Asked by Bapat
A 9-kg ball with an initial velocity of (4i + 9j) m/s collides with a wall and rebounds with a velocity of (–3i + 2j) m/s. What is the impulse exerted on the ball by the wall in the Y direction?
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Answers
Answered by
Anonymous
Impulse is force * time which is change of momentum
Po = final momentum = M( -3 i + 3 j)
Pi = initial momentum = M ( 4 i + 9 j)
change = M (-7 i - 6 j )
the y component is -6 M = -6 meters/s * 9 kg = -54 kg meters/second
Po = final momentum = M( -3 i + 3 j)
Pi = initial momentum = M ( 4 i + 9 j)
change = M (-7 i - 6 j )
the y component is -6 M = -6 meters/s * 9 kg = -54 kg meters/second
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