Conservation of Momentum EX 3

Given that ${\overrightarrow{p}}_{net}={\overrightarrow{p}}_1+{\overrightarrow{p}}_2+{\overrightarrow{p}}_3$,

we will choose a coordinate system and determine the angles with the x-axis.

Then we will compute the components of all vectors:

Momentum	x-component	y-component
$p_1$	96 cos(37°) = 83.1 $\frac{kgm}{s}$	96 sin(37°) = -48 $\frac{kgm}{s}$
$p_2$	167 cos(53°) = 100.5 $\frac{kgm}{s}$	167 cos(53°) = 133.4 $\frac{kgm}{s}$
$p_3$	$p_3$ cos(240) = -0.5 $p_3$	$p_3$ sin(240) = -0.87 $p_3$
$p_{net}$	$p_{net}$	0

This allows us to write an equation for x-component

$p_{net}=83.1+100.5-0.5p_3$

and y-component

$0=-48+133.4-0.87p_3$

Solving the second equation first, we find

$p_3=\frac{48-133.4}{-0.87}=98.2\frac{kgm}{s}$

Then we solve the first equation to find

$p_{net}=100.5+83.1-0.5(98.2)=134.5\frac{kgm}{s}$

From this equation we conclude that the total momentum has a magnitude 134.5 $\frac{kgm}{s}$ and points East since its x-component is positive. We also conclude that the magnitude of the momentum of the third particle is $98.2\frac{kgm}{s}$.