Question

A bullet of mass 10g travel­ling horizontally with a velocity of 150 ms^-1 strikes a stationary wooden block and comes to rest in 0.03 s. 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) = 10 g = 0.01 kg

Initial velocity of bullet (u) = 150 ms

Final velocity of bullet (u) = 0 [As the bullet comes to rest]

Time (t) = 0.03 s ; Acceleration (a) = ?

(1) From the first equation of motion

v = u + at = 0 = 150 + a x 0.03

⇒ a = \(\frac{-150}{0.03}\) = -5000 m/s²

(2) The distance of penetration of the bullet into the block

s = ut + \(\frac{1}{2}\)at²

= 150 x 0.03 + \(\frac{1}{2}\) x (- 5000) x (0.03)²

= 4.5 – 2.25 = 2.25m

(3) The magnitude of the force exerted by the wooden block on the bullet

F = ma = 0.01 x 5000 = 50 N

⇒ F = 50 Newton