Conservation of Momentum

A wagon with two boxes of gold having a total mass of 300kg is cut loose from horses when it is at rest by a robber. The wagon is at rest 50m up a 6.0 degree slope. The outlaw plans to have the wagon roll down the slope and across a level plane and fall into a canyon where his amigos are waiting. But in a tree 40m from the canyon edge two heros wait. One has mass 75.0kg the other has mass 60.0kg. They drop vertically into the wagon as it passes beneath them.

       *      *        50m                                                   ****
   h  *            *                                                              | |  <---tree
       *                   *                                                       | |
       *           6 degrees *                                                | |
                                                                                      | ------------- 40m -----------------------|

Hope that helps you understand.

a) if they require five seconds to grab the gold and jump out will they make it before the wagon goes over.?

h = 50 sin 6 = 5.23m
v = sqrt(2gh)
v = 10.1m/s

m_c * v = m_cf * v_f

v_f = m_c * v/ m_cf

v_f = 6.97m/s

6.97m/s * 5s = 34.8m, so it looks like they can make it  

b) when they drop into the wagon is the kinetic energy of the system (wagon plus heros) conserved?

    m_i * v_i^2 = m_f * v_f^2

        they are not equal so no.

c) if kinetic energy is not conserved does it increase, decrease and by how much?

The change in kinetic energy is a decrease of 4800 J i get closer to 10000 J so I am doing something wrong.
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Did you forget to multiply mv^2 by 1/2?
BrianGEFF719Author Commented:
>>Did you forget to multiply mv^2 by 1/2?

Does that matter if we are comparing two kinetic energies

K1 = K2
1/2m_1v_1^2 = 1/2m_2v_2^2

I just canceled the 1/2 from both sides.
canceling the 1/2 does not affect the result of =
but it can affect the magnitude of the result of -

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