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 ? (Take the axis to be to the right and the axis to be up. Express your answer in vector form.)

Answer 1

Answer:1.549 kg-m/s

Step-by-step explanation:

Given

mass of basketball =609 g

ball is moving downward right at an angle of
`32 ^(\circ)` with vertical

After hitting the floor it moves with a speed of 3 m/s

change in Momentum in x direction

`P_x=musin32\hat{i} -mvsin32\hat{i}`

where u=3m/s

v=3m/s

m=mass of ball

`P_x=0`

Change in momentum in Y-direction

`P_y=-mucos32\hat{j}-mvcos32\hat{j}=-2mvcos32`

`P_y=-1.549\hat{j} kg-m/s`