Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 5, Problem 37P
To determine

To Evaluate: The center of mass of the remaining of wreckage of an airplane of mass 1.0×104 Kg , that falls vertically downwards, crashing to the ground. The plane explodes just before impact. Investigators looking in to the cause of the crash search for the wreckage. Large fragments of mass 4.0×103 Kg , 3.0×103 Kg and 1.0×103 Kg are found respectively at the point of impact, 30 m North and 50 m East.

Expert Solution & Answer
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Answer to Problem 37P

The location of the remaining part of the wreckage is positioned at 25 m West and 45 m South

Explanation of Solution

Given info:

  Mass of airplane =1.0×104 Kg

  Mass of fragments =4.0×103 Kg

  Mass of fragments =3.0×103 Kg

  Mass of fragments =1.0×103 Kg

The fragments are found at a location 30 m North and 50 m East.

Formula used:

The x - component of position of center of mass of a system is calculated as follows:

  xcm=x×mm

Where, m is the mass of individual units of the system and x its respective distances. m gives the sum total of masses of all the units in the system.

Similarly the y -component of location for center of mass of a system is calculated as follow:

  ycm=y×mm

Where, m is the mass of the individual units of the system and y its respective distances. The total mass of all the units in the system, when added up is given by m .

Calculation:

The sum total of the masses of wreckage pieces will be equal to the mass of the airplane. Assuming mass of airplane as M and m1,m2,m3 are the masses of fragments, then

  M=m1+m2+m3+mr

Substituting M =1.0×104 Kg , m1 =4.0×103 Kg , m2 =3.0×103 Kg , m3 =1.0×103 Kg

  1.0×104 Kg=4.0×103 Kg+3.0×103 Kg+1.0×103 Kg+mr             mr = 2.0×103 Kg

Considering the center of mass to be at the origin itself:

  0=(0×m1)+(30 m×m2)+(50 m×m3)+mrx+mry

Therefore the x-component is derived as:

  m1x=m3×50 m    x=m3×50 mm1

Substituting m3=1.0×103 Kg and m1=2.0×103 Kg

  x=(50 m)×(1.0× 103 Kg)(2.0× 103 Kg)   = 25 m

The negative sign is an indication that the center of mass is positioned towards West.

Similarly the y -component is derived as follows:

  m1y=(30 m)×m2   y  =(30 m)×m2m1

Substituting m2=3.0×103 Kg and m1=2.0×103 Kg

  y=(30 m)×(3.0× 103 Kg)(2.0× 103 Kg)   = 45m

Hence the location of the center of mass is towards the South direction as indicated by the negative sign of the result.

Conclusion:

Thus, the location of the remaining portion of the wreckage is positioned at 25 m in West, and 45 m in south direction.

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