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Cart 1 Cart 2 Cart 1 Cart 2 no m/s LO After Collision Before Collision Two lab carts have the same mass and are free to move

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Answer #1

As there is no any external forces on two cart system (spring force would be internal force) so the momentum of the system would be conserved before and after the collision. So the velocity of two would be anything that follows momentum conservation.

Hence first combination could be v1= -0.2m/s and v2= 1.2 m/s

As v1+v2=1 m/s (I'm just adding velocity because all are of same mass so mass would be cancelled both side.)

Again the second combination could be v1= 0.5m/s and v2=1.5 m/s

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