Question

A 5.85gram 665m/s bullet is fired at a 2.13kg block that starts at rest. The bullet passes through the block and ends with a velocity of 449m/s. What is the final velocity of the block?

Answer 1

You can't assume that energy is conserved here, as theres all kinds of frictions involved and energy dissipation on the collision.

However, since its a simple collision, you can use conservation of momentum.

This in short says that the initial sum of momentum in a system has to equal the final sum of momentum in a system.

Momentum, which we'll call p, is equal to mass times velocity.

In this case, your block is initially at rest, so it has no momentum
p_block_initial = 0
Your bullet initially has a speed of 665m/s giving it an initial momentum of
p_bullet_initial = 665 m/s * 5.85*10^-3 kg = 3.89 kg m/s
And the bullet has a final speed of 449m/s so:
p_bullet_final = 449 m/s * 5.85*10^-3 kg = 2.23 kg m/s

The initial total momentum has to equal the final total momentum. That being said, we know that:
p_initial = p_final
p_block_initial + p_bullet_initial = p_block_final + p_bullet_final
3.89 kg m/s + 0 kg m/s = 2.23 kg m/s + p_block_final

simple arithmetic gives us that
p_block_final = 1.66 kg m/s
We also know that
p_block_final = m_block * v_block_final
We know the blocks mass, and we want to know the final velocity of the block. So we just need to divide the blocks final momentum by its mass giving us roughly:
v_block_final = 1.66/2.13 = .78 m/s

Answer 2

thank you

Answer 3

i had parts of it but i was dividing instead of subtracting