Conservation of Momentum EX 3

Given that ${\vec {p}}_{net}={\vec {p}}_{1}+{\vec {p}}_{2}+{\vec {p}}_{3}$,

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 ${\tfrac {kgm}{s}}$	96 sin(37°) = -48 ${\tfrac {kgm}{s}}$
$p_{2}$	167 cos(53°) = 100.5 ${\tfrac {kgm}{s}}$	167 cos(53°) = 133.4 ${\tfrac {kgm}{s}}$
$p_{3}$	$p_{3}$ 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

$p_{3}={\frac {48-133.4}{-0.87}}=98.2{\tfrac {kgm}{s}}$

Then we solve the first equation to find

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

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