Simple momentum problem

  • Thread starter crybllrd
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  • #1
crybllrd
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Homework Statement


A small block m=4kg and a large block m=12kg have a head on collision on a frictionless table. Before the collision, the small block is traveling 8m/s to the right and the large block is traveling 4m/s to the left.

k2xtm0.jpg


A. If the small block bounces back at 3m/s to the left, find the final velocity of the large block.

B. Suppose instead the two blocks stick together after they collide. How many Joules of energy would be lost?

Homework Equations



[tex]p=mv[/tex]

[tex]KE=\frac{1}{2}mv^{2}[/tex]

The Attempt at a Solution



I will define to the right to be the positive x direction.
First I will calculate initial momentum for both blocks.

[tex]p_{i a}=mv[/tex]
[tex]p_{i a}=(4kg)(8m/s)[/tex]
[tex]p_{i a}=32kg m/s[/tex]

[tex]p_{i b}=mv[/tex]
[tex]p_{i b}=(12kg)(-4m/s)[/tex]
[tex]p_{i b}=-48kg m/s[/tex]

[tex]p_{i}=-16kg m/s[/tex]

Now final momentum:

[tex]p_{f a}=mv[/tex]
[tex]p_{f a}=(4kg)(-3m/s)[/tex]
[tex]p_{f a}=-12kg m/s[/tex]

[tex]p_{f b}=mv[/tex]
[tex]p_{f b}=(12kg)v[/tex]
[tex]p_{f b}=12v[/tex]

[tex]p_{f}=-12kg m/s + 12v[/tex]

If momentum is conserved, then

[tex]p_{i}=p_{f}[/tex]

[tex]-16kg m/s=-12kg m/s + 12v[/tex]

[tex]v=\frac{-1}{3}[/tex]
 

Answers and Replies

  • #2
kuruman
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So far you are correct. Is there a question you wish to ask?
 
  • #3
crybllrd
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Oh, great. I had a feeling I was on the wrong track.

For part B, I need to find the energy. It is all Kinetic.


[tex]KE_{i}=\frac{1}{2}mv_{a}^{2}+\frac{1}{2}mv_{b}^{2}[/tex]

[tex]KE_{i}=\frac{1}{2}(4)(8)^{2}+\frac{1}{2}(12)(4)^{2}[/tex]

[tex]KE_{i}=128+96[/tex]

[tex]KE_{i}=224J[/tex]

Now the final kinetic energy is where I am getting confused.

[tex]KE_{f}=\frac{1}{2}mv_{a}^{2}+\frac{1}{2}mv_{b}^{2}[/tex]

How can I find the velocity of the two blocks?
 
  • #4
crybllrd
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Actually, for part B I did it another way:

[tex]p_{i}=32+-48[/tex]

[tex]p_{i}=-16[/tex]

[tex]p_{f}=16v[/tex]


[tex]p_{i}=p_{f}[/tex]

[tex]-16=16v[/tex]

[tex]v=-1m/s, or 1 m/s to the left[/tex]


[tex]KE_{i}=\frac{1}{2}mv^{2}+\frac{1}{2}mv^{2}[/tex]


[tex]KE_{i}=\frac{1}{2}(4)(8)^{2}+\frac{1}{2}(12)(4)^{2}[/tex]


[tex]KE_{i}=224J[/tex]


[tex]KE_{f}=\frac{1}{2}mv^{2}[/tex]


[tex]KE_{f}=\frac{1}{2}(16)(1)^{2}[/tex]


[tex]KE_{f}=8J[/tex]


[tex]KE_{i}-KE_{f}=E_{lost}[/tex]

[tex]KE_{i}-KE_{f}=E_{lost}[/tex]

[tex]224-8=E_{lost}[/tex]

[tex]E_{lost}=216J[/tex]
 
  • #5
kuruman
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That looks about right.
 

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