N

Question

Quizplus · Accepted Answer

The average force of the target on the bullet is equal to the change in momentum of the bullet divided by the time of impact. The change in momentum is equal to the mass of the bullet times its final velocity minus its initial velocity. The time of impact is equal to the distance traveled by the bullet divided by its average velocity. The average velocity is equal to the sum of the initial and final velocities divided by 2.Plugging in the given values, we get:$$	ext{Average force}=\frac{m(v_f-v_i)}{t}=\frac{(0.0100	ext{ kg})(-755	ext{ m/s}-0	ext{ m/s})}{(14.5	imes10^{-2}	ext{ m})/(0.5(755	ext{ m/s}+0	ext{ m/s})}$$$$=\boxed{2.07	imes10^5	ext{ N}}$$

A 100-G Bullet Traveling Horizontally at 755 M/s Strikes a Stationary