Question

A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before coming to rest. Assume that the acceleration of the bullet is constant. What force is exerted by the wood on the bullet

Answer 1

Answer:

The force exerted by the wood on the bullet is 399.01 N

Explanation:

Given;

mass of bullet, m = 0.0021 kg

initial velocity of the bullet, u = 497 m/s

final velocity of the bullet, v = 0

distance traveled by the bullet, S = 0.65 m

Determine the acceleration of the bullet which is the deceleration.

Apply kinematic equation;

V² = U² + 2aS

0 = 497² - (2 x 0.65)a

0 = 247009 - 1.3a

1.3a = 247009

a = 247009 / 1.3

a = 190006.92 m/s²

Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;

F = ma

F = 0.0021 x 190006.92

F = 399.01 N

Therefore, the force exerted by the wood on the bullet is 399.01 N