Question

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.)

Answer 1

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

Step-by-step 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\\`