Question

A basketball has a mass of 570g. Heading straight downward, in the y-direction, it hits the floor with the speed of 5 m/s and rebounds straight up with nearly the same speed. What was the momentum change $∆\stackrel{\rightharpoonup }{p}$?

Short answer

Answer

The momentum change will be ( 0,5.7,0 ) kg m/s.

Step-by-step solution

Identification of given data

The given value is below as,

• Mass of the basketball m= 570 g =0.57 kg.

• Value of initial and final velocities,$\stackrel{\rightharpoonup }{v_i}=\left(0,-5,0\right)m/s,\stackrel{\rightharpoonup }{v_f}=\left(0,5,0\right)m/s$

Momentum 

Momentum is nothing but the product of the mass of the object and the velocity of the object.

Calculation for the momentum change

Initial momentum can be calculated as:

$\stackrel{\rightharpoonup }{p_i}=m\stackrel{\rightharpoonup }{v_i}$

Initial momentum can be calculated as:

$\stackrel{\rightharpoonup }{p_f}=m\stackrel{\rightharpoonup }{v_f}$

Change in the momentum can be written as follows:

$∆\stackrel{\rightharpoonup }{p}=\stackrel{\rightharpoonup }{p_f}-\stackrel{\rightharpoonup }{p_i}$

Putting the values and we get,

$$\begin{gathered} ∆\stackrel{\rightharpoonup }{p}=\stackrel{\rightharpoonup }{m{v}_f}-\stackrel{\rightharpoonup }{m{v}_i} \\ =0.57kg\times \left(0,5,0\right)m/s-0.57kg\times \left(0,-5,0\right)m/s \\ =2\times 0.57kg\times \left(0,5,0\right)m/s \\ =\left(0,5.7,0\right)kg.m/s \end{gathered}$$

Therefore, the change in momentum in the tennis ball is$=\left(0,5.7,0\right)kg.m/s$ .