Question

A basketball has a mass of 575 g. Moving to the right and heading downward at an angle of 31° to the vertical, it hits the floor with a speed of 4 m/s and bounces up with nearly the same speed, again moving to the right at an angle of 31° 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

Taking the x axis to the right and the y axis to be up, the total change of momentum is
`\Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}`

Step-by-step explanation:

The momentum
`\vec{p}` is given by:

`\vec{p} = m \ \vec{v}`

where m is the mass and
`\vec{v}` is the velocity. Now, taking the suffix i for the initial condition, and the suffix f for the final condition, the change in momentum will be:

`\Delta \vec{ p} = \vec{p}_f - \vec{p}_i`

`\Delta \vec{ p} = m \ \vec{v}_f - m \ \vec{v}_i`

`\Delta \vec{ p} = m (\ \vec{v}_f -  \ \vec{v}_i)`

As we know the mass of the ball, we just need to find the initial and final velocity.

Knowing the magnitude and direction of a vector, we can obtain the Cartesian components with the formula

`\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )`

where
`| \vec{A} |` is the magnitude of the vector and θ is the angle measured from the x axis.

Taking the x axis to the right and the y axis to be up, the initial velocity will be:

`\vec{v}_i = 4 (m)/(s) ( \ cos ( - (90  \ °- 31 \°)) , sin( - (90  \ ° - 31\°) ) ) =`

where minus sign appears cause the ball is going downward, and we subtracted the 31 ° as it was measured from the y axis

So, the initial velocity is

`\vec{v}_i = 4 (m)/(s) ( \ cos ( - 59 \°) , sin( - 59 \°)) =`

`\vec{v}_i =  ( \ 2.0601  \ (m)/(s) , - 3.4286 (m)/(s)) =`

The final velocity is

`\vec{v}_i = 4 (m)/(s) ( \ cos ( 90  \ °- 31 \°) , sin( 90  \ ° - 31\°)) =`

`\vec{v}_i = 4 (m)/(s) ( \ cos ( 59 \°) , sin(  59 \°)) =`

`\vec{v}_i =  ( \ 2.0601  \ (m)/(s) ,  3.4286 (m)/(s)) =`

So, the change in momentum will be

`\Delta \vec{ p} = m (\ \vec{v}_f -  \ \vec{v}_i)`

`\Delta \vec{ p} = 0.575 \ kg (\  ( \ 2.0601  \ (m)/(s) ,  3.4286 (m)/(s) -  ( \ 2.0601  \ (m)/(s) , - 3.4286 (m)/(s)))`

`\Delta \vec{ p} = 0.575 \ kg (\  ( \ 2.0601  \ (m)/(s) -  \ 2.0601  \ (m)/(s),  3.4286 (m)/(s) +  3.4286 (m)/(s))`

`\Delta \vec{ p} = 0.575 \ kg (\  0 , 2 * 3.4286 (m)/(s) )`

`\Delta \vec{ p} = 0.575 \ kg * 2 * 3.4286 (m)/(s) \hat{j}`

`\Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}`