Question

A bullet of mass 0.010 kg and speed of 200 m/s is brought to rest in a wooden block after penetrating a distance of 0.10 m. This process takes 0.02 second. The magnitude of the impulse delivered to the bullet by the block during this time is

Answer 1

Initial speed of the bullet = u = 200 m/s

Final speed of the bullet = v = 0 m/s

Time of travel = t = 0.02 sec.

From laws of motion, we get  

v = u + at  

200 = -0.02 a  

a = -10000 m/s^2 (This is deceleration hence negative sign)

From Newton’s law, we know that

Force = mass * acceleration  

F=ma  

(0.01)(10000) = 100N  

Impulse = Force * time = Ft = 100 * 0.02 = 2 Ns