Momentum and Kinetic Energy

  • #1
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Homework Statement


On a frictionless, horizontal air table, puck A (with mass 0.250 kg ) is moving toward puck B (with mass 0.360 kg ), that is initially at rest. After the collision, puck A has a velocity of 0.125 m/s to the left, and puck B has velocity 0.655 m/s to the right.

I already solved for the puck's initial velocity, which is 0.818m/s. Now I have to find the change in the total kinetic energy of the system that occurs during the collision.

Homework Equations


KE=(1/2)mv^2

The Attempt at a Solution


KE(initial)=(1/2)(0.250kg)(0.818m/s)^2=0.0805J
KE(final)=(1/2)(0.250kg)(0.125m/s)^2+(1/2)(0.360kg)(0.655m/s)^2=0.07917
ΔK=0.0805J-0.07917J=0.00487J.
I tried this both negative and positive, but both are wrong. I don't know what I did wrong.
 

Answers and Replies

  • #2
Orodruin
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I already solved for the puck's initial velocity, which is 0.818m/s.
Please show your work.
 
  • #3
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Please show your work.
mv=m1v1+m2v2
v=(-(0.125m/s)(0.250kg)+(0.655m/s)(0.360kg))/0.250kg=0.818m/s. This is definitely right because I already submitted the answer and got it correct.

Should the velocity of Puck A be negative when solving for kinetic energy?
 
  • #4
Orodruin
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Is whatever program you are feeding the answers into sensitive to the number of significant digits?
 
  • #5
Orodruin
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KE(initial)=(1/2)(0.250kg)(0.818m/s)^2=0.0805J
This is not numerically correct.
 

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