Fundamentals of Physics Extended
10th Edition
ISBN: 9781118230725
Author: David Halliday, Robert Resnick, Jearl Walker
Publisher: Wiley, John & Sons, Incorporated

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Textbook Question
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Chapter 7, Problem 1Q

Rank the following velocities according to the kinetic energy a particle will have with each velocity, greatest first: (a) v = 4 i ^ + 3 j ^ , (b) v = −4 i ^ + 3 j ^ , (c) v = −3 i ^ + 4 j ^ , (d) v = 3 i ^ − 4 j ^ , (e) v = 5 i ^ , and (f) v = 5 m/s at 30° to the horizontal.

To determine

To rank:

The velocity according to the greatest kinetic energy at first.

Solution:

All velocities have the same kinetic energy.

### Explanation of Solution

1) Concept:

To rank kinetic energy, we have to find first find the resultant velocity, and then use the formula for kinetic energy in which the mass is the same for all velocities. Therefore, kinetic energy is ranked only on the magnitude of velocity.

2) Formula:

i) v=vx2+vy2

ii) KE=0.5×m×v2

3) Given:

i) v=4i^+3j^

ii) v=-4i^+3j^

iii) v=-3i^+4j^

iv) v=-3i^-4j^

v) v=5i^

vi) v=5   at  30° to horizontal v=5cos30i^+5sin30j^

4) Calculation:

First find the resultant velocity as follows:

v=vx2+vy2

For v=4i^+3j^

v=42+32

v=5 m/s

For v=-4i^+3j^

v=(-4)2+32

v=5 m/s

For v=-3i^+4j^

v=-32+42

v=5 m/s

For v=-3i^-4j^

v=-32+(-42)

v=5 m/s

For v=5i^

v=52

v=5 m/s

For v=5cos30i^+5sin30j^

v=52(cos230+sin230)

v=5 m/s

Now kinetic energy can be found by using the following formula:

KE=0.5×m×v2

Assume m=1 kg

So

KE=0.5×v2

As the resultant velocity  v=5m/s is the same for all cases, kinetic energy is also the same for all.

Conclusion:

Resultant velocity is found by using Pythagoras theorem; from where the kinetic energy could be calculated.

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