Kinetic energy lost during a collision?

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SUMMARY

The discussion focuses on a completely inelastic collision involving a 1 kg object traveling at 1 m/s colliding with a 2 kg object at rest. The final velocity after impact is calculated using the momentum conservation equation, yielding a result of 1/3 m/s. The kinetic energy before the collision is 0.5 J, while the kinetic energy after the collision is 0 J, resulting in a total kinetic energy loss of 0.5 J. The calculations emphasize the importance of consistent variable naming in physics equations.

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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??
 
Last edited:
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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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