Perfectly Inelastic Ball Collision: Final Velocities of Both Balls

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In a perfectly inelastic collision between Ball 1 (100 g at 10 m/s) and Ball 2 (300 g at rest), momentum conservation leads to both balls moving together at a final velocity of 2.5 m/s. Although energy conservation was attempted, it does not apply in inelastic collisions, which is why a calculation yielding 5 m/s is incorrect. The correct approach involves using the momentum formula, as kinetic energy is not conserved during such collisions. The discussion emphasizes the importance of understanding the distinction between momentum and energy conservation in collision scenarios. This highlights the necessity of using appropriate principles for accurate results in physics problems.
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Ball 1, with a mass of 100 g and traveling at 10 m/s, collides head-on with Ball 2, which has a mass of 300 g and is initially at rest. What are the final velocities of each ball if the collision is perfectly inelastic?


We're suppose to be talking about Energy, but I figured I could use:

mv+mv (initial)=mv (final)



I got 2.5 m/s for both balls, because if the objects are inelastic they will be traveling at the same speed after collision and momentum is conserved.
Is this right?
 
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That would be correct.
 
but now I just tried to use conservation of Energy and got a different answer. I tried
1/2 mv^2 + 1/2 mv^2 = 1/2mv^2 (final for system)

I got both balls traveling at 5 m/s ??
 
PnotConserved said:
but now I just tried to use conservation of Energy and got a different answer. I tried
1/2 mv^2 + 1/2 mv^2 = 1/2mv^2 (final for system)

I got both balls traveling at 5 m/s ??

Kinetic energy is not conserved in an inelastic collision.
 
I tried the V' substitution before but was wrong.

The expression that the book describes it to be:
---> the square root of [(m+M)kd^2] / m^2
 
You might want to find out the kinetic energy of the balls before the collision, because you are talking about a change in energy.
 
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