Matt's Momentum After Jumping from Skateboard

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The discussion centers on a physics problem involving conservation of momentum as Matt jumps off his skateboard. Matt is initially gliding at 4.00 m/s, and upon jumping backward, he kicks the skateboard forward at 8.00 m/s. The relevant equation for momentum conservation is p_final = p_initial, which leads to calculating the final velocities of both Matt and the skateboard. The initial momentum is calculated using their combined masses and velocities, resulting in a final velocity of approximately 4.49 m/s for the system. The conversation emphasizes the application of momentum conservation principles to solve the problem effectively.
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Matt is gliding on his skateboard at 4.00 m/s. He suddenly jumps backward off the skateboard, kicking the skateboard forward at 8.00 m/s. Matt's mass is 50 kg and the skateboard's mass is 7 kg. How fast is Matt going as his feet hit the ground?

I have no idea where to begin. Should I first find the change in momentum?
 
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All you have to do is use conservation of momentum:

p_{final} = p_{initial}
 
Is this right

BLaH! said:
All you have to do is use conservation of momentum:

p_{final} = p_{initial}

v_f(50kg + 7kg) = 50kg*4m/s + 7kg*4m/s
v_f = 4.49m/s ?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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