Elastic Collision and max height on an Incline

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The discussion centers on calculating the maximum height a block of mass m1 (8.00 kg) reaches after an elastic collision with another block (m2 = 14.0 kg) on a frictionless incline. The initial velocity of m1 before the collision is calculated using gravitational potential energy converted to kinetic energy, yielding approximately 9.9 m/s. After the collision, the final velocity of m1 is determined to be -2.7 m/s. The height reached post-collision is calculated using the work-energy principle, but the initial answer is deemed incorrect. A suggestion is made to directly convert kinetic energy to potential energy for a more accurate result, indicating that this method may yield a final answer that is double the previous calculation.
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Consider a frictionless track ABC as shown in Figure P8.23. A block of mass m1 = 8.00 kg is released from A. It makes a head-on elastic collision at B with a block having a mass of m2 = 14.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.

p8-23.gif


To find the initial velocity of mass 1 right before the collision:

Ug = KE
mgh = .5mv2
gh = .5v2
(9.8)(5) = .5v2
v = 9.9

Find the final velocity of mass 1 right after the collision:

vf1 = (m1-m2/m1+m2)vi1
vf1 = (8-14/8+14)(9.9)
vf1 = -2.7

Find the height at which the mass will reach with the initial velocity 2.7:

Wnet = \DeltaKE
KE - Ug = \DeltaKE
.5mv2 - mgh = .5mvf2 - .5mvi2
.5vi2- gh = .5vf2 - .5vi2
.5(2.7) - (9.8)h = .5(0)2 - .5(2.7)2
h = .7438

Final answer is incorrect. Any ideas?
 
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If you know the initial velocity, rather than evaluating the hight using the net work, try just changing kinetic energy to potential energy in the same way that you converted them at the beginning.

In terms of total work, you will find that your final answer turns out to be double the answer using this method.

I hope this helps a little.
 
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