• Support PF! Buy your school textbooks, materials and every day products Here!

Momentum conservation (ballistic pendulum)

  • #1
5
0

Homework Statement


A .01kg bullet is fired into a 1.2kg block hanging from a 1m wire. The bullet exits the block with a speed of 200m/s and the block swings to a height of .2 meters. What is the original velocity of the bullet? What percentage of the original energy of the bullet is no longer in mechanical forms of energy?

Homework Equations


I know that initial momentum = final momentum but I can't seem to find the velocity of the block after the collision.

The Attempt at a Solution


I had this on a test and tried energy conservation. Obviously this isn't correct but I cant find the momentum of the block. I've been working this for about 5 hours now so i'm clearly not getting something so a step by step explanation would be great.
[/B]
GPE of block = (mgh) = (1.2 x 9.8 x 0.2m) = 2.353 Joules. This has resulted from KE of the bullet having swung the block.
The bullet exited the block. Its KE after exit = 1/2 (m*v^2) = 1/2 (0.01*200^2) = 200 Joules.
The total energy has come from the bullet, = (200 + 2.353) = 202.353 Joules.
Original bullet V = sqrt.(2KE/m) = sqrt.(404.706/0.01) = 201.2 m/sec.
 

Answers and Replies

  • #2
Doc Al
Mentor
44,942
1,205
GPE of block = (mgh) = (1.2 x 9.8 x 0.2m) = 2.353 Joules.
That's the energy of the block after the bullet has passed through it. Use this to determine the speed of the block.
 
  • #3
5
0
Ok, so this is what I ended up with:

mgh = 1/2mv2
v=sqrt(2gh) = 1.98m/s

Pi = Pf
mpvpi = mbvb + mpvpf
vpi = (mbvb + mpvpf) / mp
vpi = 437.6 m/s

Thanks for the help!
 

Related Threads on Momentum conservation (ballistic pendulum)

Replies
3
Views
4K
Replies
1
Views
1K
Replies
3
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
7
Views
3K
Replies
2
Views
7K
  • Last Post
Replies
1
Views
4K
Replies
3
Views
3K
  • Last Post
Replies
2
Views
1K
Top