Physics questions

Question

A bullet of mass 50.0 g strikes a wooden block of mass 5.0 kg and becomes embedded in the block. The block and bullet then flies off at 10.0 m/s. What was the original velocity of the bullet in m/s?

A basketball with a mass of 0.50 kg is thrown horizontally against a wall with a velocity of 15 m/s. If the ball rebounds with a velocity of 13 m/s, what is the impulse of the collision?

Answer 1

A. Use conservation of momentum.

Calculate the momentum at time 1 (which was prior to the bullet striking the block). The only thing with momentum is the bullet and its momentum is its mass times its velocity. 50.0 g = 0.0500 kg and we do not know the velocity so the momentum at time 1 (P₁) can be written as:

P₁ = 0.0500v

Calculate the momentum at time 2 (which was when the bullet and block are moving at 10.0 m/s). The momentum is the mass times its velocity or 5.0500 kg times 10.0 m/s. So the momentum at time 2 (P₂) can be written as:

P₂ = (5.0500)(10.0) = 50.5 kg•m/s

Momentum is conserved so P₁ = P₂ meaning 0.0500v = 50.5 (divide both sides by 0.0500 to solve).

B. Use the Momentum-Impulse Theorem meaning the change in momentum equals the impulse. Remember the momentum is a vector having the same direction (or sign) as velocity.

The momentum at time 1, P₁, (before the collision) is mv = (0.50)(15) = 7.5 kg•m/s.

The momentum at time 2, P₂, (after the collision) is (0.50)(-13) because the velocity is in the opposite direction. or -6.5 kg•m/s.

The change in momentum is P₂ - P₁ = -6.5 - 7.5 = -14 kg•m/s

Impulse equals the change in momentum so the impulse is -14 kg•m/s = -14 N•s. The fact that it is negative means the applied force of the impulse occurred in the negative direction (in the direction the ball was traveling at time 2).