An 1000kgautomobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a constant acceleration of4.0m/s2 . At the same instant a 2000kgtruck, 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 att=3.0s ? (b) What is the speed of the com then?

Short Answer

Expert verified

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

(b) Speed of center of mass is9.3m/s

Step by step solution

01

Listing the given quantities

Mass of automobile, M1=1000kg

Acceleration of automobile, a=4.0m/s2.

Mass of truck, M2=2000kg

Speed of truck,localid="1657266200854" v2=8.0m/s.

02

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:

xcom=m1x1+m2x2+....m1+m2+m3+....v=distancetimes=vot+12at2

03

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,

x1=vot+12at2x1=0+12(4.0)(3.0)2 x1=18m

The speed oftheautomobile is

v1=at=(4.0)(3.0)=12.0m/s

At the same time the position ofthetruck is

x2=vt=(8.0)(3.0)=24m

The speed of truck is ,

v2=8.0m/s.v2=8.0m/s.

04

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

The location of center of mass is

xcom=m1x1+m2x2m1+m2=(1000kg)(18m)+(2000kg)(24m)(1000kg+2000kg)=22m

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

05

(b) Calculation of speed of center of mass 

The speed of center of mass,

vcom=m1v1+m2+v2m1+m2=(1000kg)(12m/s)+(2000kg)(8m/s)(1000kg+2000kg)=9.3m/s

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

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