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In a physics lab, a cube slides down a frictionless

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 29P Chapter 7

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 29P

In a physics lab, a cube slides down a frictionless incline as shown in Fig. 7-50 and elastically strikes another cube at the bottom that is only one-half its mass. If the incline is 35 cm high and the table is 95 cm off the floor, where does each cube land? [Hint Both leave the incline moving horizontally.)

Step-by-Step Solution:

Solution 29P:

        We have to determine the horizontal distance of landing of both the cubical block.

Step 1 of 6</p>

Concept:

        Law of conservation of energy: For an isolated system, the total amount of energy always remains constant. However, it can be converted from one form to other.

Law of conservation of linear momentum: When no external force acts on the system, the final momentum of the system is always equal to the initial momentum.

Step 2 of 6</p>

As the cube of mass  slide down an inclined plane from rest its kinetic energy increase at the cost of fall in potential energy. Using law of conservation of energy,

Gain in kinetic energy of cube = Loss in potential energy of cube

        

                                        …(1)

Where  Fall in height

        velocity of sliding cube at the instant it strikes the other cube of mass at rest.

Step 3 of 6</p>

As both the cubes collide, let their speeds after collision be respectively. Hence from principle of conservation of linear momentum, we get,

        Initial momentum of system = Final momentum of system

                     

                                           

       …(2)

Also since no external force is acting on the system, therefore the relative velocity of both the cubes does not change.

                                                           

                      …(3)

Putting value of from equation (3) in equation (1), we get,

                                                      …(4)

From equation (4) and (3), we get,

                                                      …(5)

Step 4 of 6

Chapter 7, Problem 29P is Solved
Step 5 of 6

Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

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