Elastic and Inelastic Collisions

  • #1
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Homework Statement


A bullet with a mass of 3.25 g is fired horizontally at two blocks resting on a smooth and frictionless table top as shown in the Figure. The bullet passes through the first 1.25 kg block, and embeds itself in a second 1.80 kg block. Speeds v1 = 2.40 m/s and v2 = 2.60 m/s, are thereby imparted on the blocks. The mass removed from the first block by the bullet can be neglected.

a) Find the speed of the bullet immediately after emerging from the first block.
b) Find the initial speed of the bullet.

http://psblnx03.bd.psu.edu/res/fsu/capalibrary/16Momentum/Graphics/prob10a.gif

The Attempt at a Solution


I don't get how to solve this problem without knowing the speed of the bullet.
 
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Answers and Replies

  • #2
Doc Al
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I don't get how to solve this problem without knowing the speed of the bullet.
You're asked to figure out the speed of the bullet. Hint: What's conserved during these collisions?
 
  • #3
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You're asked to figure out the speed of the bullet. Hint: What's conserved during these collisions?
I know it's momentum, but the equation doesn't make sense, blah I forgot to put it up there.

object 1 is block 1 and object 2 is the bullet.

m1v1 + m2v2 = m1v1 + m2v2

What would you call the initial condition? The block would be zero, but what about the bullet? The final condition is all there.

m2v2 = m1v1 + m2v2
 
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  • #4
Doc Al
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What would you call the initial condition? The block would be zero, but what about the bullet? The final condition is all there.

m2v2 = m1v1 + m2v2
Use different symbols to represent the speed of the bullet before and after the collision. (Such as v2 and v2'.) You'll use this equation later. Hint: Analyze the collision with block 2 first.
 
  • #5
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I did that before but left divided by the wrong mass (.003 kg instead of .00325 kg) which lead me to believe it was wrong. Thanks, solved.
 

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