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A basketball player passes a ball to a teammate at a velocity of 6 m/s. The ball has a mass of 0. 51 kg. If the original player has a mass of 59 kg 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. â€"0. 05 m/s â€"0. 5 m/s â€"0. 6 m/s â€"6 m/s.

Respuesta :

The velocity of the player is - 0.05 m/s.

From the law of conservation of linear momentum, the momentum after collision is equal to the momentum before collision. We can see that the initial momentum of the system is zero.

Hence;

0 = (0. 51 kg ×  6 m/s) +  59 kg v

0 = 3.06 + 59v

v = -3.06/59

v = - 0.05 m/s

The negative sign means that the velocity of the player is opposite that of the ball.

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