Question

A bullet of mass 0.006kg travelling at 180m/s penetrates deeply into a target and is brought to rest at 0.01 sec. Calculate the:

a) distance of penetration of the target.
b) force exerted on the target.

Answer 1

Answer:

a) 0.9 m

b) -108 N

Explanation:

acceleration =(final velocity - initial velocity)/ time

=

$\frac{0 - 180}{0.01}$

= -18000 m/s²

v² = u² + 2aS

0 = 180² + 2(-18000)S

36000S = 32400

S =

$\frac{32400}{36000}$

= 0.9 m

Force = ma = 0.006 x -18000

= -108 N ( since force is applied opposite to the direction of displacement of bullet)