To solve the problem it is necessary to apply the concepts related to the conservation of linear Moment, that is to say
[tex]P=mv[/tex]
Where,
m = Mass
v = Velocity
P = Linear momentum
For the given data we have to:
[tex]v_1 = 180\frac{km}{h}(\frac{1h}{3600s})(\frac{1000m}{1km})[/tex]
[tex]v_1 = 50m/s[/tex]
The components of this force would be given by,
[tex]v_z = 50*sin37 = 30.09m/s[/tex]
[tex]v_{x,y} = 50*cos37=39.93m/s \rightarrw[/tex]According to the definition given at the end of the problem, this component corresponds to that expressed for x and y.
Applying the previous equation we have,
[tex]\vec{P} = m(39.93cos\theta \hat{i}+39.93sin\theta \hat{j}-30.09\hat{k})[/tex]
Note: The component at this direction must also decomposed
The mass is 143g=0.143kg, then:
[tex]\vec{P} = (0.143)(39.93cos(22)\hat{i}+39.93sin(22) \hat{j}-30.09\hat{k})[/tex]
Therefore the final vector is:
[tex]\vec{P} = 5.29\hat{i}+2.138\hat{j}-4.29\hat{k})[/tex]
