Kinetic energy lost during a collision?

AI Thread Summary
In a completely inelastic collision involving a 1 kg object traveling at 1 m/s and a 2 kg object at rest, the final velocity after impact is calculated to be 1/3 m/s. The kinetic energy before the collision is 0.5 J, while the kinetic energy after the collision is 0 J, indicating a total kinetic energy loss of 0.5 J. There was confusion regarding the calculation of kinetic energy lost, with an incorrect interpretation suggesting energy was gained. It was advised to ensure consistent variable naming between momentum and kinetic energy equations. The discussion emphasizes the importance of clarity in calculations and variable usage in physics problems.
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Homework Statement


A 1kg object traveling at 1 m/s collides head-on with a 2 kg object initially at rest. If the collision is completely inelastic, determine a) the velocity of the objects after impact and b) how much kinetic energy was lost during the collision.

Homework Equations


Part a) m1v1 + m2v2 = (m1+m2)v'
Part b) ke before =1/2m1v1^2, ke after= 1/2 (m1 + m2)v2^2

The Attempt at a Solution


Part a)
m1v1 + m2v2 = (m1+m2)v'
1 + 0 = 3(v')
v' = 1/3

Part b)
KE before: 1/2 m1v1^2
.5 (1)(1^2)
=.5
KE after: 1/2 (m1+m2)v2^2
.5(3)(0) = 0?

KE Lost:
(kinetic before - kinetic after)/kinetic before
(.5-0)/.5 = 1??
 
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nips said:
did i do this right?

You have the system gaining energy! That can't be right. Check your figures.

Plus, you found the fractional loss, and you were asked for just the loss.
 
nips said:

Homework Statement


A 1kg object traveling at 1 m/s collides head-on with a 2 kg object initially at rest. If the collision is completely inelastic, determine a) the velocity of the objects after impact and b) how much kinetic energy was lost during the collision.

Homework Equations


Part a) m1v1 + m2v2 = (m1+m2)v'
Part b) ke before =1/2m1v1^2, ke after= 1/2 (m1 + m2)v2^2

Explain what are v1, v2 and v'.
 
The velocities you're using for kinetic energy are wrong; try to use the same variable names as with your momentum equation.
 
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