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?

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

To find the speed of the bullet, we can use the principle of conservation of mechanical energy. The potential energy at the maximum angle is equal to the initial kinetic energy of the bullet.

1. Calculate the potential energy at the maximum angle:
The potential energy is given by the formula: PE = mgh
where m is the mass of the block (2.1 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (0.0873 m).
Substituting the values into the formula, we get:
PE = (2.1 kg) * (9.8 m/s^2) * (0.0873 m) = 1.811 J

2. Use the principle of conservation of mechanical energy:
The initial kinetic energy of the bullet is equal to the potential energy when the block reaches the maximum angle.
The kinetic energy is given by the formula: KE = 0.5 * m * v^2
where m is the mass of the bullet (0.015 kg) and v is the velocity we want to find.
Equating the potential energy to the initial kinetic energy, we have:
1.811 J = 0.5 * (0.015 kg) * v^2
Simplifying the equation, we get:
v^2 = (2 * 1.811 J) / (0.015 kg)
v^2 = 241.47 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 15.53 m/s

Therefore, the speed of the bullet is approximately 15.53 m/s.

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.