Question

An 11.0 g bullet is moving to the right with speed 210 m/s when it hits a target and travels an additional 22.6 cm into the target. What are the magnitude (in N) and direction of the stopping force acting on the bullet? Assume the stopping force is constant.

Answer 1

Stopping force on the bullet is 1073.23 N to the left.

Step-by-step explanation:

Mass of bullet, m = 11.0 g = 0.011 kg

Initial speed of bullet, u = 210 m/s

Final speed of bullet, v = 0 m/s

Displacement of bullet, s = 22.6 cm = 0.226 m

We have equation of motion v² = u² + 2as

Substituting

0² = 210² + 2 x a x 0.226

a = -97566.37 m/s²

We have

Force, F = Mass x Acceleration

F = 0.011 x -97566.37 = -1073.23 N

Negative means opposite to the direction of motion of bullet.

Stopping force on the bullet is 1073.23 N to the left.