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3 years ago

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A basketball has a mass of 609 g. Moving to the right and heading downward at an angle of 32° to the vertical, it hits the floor

with a speed of 3 m/s and bounces up with nearly the same speed, again moving to the right at an angle of 32° to the vertical. What was the momentum change Δp? (Take the +x axis to be to the right and the +y axis to be up. Express your answer in vector form.)

1 answer:

Vladimir [108]3 years ago

8 0

Answer:

$\Delta p=(0,3.10)kg*m/s\\$

Explanation:

Momentum change:

$\Delta p=p_{f}-p_{o}\\$ : vector

p=mv

$p_{o}=(p_{ox, p_{oy}}}=(m*v*sin(\theta),-m*v*cos(\theta) )\\$ : the ball move downward with an angle theta to the vertical

$p_{f}=(p_{fx, p_{fy}}}=(m*v*sin(\theta),+m*v*cos(\theta) )\\$ :the ball move upward with the same angle theta to the vertical, with same speed

So:

$\Delta p=p_{f}-p_{o}=(0,2m*v*cos(\theta))=(0,2*0.609*3*cos(32))=(0,3.10)kg*m/s\\$

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