Answer to Question #282009 in Physics for Grace

Question

A cheering squad is made up of three members: Lynne of 50kg moving at 3 m/sNorth. Angie of mass 58kg moving eastward at 3 m/s and Mercy of mass 6 kg moving at 3 m/s south. What is the combined momentum of the squad?

Answer 1

Given:

$m_1=50\:\rm kg$

$m_2=58\:\rm kg$

$m_3=6\:\rm kg$

$v_{1x}=0\:{\rm m/s},\quad v_{1y}=3\:\rm m/s$

$v_{2x}=3\:{\rm m/s},\quad v_{2y}=0\:\rm m/s$

$v_{3x}=0\:{\rm m/s},\quad v_{3y}=-3\:\rm m/s$

The momentum of combined system

$$p_x=m_1v_{1x}+m_2v_{2x}+m_3v_{3x}\\ =58*3=174\:\rm kg*m/s$$

$$p_y=m_1v_{1y}+m_2v_{2y}+m_3v_{3y}\\ =50*3+6*(-3)=132\:\rm kg*m/s$$

Magnitude of momentum

$$p=\sqrt{p_x^2+p_y^2}=\sqrt{174^2+132^2}=218\:\rm kg*m/s$$

Direction of momentum

$$\theta=\tan^{-1}\frac{p_y}{p_x}=\tan^{-1}\frac{132}{174}=37^\circ\rm\quad N\:of\: E$$