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.

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?

3 answers

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