Question

A basketball has a mass of 570 g. Moving to the right and heading downward at an angle of 30 to the vertical, it hits the floor with the speed of 5 m/s and bounces up with nearly the same speed, again moving to the right an angle of 30 degree to the vertical. What was the momentum change$∆\stackrel{\rightharpoonup }{p}$?

Short answer

Answer

The momentum change will be ( 0,4.93,0) kg .m/s

Step-by-step solution

Identification of given data

The given data listed below as,

• Mass of the basketball m= 570 g = 0.57 kg,.
• Velocity, v = 5m/s.
• $\theta =30°$

The momentum change 

It bounces up with the same speed, again moving to the right at an angle of 30 degrees to the vertical. We need to find the change in momentum.

 Step 3: Calculation for the momentum change

For the change in momentum, this will become:

$$\begin{gathered} ∆p=mv\cos \theta -\left(-mv\cos \theta \right) \\ ∆p=2mv\cos \theta \end{gathered}$$

Substituting the values and we get,

$$\begin{gathered} ∆p=2\times 0.57kg\times m/s\times \cos 30° \\ =4.93kg.m/s \end{gathered}$$

The change will be in the y-direction only.

Hence, the momentum change will be.

$\left(0,4.93,0\right)kg.m/s$