Question

A .223 rifle bullet, traveling at 370.m/s, hits a block of pine wood, and goes in, to a depth of 0.130m. The mass of the bullet is 1.75g (mass = 0.00175kg). Assume a constant slowing force. (A) What is the acceleration that the bullet experiences as it comes to a stop? (B) How much time is required for the bullet to stop? (C) What force, in Newtons, does the wood exert on the bullet?

Answer 1

Answer:

Explanation:

A)

v² = u² + 2as

a = (v² - u²) / 2s

a = (0² - 370²) / (2(0.130))

a = -526,538 m/s²

B)

t = v/a

t = 370 / 526538

t = 0.0007027... s or 0.7 ms

C)

F = ma

F = 0.00175(526,538) = 921.442307... = 921 N