How Does Height Affect Momentum in a Frictionless Track Collision?

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SUMMARY

The discussion focuses on calculating the final velocity of two colliding blocks on a frictionless track, specifically a block of mass m1=8.00kg released from a height of 5.00m colliding with a stationary block of mass m2=10.0kg. The relevant equations include momentum conservation (M1V1 + M2V2 = (M1 + M2)(Vi)) and energy conservation (1/2(MVf^2) - 1/2(MVi^2) + (MGHf - MGHi)). Participants agree that using the momentum equation is more straightforward than the kinetic energy approach, and kinematics can be applied to find the initial velocity of m1 before the collision.

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fishert16
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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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