Ginve data:
The mass of truck is m=2650 kg.
The initial speed of truck is u=39 km/h (east).
The final speed of truck is v=62 km/h (south).
Part (e)
The velocity of truck in unit vector notation is,
[tex]v=39(i)+62(-j)[/tex]
According to third law of Newton, the impulse and moment of a moving body is same, therefore we can write,
[tex]\begin{gathered} p=mv \\ p=2650kg((39\frac{km}{h}\times\frac{1\text{ m/s}}{\frac{3.6km}{h}})(i)+(62\frac{km}{h}\times\frac{1\text{ m/s}}{\frac{3.6km}{h}})(-j)) \\ p=2650kg((\frac{10.83m}{s})(i)+(\frac{17.22m}{s})(-j)) \\ p=\frac{28708.3kgm}{s}(i)+\frac{45638.8kgm}{s}(-j) \end{gathered}[/tex]
Thus, the x-component of impulse is 28708.2 kgm/s.
Part (g)
As calculated above the y-component of impulse is 45638.8 kgm/s.
Part (i)
The magnitude of impulse will be,
[tex]\begin{gathered} p=\sqrt[]{28708.3^2+45638.8^2} \\ p=\frac{53917.3kgm}{s} \end{gathered}[/tex]
Thus, themagnitude of impulse is 53917.3 kgm/s.
Part (j)
The direction of impulse will be,
[tex]\begin{gathered} \tan \theta=\frac{45638.8}{28708.3} \\ \theta=57.8\degree \end{gathered}[/tex]
The direction counterclock wise from x-axis will be,
[tex]\begin{gathered} \theta=270\degree+57.8\degree \\ \theta=327.8\degree \end{gathered}[/tex]
