Question

A bullet with mass 0.013Kg travels at 550 m/s. It punches though a block of wood in 0.004 s and emerges going a mere 100 m/s. A.)What is the change in momentum of the bullet? B.) What impulse does the wood exert on the bullet? C.) What's the average frictional force exerted on the bullet?

Answer 1

A) The change in momentum of the bullet is given by the difference between the initial momentum and the final momentum. The initial momentum is p1 = m1v1 = 0.013 kg x 550 m/s = 7.15 kg m/s. The final momentum is p2 = m1v2 = 0.013 kg x 100 m/s = 1.3 kg m/s. Therefore, the change in momentum is Δp = p2 - p1 = -5.85 kg m/s.

B) The impulse exerted on the bullet by the wood is equal to the change in momentum of the bullet. Therefore, the impulse is also -5.85 kg m/s.

C) The average frictional force exerted on the bullet can be found using the impulse-momentum theorem, which states that the impulse is equal to the average force multiplied by the time interval over which it acts. Therefore, the average force is F = Δp / Δt, where Δt is the time interval over which the force acts. In this case, Δt is given as 0.004 s. Therefore, the average frictional force is F = (-5.85 kg m/s) / (0.004 s) = -1462.5 N. Since the force is negative, we can conclude that the frictional force acts in the opposite direction to the motion of the bullet.