Answer to Question #205375 in Mechanics | Relativity for Jessy

Question

A 1200-kg SUV is moving along a straight highway at 12.0 m/s. Another car, with mass 1800 

kg and speed 20.0 m/s, has its center of mass 40.0 m ahead of the center of mass of the SUV 

(Fig. E8.54). Find (a) the position of the center of mass of the system consisting of the two 

cars; (b) the magnitude of the system’s total momentum, by using the given data; (c) the 

speed of the system’s center of mass; (d) the system’s total momentum, by using the speed 

of the center of mass. Compare your result with that of part (b).

Answer 1

$\text{We introduce a coordinate system consisting of a straight line and the}$

$\text{origin of the coordinate system is the position of the center of mass}$

$\text{of the SUV}$

$\text{for SUV:}$

$m_1 =1200;r_1=0;v_1=12$

$p_1=m_1v_1=12*1200=14400$

$\text{for another car:}$

$m_2 =1800;r_2=40;v_2=20$

$p_2=m_2v_2=1800*20=36000$

$a)\text{position of the center of gravity of the system:}$

$r=\frac{m_1r_1+m_2r_2}{m_1+m_2}=\frac{1200*0+1800*40}{1200+1800}=24$

$\text{Answer: position of the center of gravity of the system 24 m in front of the SUV}$

$b) p = p_1+ p_2=14400+36000 =50400$

$\text{Answer: } p= 50400$

$с)v= \frac{p_1+p_2}{m_1+m_2}=\frac{50400}{3000}=16.8$

$\text{Answer: }v=16.8$

$d) p'=v*(m_1+m_2)=16.8*3000 =50400$

$p'=p$

$\text{Answer: }p'=p=50400$