A tennis ball of mass m travelling with velocity (v_x, 0,0) hits a ball and rebounds with velocity (–v_x,0,0).

(a) what was the change in momentum of the tennis ball?

(b) What was the change in magnitude of the momentum of the tennis ball?

• Final and initial velocity are ,$v_f=\left(-v_x,0,0\right),v=\left(v_x,0,0\right)$.

In the elastic collision, the total momentum is conserved.

Momentum is the product of the mass of the object and the velocity of the object

Initial momentum can be calculated as:

$p_i=m{v}_i$

Initial momentum can be calculated as:

$p_f=m{v}_f$

Change in the momentum can be written as follows:

$∆p=p_f-p_i$

Putting the values and we get,

$$\begin{gathered} ∆p=m{v}_f-m{v}_i \\ =m\left(-v_x,0,0\right)-m\left(v_x,0,0\right) \\ =\left(-2m{v}_x,0,0\right) \end{gathered}$$

Therefore, the change in momentum in the tennis ball is$=\left(-2m{v}_x,0,0\right)$ .

Change in the magnitude of the momentum = magnitude of final momentum – the magnitude of initial momentum

That can be written as:

$\left|∆p\right|=\left|p_f\right|-\left|p_i\right|$

Substituting values in the above expression and we get,

$$\begin{gathered} \left|∆p\right|=\sqrt{m{\left(v_x\right)}^2+0+0}-\sqrt{m{\left(-v_x\right)}^2+0+0} \\ =\sqrt{m{\left(v_x\right)}^2+0+0}-\sqrt{m{\left(-v_x\right)}^2+0+0} \\ =0 \end{gathered}$$

Hence, the change in the magnitude of the momentum is 0.