Inelastic Collision Question

  • Thread starter wessleym
  • Start date
  • #1
2
0
Here's a question for anyone out there who would really like to less than moderately challenge their physics skills:
There is a 40 kilogram box hanging from the ceiling, at rest. A bullet with a mass of 0.01 kilogram hits it, forcing it to swing back on the ropes. As it swings back, it gets 0.02 meters off of the ground. Assume the box's bottom remains parellel with the ground. How fast was the bullet moving? Thanks to anyone who can figure this one out!
 

Answers and Replies

  • #2
54
0
(Sounds like a homework question)

You can do this in two parts, moving backwards in events. First, you can use conservation of energy to determine the velocity of the box at the instant it absorbed the bullet ( mgh = (1/2)mv^2 ). Here, "m" is the mass of the box-bullet combo.

Once you have "v" of the bullet-box combo, you can use conservation of momentum to determine what the original velocity of the bullet was
( (mass bullet-box combo)(velocity) = (mass bullet)(velocity bullet) + (mass box)(velocity box) )

Of course the initial velocity of the box is zero, so it all works out nicely.


* All of this assumes an "instant" collision between the bullet and the box.
 

Related Threads on Inelastic Collision Question

  • Last Post
Replies
5
Views
2K
Replies
6
Views
668
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
4
Views
638
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
0
Views
656
  • Last Post
Replies
0
Views
851
Top