Bullet fired into a block that and swings like a pendulum

[kerbyjonsonjr]

Homework Statement

A 2.5 kg wood block hangs from the bottom of a 1.0 kg, 1.2 m long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 12 g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 33 angle.

Homework Equations

KE=1/2mv²
gravitational potential=mgh

The Attempt at a Solution

Since the kinetic energy of the bullet all ends up as potential energy at the top of the swing I set kinetic energy equal to the sum of all the gravitational potential energies. The height that the rod goes would be from its center of mass so it would be cos33=h/.6 which equals .5m. Then the height of the bullet and the block would be cos33=h/1.2 which equals 1 m. Then I used the equation 1/2mv₂= mgh + (m+m)gh and that plugged in with numbers looks like 1/2(.012)v²=1(9.81)(.5) +(2.5 +.012)(9.81)(1) and then I solved for v and got v= 70.2 m/s

I do not know where I went wrong. If anybody could help me that would be great and I would really appreciate it.

 

[tiny-tim]

hi kerbyjonsonjr!

this is a perfectly inelastic collision … energy is not conserved, and you have to use conservation of momentum or angular momentum (momentum and angular momentum are always conserved in collisions) …

only after you find the initial speed can you then use conservation of energy

 

[Doc Al]

Since the kinetic energy of the bullet all ends up as potential energy at the top of the swing

Ah, but it doesn't. The collision between bullet and block is perfectly inelastic. Mechanical energy is not conserved during the collision, but something else is. What?

You'll need to treat the behavior of bullet and pendulum in two phases:
(1) The collision itself.
(2) The rising of the pendulum after the collision.

Different things are conserved during each phase.

 

[44r0n0wnz]

Al is right. In the first part, energy is not conserved because the collision is inelastic. After the collision, the block moves right after the bullet stops, making this part of it elastic. Solve by using conservation of momentum for the first part and conservation of energy for the second part.