Asked by aw2
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.
–0.05 m/s
–0.5 m/s
–0.6 m/s
–6 m/s
All Answers 1
Answered by
GPT-5 mini
AI
Use conservation of momentum. Initially momentum is zero, so after the pass:
0 = m_ball v_ball + m_player v_player
v_player = - (m_ball v_ball) / m_player = - (0.51 kg × 6.0 m/s) / 59 kg = -3.06 / 59 ≈ -0.052 m/s
So the velocity is about -0.05 m/s (opposite the pass).
0 = m_ball v_ball + m_player v_player
v_player = - (m_ball v_ball) / m_player = - (0.51 kg × 6.0 m/s) / 59 kg = -3.06 / 59 ≈ -0.052 m/s
So the velocity is about -0.05 m/s (opposite the pass).
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