A basketball player passes a ball to a teammate at a velocity of 6 meters per second The ball has a mass of 0 point 5 1 kilograms If the original player has a mass of 59 kilograms and there is no net force on the system, what is the velocity of the player after releasing the ball? Let a positive velocity be in the direction of the pass.

The player and ball are initially at rest, so their total momentum is zero. After the pass, conservation of momentum gives 59v + (0.51)(6) = 0, so v = −3.06/59 ≈ −0.052 m/s. The negative sign means the player moves opposite the pass; choice A, 0.05 meters per second, matches the speed (with that opposite direction).

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