How Does Height Affect Momentum in a Frictionless Track Collision?

[fishert16]

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.

 

[robbondo]

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.

 

[m2003]

Your approach is correct. The first equation, known as the conservation of momentum equation, states that the total momentum before the collision is equal to the total momentum after the collision. In this case, the initial momentum (M1V1) of the first block is equal to the final momentum [(M1+M2)(Vf)] of the combined blocks.

The second equation, known as the conservation of energy equation, states that the total initial energy (kinetic and potential) is equal to the total final energy (kinetic and potential). In this case, the initial energy is the kinetic energy of the first block (1/2M1V1^2) plus its potential energy (MGH1), and the final energy is the kinetic energy of the combined blocks (1/2(M1+M2)Vf^2) plus their potential energy (MGH2).

By setting these two equations equal to each other, you can solve for Vf. Make sure to use the correct values for masses, velocities, and heights in the equations.