Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 5, Problem 34P

(a)

To determine

To Evaluate: The position of center of mass at t=0 , wherein the location of two particles, which are initially at rest, is shown in Fig. 5-22, and each particle is subject to a constant force, as indicated.

  Physics Fundamentals, Chapter 5, Problem 34P , additional homework tip  1

(a)

Expert Solution
Check Mark

Answer to Problem 34P

The position of center of mass of particles is 0 for the value of t=0 .

Explanation of Solution

Given:

The time t=0 . The location of the particles and the force acting on them has been shown in the Fig. 5-22.

Formula used:

The x-coordinate of center of mass of particles shown in the figure can be calculated by the following formula:

  xcm=(x1m1+x2m2)(m1+m2)

Where m1 and m2 are the masses of particles 1 and 2 respectively while x1 and x2 are the positions of the particles 1 and 2 respectively.

Similarly the y-coordinates can also be calculated:

  ycm=(y1m1+y2m2)(m1+m2)

Where m1 and m2 are the masses and y1 and y2 are the positions of particles 1 and 2 respectively.

Calculation:

When t=0 , the position of center of mass of particles is calculated as:

  xcm=(x1m1+x2m2)(m1+m2)

Substituting m1=1.00 Kg, m2=2.00Kg, x1=20.0 cm and x2=10.0 cm

  xcm=(20.0 cm×1.00 Kg)+(-10.0 cm×2.00 Kg)(1.00 Kg+2.00 Kg)      = 0 cm

Conclusion:

At t=0 , the position of center of mass of the particles is 0 .

(b)

To determine

To Evaluate: The resultant force acting on the system, wherein the location of two particles, which are initially at rest, is shown in Fig. 5-22, and each particle is subject to a constant force, as indicated.

  Physics Fundamentals, Chapter 5, Problem 34P , additional homework tip  2

(b)

Expert Solution
Check Mark

Answer to Problem 34P

System’s resultant force is calculated as 14.14 N .

Explanation of Solution

Given:

The magnitude of the force acting on 1 Kg particle, in the positive direction, along the x-axis is 10.0N

   Fx=10.0 N

The magnitude of force acting on 2.00 Kg particle, in the positive direction, along the y-axis is 10.0 N .

   Fy=10.0 N

Formula used:

Formula for the resultant force (F) is given by:

  F=Fx2+Fy2

Where Fx is the force acting in the x direction and Fy is the force acting in the y direction.

Calculation:

The resultant force is being calculated as:

  F=Fx2+Fy2    = (10.0 N)2+ (10.0N)2    = 14.14 N

Conclusion:

Thus, the system’s resultant force = 14.14 N

(c)

To determine

To Evaluate: The location of center of mass at t=5.00 s , wherein the location of two particles, which are initially at rest, is shown in Fig. 5-22, and each particle is subject to a constant force, as indicated.

  Physics Fundamentals, Chapter 5, Problem 34P , additional homework tip  3

(c)

Expert Solution
Check Mark

Answer to Problem 34P

At t=5.0 s the location for center of mass is (41.67 m, 41.67 m)

Explanation of Solution

Given:

The magnitude of the force acting on 1 Kg particle, in the positive direction, along the x-axis is 10.0N

   Fx=10.0 N

The magnitude of force acting on 2.00 Kg particle, in the positive direction, along the y-axis is 10.0 N .

   Fy=10.0 N

Formula used: It is known that Force = Mass×Acceleration

  x=vt+12at2

Where x is the displacement, v is the velocity, a is the acceleration and t is the time.

The x-coordinate of center of mass of particles shown in the figure can be calculated by the following formula:

  xcm=(x1m1+x2m2)(m1+m2)

Where m1 and m2 are the masses of particles 1 and 2 respectively, while x1 and x2 are the positions of the particles 1 and 2 respectively.

Similarly the y-coordinates can also be calculated:

  ycm=(y1m1+y2m2)(m1+m2)

Where m1 and m2 are the masses and y1 and y2 are the positions of particles 1 and 2 respectively.

Calculation:

As a constant force is acting on each particle, the acceleration of particle m1 is calculated as follows:

  F    =m1×a1 a1=Fm1

Substituting F=10.0 N and m1=1.00 Kg

  a1=10.0 N1.0 Kg    = 10.0 m/s2

Position of particle m1 is calculated as follows:

  x1=(v1t)+(12a1t2)

When the particle is at rest the velocity v1=0 and

  x1=(12a1t2)

Substituting a1=10.0 m/s2 and t=5.00 s

  x1=(12×10.0 m/s2)×(5.00 s)2    = 125 m

Similarly the acceleration of m2 is calculated as follows:

  F    =m2×a2 a2=Fm2

Substituting F=10.0 N and m2=2.00 Kg

  a2=10.0 N2.0 Kg    = 5.0 m/s2

And position of m2 is calculated as follows:

  y1=(12a2t2)

Substituting values of a2=5.00 m/s2 and t=5.00 s

  y1=(12×5.0m/s2)(5.0 s)2    = 62.5 m

At t=5.0 s , the x-coordinate of center of mass of two particles is calculated as follows:

  xcm=(x1m1+x2m2)(m1+m2)

Substituting m1=1.00 Kg, x1=125 m, m2=2.00 Kg and x2=0

  xcm=(125 m × 1.00 Kg)+(0 × 2.00 Kg)(1.00 Kg + 2.00 Kg)      = 41.67 m

Similarly the y-coordinate of center of system’s mass is calculated at t=5.0 s

  ycm=(y1m1+y2m2)(m1+m2)

Substituting m1=1.00 Kg, y1=0, m2=2.00 Kg and y2=62.5 m

  ycm=(0×1.00 Kg)+(62.5 m×2.00 Kg)(1.00 Kg + 2.00 Kg)     = 41.67 m

The x and y coordinates of the center of mass for t=5.0 s is (41.67 m, 41.67 m)

Conclusion:

Thus, the location of center of mass at t=5.0 s is (41.67 m, 41.67 m) .

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