Question

An $\mathbf{1000}\mathbf{kg}$automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a constant acceleration of$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ . At the same instant a $\mathbf{2000}\mathbf{kg}$truck, travelling at a constant speed of , overtakes and passes the automobile. (a) How far is the com of the automobile–truck system from the traffic light at$\mathbf{t}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{s}$ ? (b) What is the speed of the com then?

Short answer

(a) The distance of center of mass of system from traffic signal is 22m

(b) Speed of center of mass is$9.3m/s$

Step-by-step solution

Listing the given quantities

Mass of automobile, $M_1=1000kg$

Acceleration of automobile, $a=4.0m/s^2$.

Mass of truck, $M_2=2000kg$

Speed of truck,localid="1657266200854" $v_2=8.0m/s.$

Understanding the concept of center of mass and kinematic equations

Using the kinematics equation, we can find the location of the automobileas well as the truck at the given time. From that, we can find the center of mass of the system and also the speed of the center of mass.

Formula:

$$\begin{gathered} x_{com}=\frac{m_1{x}_1+m_2{x}_2+....}{m_1+m_2+m_3+....} \\ v=\frac{dis\tan ce}{time} \\ s=v_{o}t+\frac{1}{2}a{t}^2 \end{gathered}$$

Calculation of required terms

The automobile and truck are initially at rest.

So, at the given time, $t=3.0sec$ the location of the automobile can be found by using the kinematic equation,

$$\begin{gathered} x_1=v_{o}t+\frac{1}{2}a{t}^2 \\ {x}_1=0+\frac{1}{2}(4.0){(3.0)}^2{x}_1=18m \end{gathered}$$

The speed oftheautomobile is

$v_1=at=(4.0)(3.0)=12.0m/s$

At the same time the position ofthetruck is

$x_2=vt=(8.0)(3.0)=24m$

The speed of truck is ,

$v_2=8.0m/s.$$v_2=8.0m/s.$

(a) Calculation of distance of center of mass of system from traffic signal

The location of center of mass is

$$\begin{gathered} x_{com}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2} \\ =\frac{\left(1000kg\right)\left(18m\right)+\left(2000kg\right)\left(24m\right)}{(1000kg+2000kg)} \\ =22m \end{gathered}$$

Thus, the distance of the center of mass from traffic signal is 22m.

(b) Calculation of speed of center of mass 

The speed of center of mass,

$$\begin{gathered} v_{com}=\frac{m_1{v}_1+m_2+v_2}{m_1+m_2} \\ =\frac{\left(1000kg\right)(12m/s)+\left(2000kg\right)(8m/s)}{(1000kg+2000kg)} \\ =9.3m/s \end{gathered}$$

Thus, the velocity of the center of mass is 9.3m/s.