How Does Height Affect Momentum in a Frictionless Track Collision?

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In a frictionless track collision scenario, a block of mass 8.00 kg is released from a height of 5.00 m and collides with a stationary block of mass 10.0 kg. The discussion focuses on determining the final velocity of the combined blocks after the collision. Participants suggest using two main equations: one for momentum conservation and another for energy conservation. The consensus leans towards using the momentum equation for simplicity, while also noting that kinematics can provide the initial velocity needed for calculations. Ultimately, the problem can be approached effectively through either momentum or kinetic energy methods.
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Consider a frictionless track as shown in the diagram below. A block of mass m1=8.00kg is released from point A at a height of 5.00m. It collides with a block of mass m2=10.0 kg, initially at rest, at point B. The two blocks stick together. Determine the final velocity of the blocks


Relevant equations

M1V1+M2V2 = (M1+M2)(Vi)
[1/2(MVf^2)-1/2(MVi^2) + (MGHf - MGHi)

I haven't attempted this problem, I wanted to make sure that it was the second equation and If I am right I think I should just plug into the second equation and solve for Vf.
 
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I think it would be easier to use the first equation. Just use kinematics to determine the initial velocity of the first mass, then you have v1 for the first mass and m1 and m2, just solve for vi.

P.S. the kinematics equation your looking for is vf=v0+ad, accerleration is just gravity, and the distance is 5m.

Also, it's possible to solve this equation using kinetic energy, but the first equation is much easier in my opinion.
 
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