Question

A bullet of mass 10 gram travelling horizontally with a velocity of 150 metre per second strikes a stationary wooden block and comes to rest in 0.03 second. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted buy the wooden block on the bullet.

Answer 1

we know that

v = u + at

or deceleration

a = (v - u) / t

here,

u = 150 m/s

v = 0

t = 0.03 s

so,

a = (0 - 150) / 0.03

thus,

a = -5000 m/s^2

now, the distance penetrated will be

s = (v^2 - u^2) / 2a

or

s = (0 - 150^2) / (2 x -5000) = 22500 / 10000

thus,

s = 2.25m

..

the retarding force exerted will be

F = ma

here,

m = 10g = 0.01kg

so,

F = 0.01 x 5000

thus,

F = 50N