With respect to ground and defining left (-), right (+):
1) Before fired ball had <- momentum = P1= m x v1 = -m x 50mph
2) After fired ball has P1 momentum plus -> P2 = m x v = +m x 50mph
In the defined frame: net momentum = P1+P2 = 0
OP stands correct
#physics
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But you could fire it at 100mph and drive at 1mph, momentum will still be conserved. This is just a special case.
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@WarrenSmithFOX you need to tee this vision up to cross to during forward pass debates -
I think the physics might be lost on the disbelievers.
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Thanks. Twitter will use this to make your timeline better. UndoUndo
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To be a ball, would you survive being fired upward out of a falling elevator at the same speed? Or is this only relevant on a horizontal axis?
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You can cancel out your momentum when falling from an elevator with the same principle. BUT the force you would apply on the person to cancel would be the same as the impact, which means RIP
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I like you added "earth reference frame" just to head off that little bit of pedantry before it started.
Thanks. Twitter will use this to make your timeline better. UndoUndo
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Thus conclusively proving the validity of the ‘moment of perfect stillness’.pic.twitter.com/UhCNKBkfyp
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I think this may be the cheesiest reply I've ever seen from
@MrSteveMatchett
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