Question

A bullet of mass 10 g travelling horizontally with a velocity of 150 ms^-1 strikes a stationery wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer 1

Mass of bullet m = \(\frac{10}{1000}\) = kg
= 0.001 Kg.
Initial velocity, u = 150ms^-1
Final velocity v = 0 (since the bullet comes to rest)
Time t = 0.03 s.
From equation of motion
v = u + at
0 = 150 + a × 0.03
∴ a = \(\frac{-150}{0.03}\)
= -5000 ms^-2
Magnitude of the force applied by the bullet on the block F = ma
= 0.01 × -5000 = -50N.