A 15 g bullet strikes and embeds in a 2.1 kg block suspended at the end of a 1.2 m string. After the collision the string rises to a maximum angle of 22 degrees to the vertical.

Question

what is the speed of bullet?

what I did
MASS OF BULLET = 0.015 KG

i found the height of when it reached the max angle = 0.0873 m

i calculated the potential energy of when it reached the max angle = mgh= 1.811 j

I don't know what to do next?

Answer 1

This is a perfectly inelastic collision, so we know that energy is conserved. (Good job identifying the problem!)

Assuming you choose your "0" PE point to be the bullet's initial height:
E(i) = E(f)
.5m(bullet)v(bullet)^2 = m(tot)gh(system), where h is the height above the bullet's initial line of travel.

You're given the masses, so do some algebra from there to solve for v.

Answer 2

In an elastic collision energy is conserved.
NOT if it is inelastic.
What is conserved in both cases is momentum.
The total kinetic energy at the bottom = (1/2) m v^2
where m = 2.1 + .015 = 2.12 kg
so your PE at top = (1/2)(2.12)v^2
solver for that v at the bottom.
Then use conservation of momentum
.015 u + 2.1 *0 = 2.12 v
solve for u

Answer 3

Thank you Damon got the the right answer.Alex you are still getting participation marks hehe