Question

15. A bullet of mass 5g travelling at a speed of 120m/s penetrates deeply into a fired target and is brought to rest in .01 s. Calculate a) the distance of penetration in the target b) average force exerted on the bullet

Answer 1

Let 's' be the distance through which the bullet penetrates.

Time taken = 0.015 s

Initial velocity = 120 m/s

Final velocity = 0

Acceleration = a

Using, v = u + at

=> 0 = 120 + 0.015a

=> a = -8000 m/s2

Now,

v2 = u2 + 2as

=> 0 = 1202 + 2(-8000)s

=> s = 0.9 m

Average force on the bullet = ma = (0.005 kg) X (-8000 m/s2) = -40 N

The negative sign implies that the force is against the direction of motion of the bullet