The diagram below shows a blue cart chasing a yellow cart on a frictionless track. The velocities and masses of the carts are given in the diagram. When the blue cart catches up with then yellow cart, the carts stick together and move as one.
What is the momentum of the two-cart system after the collision?

3.0 (k g) (m) / s

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4.0 (k g) (m) / s

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5.0 (k g) (m) / s

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8.0 (k g) (m) / s

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We do not have enough information to determine the momentum of the two-cart system after the collision

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Solution

The correct option is E $8.0 (k g) (m) / s$
Since no external force acts on the system in horizontal direction, the total momentum of system before collision is equal to the total momentum of system after the collision.
Hence $P = m_{1} v_{1} + m_{2} v_{2}$
$= 3 k g m / s + 5 k g m / s = 8 k g m / s$

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The correct option is E 8.0(kg)(m)/sSince no external force acts on the system in horizontal direction, the total momentum of system before collision is equal to the total momentum of system after the collision.Hence P=m1v1+m2v2=3kgm/s+5kgm/s=8kgm/s