Several forces´s impulse

Data

• Mass of the body: m=35 kg
• Performance time of the forces: t = 3 minutes = 3·60 s = 180 s
• Forces: ${\overrightarrow{F}}_1=125·\overrightarrow{i}+200·\overrightarrow{j}\hspace{0.17em}N ;\hspace{0.17em}{\overrightarrow{F}}_2=-400·\overrightarrow{i} N ;\hspace{0.17em}{\overrightarrow{F}}_3=-150·\overrightarrow{i}+180·\overrightarrow{j}N$
• Rest part ( ${\overrightarrow{v}}_i=0$ )

Resolution

We start calculating the resultant force. For that we simply have to sum the forces provided.

$${\overrightarrow{F}}_T={\overrightarrow{F}}_1+\hspace{0.17em}{\overrightarrow{F}}_2+\hspace{0.17em}{\overrightarrow{F}}_3=125·\overrightarrow{i}+200·\overrightarrow{j}\hspace{0.17em}-400·\overrightarrow{i} -150·\overrightarrow{i}+180·\overrightarrow{j}=-425·\overrightarrow{i}+380·\overrightarrow{j} N$$

Now we can calculate the impulse requested multiplying the resultant force obtained by the interval of time during the forces are acting:

$$\overrightarrow{J}=\overrightarrow{F}·∆t=\left(-425·\overrightarrow{i}+380·\overrightarrow{j}\right)·180=-76500·\overrightarrow{i}+68400·\overrightarrow{j} N·s$$

As we know, because the impulse-momentum theorem, the previous value coincides with the variation of linear momentum requested. We find the velocity requested turning to the definition of that linear momentum:

$$\overrightarrow{J}=∆\overrightarrow{p}={\overrightarrow{p}}_f-{\overrightarrow{p}}_i=m·{\overrightarrow{v}}_f-\cancel{m·{\overrightarrow{v}}_i}\Rightarrow {\overrightarrow{v}}_f=\frac{\overrightarrow{J}}{m}=\frac{-76500·\overrightarrow{i}+68400·\overrightarrow{j}}{35}=2185.7·\overrightarrow{i}+1954.28·\overrightarrow{j} m/s$$