Question

A basketball has a mass of 601 g. Moving to the right and heading downward at an angle of 29 degree to the vertical, it hits the floor with a speed of 6 m/s and bounces up with nearly the same speed, again moving to the right at an angle of 29 degree to the vertical. What was the momentum change Delta vector p? (Take the +x axis to be to the right and the +y axis to be up. Express your answer in vector form.) Delta vector p = <0, 6.31, 0> <0, 6.31, 0> kg.m/s

Answer 1

Answer:

Change in momentum, $\Delta p=6.3\ kg-m/s$

Explanation:

It is given that,

Mass of the basketball, m = 601 g = 0.601 kg

The basketball makes an angle of 29 degrees to the vertical, it hits the floor with a speed, v = 6 m/s

It bounces up with the same speed, again moving to the right at an angle of 29 degree to the vertical. We need to find the change in momentum. It is given by :

$\Delta p=mv\ cos\theta-(-mv\ cos\theta)$

$\Delta p=2mv\ cos\theta$

$\Delta p=2\times 0.601\times 6\times \ cos(29)$

$\Delta p=6.3\ kg-m/s$

So, the change in momentum of the basketball is 6.3 kg-m/s. Hence, this is the required solution.