Answer to Question #208219 in Mechanics | Relativity for Mayito

Question

Two masses, 0.600 kg and 0.300 kg, begin uniform motion at the same speed, 0.800 m/s, from the origin at t = 0 and travel in the directions shown in Figure

(a) Find the velocity of the center of mass in unit–

vector notation. (b) Find the magnitude and direction of the velocity of the center of mass. (c) Write the position vector of the center of mass as a function of time.

Answer 1

Part a

$\sum Vx=- 0.6*0.8cos45 +0.3*0.8cos 45= -0.1697 m/s$

$\sum Fy= 0.6*0.8sin45 +0.3*0.8sin 45= 0.5091 m/s$

Therefore -0.1697i +0.5091j m/s

Part b

Magnitude $\sqrt{(-0.1697)^2+(0.50910^2}= 0.5366 m/s$

Directtion $tan \theta =\frac{0.5091}{-0.1691} \implies \theta = 71.6^0$

Part c

Path of the center of mass in a two-particle system

$r(t)=[ \frac{m_1r_1(t)+m_2r_2(t)}{m_1+m_2}]$