Question

A 22-g bullet traveling 250 m/s penetrates a 1.7 kg block of wood and emerges going 120 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

Answer 1

The velocity of the block after the bullet emerges can be determined using the conservation of momentum, which states that the total momentum before the event is equal to the total momentum after the event.

Step-by-step explanation:

The question involves a physics concept known as the conservation of momentum. Since momentum is conserved, the momentum before the collision is equal to the momentum after the collision. We can use this principle to find the speed of the wooden block after the bullet passes through it.

Initial momentum of system = (mass of bullet * velocity of bullet before) + (mass of block * velocity of block before)

Final momentum of system = (mass of bullet * velocity of bullet after) + (mass of block * velocity of block after)

Given:

• Mass of bullet (m1) = 0.022 kg
• Velocity of bullet before (vb1) = 250 m/s
• Velocity of bullet after (vb2) = 120 m/s
• Mass of block (m2) = 1.7 kg
• Velocity of block before (v2) = 0 m/s (stationary)

Let's denote the unknown velocity of the block after the collision as v2f.

So, (m1 * vb1) + (m2 * v2) = (m1 * vb2) + (m2 * v2f)

(0.022 kg * 250 m/s) + (1.7 kg * 0 m/s) = (0.022 kg * 120 m/s) + (1.7 kg * v2f)

Solve for v2f to find the velocity of the block after the bullet emerges.