Impulse from linear momentum variation

Data

• Mass of the ball: m = 50g = 0.05 kg
• Velocity before and after the crash: ${\overrightarrow{v}}_i=-20·\overrightarrow{i}-20\overrightarrow{j} m/s ;\hspace{0.17em}{\overrightarrow{v}}_f=15·\overrightarrow{i}+20\overrightarrow{j} m/s$
• Force´s time of performance t = 0.05s

Resolution

Impulse-momentum theorem  says the impulse coincides with the variation of the linear momentum, so:

$$\begin{gathered} \overrightarrow{J}=∆\overrightarrow{p}={\overrightarrow{p}}_f-{\overrightarrow{p}}_i=m·{\overrightarrow{v}}_f-m·{\overrightarrow{v}}_i=m·\left({\overrightarrow{v}}_f-{\overrightarrow{v}}_i\right)= \\ =0.05·\left(15·\overrightarrow{i}+20\overrightarrow{j}-\left(-20·\overrightarrow{i}-20\overrightarrow{j}\right)\right)=35·\overrightarrow{i}+40·\overrightarrow{j}N·s \end{gathered}$$

On the other hand, from definition of impulse, we can calculate the force exerted by the racquet:

$$\overrightarrow{J}=\overrightarrow{F}·∆t\Rightarrow \overrightarrow{F}=\frac{\overrightarrow{J}}{∆t}=\frac{35·\overrightarrow{i}+40·\overrightarrow{j}}{0.05}=700·\overrightarrow{i}+800·\overrightarrow{j} N$$