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A 40 kg boy is riding a 2.5 kg skate board at a velocity of 5.3 m/s across a level side walk, jumps forward to leap over a wall. Just after leaving contact with board, the boys velocity relative to the side walk is 6 m/s, 9.5 degree above horizontal. Ignore any friction between the skateboard and the side walk. What is the skateboard's velocity relative to the side walk at this instant?

I solved this and got the correct answer -4.62 m/s((Mboy+Mskate)5.3m/s = (Mboy)(Vboy-x comp) + (Mskate)(Vskate)) but I am not convinced that conservation of momentum can be applied in this case since the boy jumping at an angle to horizontal is applying an external force on the skateboard. Can someone explain this to me? thank you

rasdocus
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2 Answers2

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Conservation of momentum is a vector equation and can be valid in one direction but not in another. There are external forces on the system but they occur in the vertical direction; conservation of momentum in the horizontal direction holds because there are no external forces in this direction.

Look up how conservation of momentum is derived using Newton's Second and Third Laws (e.g., pp 6-9 of these notes), to see why this is true.

Ben H
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You apply momentum conservation to the system boy-board and not to the board alone. The force between the boy and board is internal to the system. Neither the board's nor the boy's momentum is conserved but the overall (system's) momentum is conserved.

nasu
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  • In that case can momentum conservation be applied in y axis as well? does it mean that board will get velocity in y axis? – rasdocus Feb 12 '23 at 04:00
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    No, on the y axis there are forces external to the system. When he pushes against the board, the normal force from the ground is larger than the system's weight and there is a net force along the vertical axis. Only along the horizontal is assumed that there are no external forces (the problem says to ignore any friction for a reason; for this reason). – nasu Feb 12 '23 at 04:28