Question

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.

Answer 1

We can solve the problem by using conservation of momentum.

The player + ball system is an isolated system (there is no net force on it), therefore the total momentum must be conserved. Assuming the player is initially at rest with the ball, the total initial momentum is zero:

$p_i = 0$

The total final momentum is:

$p_f = p_p + p_b$

where $p_p$ is the momentum of the player and $p_b$ is the momentum of the ball.

The momentum of the ball is: $p_b = mv=(0.51 kg)(6 m/s)=3.06 kg m/s$

While the momentum of the player is: $p_p = Mv_p$, where M=59 kg is the player's mass and vp is his velocity. Since momentum must be conserved,

$p_f = p_i = 0$

so we can write

$p_f = Mv_p + p_b =0$

and we find

$v_p = -\frac{p_b}{M}=-\frac{3.06 kg m/s}{59 kg}=-0.052 m/s$

and the negative sign means that it is in the opposite direction of the ball.

Answer 2

Answer:

A) -0.05 m/s

Explanation:

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