Inelastic Collision Question

  • Thread starter wessleym
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  • #1
wessleym
2
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I already posted this in the General room, so sorry. It should have been in here in the first place. 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
jamesrc
Science Advisor
Gold Member
477
1
Use the conservation of momentum to find the initial velocity of the bullet/box system in terms of the initial bullet velocity. (Just assume a purely horizontal bullet velocity). You're working under the assumption that the bullet embeds itself and gets stuck in the block (an inelastic collision), so you should set the initial momentum of the bullet equal to the post collision velocity of the bullet/box.

You'll need another equation to solve the problem, so use what you know about consevation of energy. The system starts with some initial kinetic energy (a function of the velocity found above) and potential energy (though you should set the potential energy equal to 0 at the initial height). The problem has given you information about the point where the kinetic energy is 0 and all the energy is gravitational potential energy. Set the total energy at those two positions equal and you'll have another equation. Solve the system of equations and you're done.
 
  • #3
PrudensOptimus
641
0
mBox = 40kg
mBullete = 0.01 kg

Δx = 0.02m = Δy


WNC = ΔKE + ΔPE --> Work = 0 in collisions

0.5mBullete(v^2 - v0^2) = -(-mgΔx)

v = 28.3m/s


mBulletev = (mbullete + mbox)v'

v' = 0.0071 m/s --- answer. after it hits the box the box stopps it basically.
 

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