A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the bullet's mechanical energy? (b) what is the magnitude of the average force from the wall stopping it?

Question

Physics

Megan Wilkins

9 May, 13:57

A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the bullet's mechanical energy? (b) what is the magnitude of the average force from the wall stopping it?

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Answers (1)

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Wendy Hale · Answer 1

M = 30 g = 0.03 kg, the mass of the bullet

v = 500 m/s, the velocity of the bullet

By definition, the KE (kinetic energy) of the bullet is

KE = (1/2) * m*v²

= 0.5 * (0.03 kg) * (500 m/s) ² = 3750 J

Because the bullet comes to rest, the change in mechanical energy is 3750 J.

The work done by the wall to stop the bullet in 12 cm is

W = (1/2) * (F N) * (0.12 m) = 0.06F J

If energy losses in the form of heat or sound waves are ignored, then

W = KE.

That is,

0.06F = 3750

F = 62500 N = 62.5 kN

Answer:

(a) 3750 J

(b) 62.5 kN