Modern Physics for Scientists and Engineers
Modern Physics for Scientists and Engineers
4th Edition
ISBN: 9781133103721
Author: Stephen T. Thornton, Andrew Rex
Publisher: Cengage Learning
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Chapter 2, Problem 30P
To determine

Coordinates (x, t) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Expert Solution & Answer
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Answer to Problem 30P

The coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Explanation of Solution

Let’s coordinates at the timing of bullet firing are as follows:

In system K (x1, t1) and (x1, t1) in system Kʹ.

Let’s coordinates at the timing of target attack are as follows:

In system K (x2, t2) and (x2, t2) in system Kʹ.

System Kʹ is moving along the x-axis of the system K, therefore coordinates on x-axis in both system will change. Thus (y= y) and (z= z)

In system K coordinates at time of bullet firing are as follows:

x1=1 m, t1=3 ns                                                                                                        (I)

In system K coordinates at time of target attacking are as follows:

x2=1 m+120 m                                                                                                 =121 m        (II)

And

t2=t1+distnace tarvel by bullet speed of the bullet

Substitute the distance travelled equal to 120 m and speed of the bullet 0.98c in above equation to get the time coordinate at time of target attack in system K. Thus,

t2=3×109 s+120 m0.98×3×108 m / s                                                                    =3×109 s+408×109 s=411 ns        (III)

Use Lorentz transformations to get the coordinates in the system Kʹ. Thus,

t1=γ(t1vx1c2)                                                                                                t2=γ(t2vx2c2)x1=γ(x1vt1)x2=γ(x2vt2)        (IV)

Where, γ is the relativistic factor,

y=11v2/c2

Where, v is the velocity of the system Kʹ and c is the velocity of the light.

Substitute v=0.8c in y=11v2/c2 to get relativistic factor. Thus

y=11(0.8c)2/c2                                                                                          =110.64=106=53        (V)

Substitute t1=3 ns, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation t1=γ(t1vx1c2) to get t1. Thus,

t1=53(3 ns 0.8c×1 mc2) =53(3 ns 0.8×1 m3×108 m / s) =53(3 ns 2.67×109 s)=53(3 ns 2.67 ns)

Simplify the above equation

t1=53(3 ns 2.67 ns)                                       =53×0.33 ns=0.55 ns

Substitute t2=411 ns, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation t2=γ(t2vx2c2) to get t2. Thus,

t2=53(411 ns 0.8c×121 mc2) =53(411 ns 0.8×121 m3×108 m / s) =53(411 ns 322.67×109 s)=53(411 ns 322.67 ns)

Simplify the above equation

t1=53(411 ns 322.67 ns)=53×88.33 ns=147.2 ns

Substitute t1=3 ns=3 ×109 s, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation x1=γ(x1vt1) to get x1. Thus,

x1=53(1 m0.8×3×108 m / s×3×109 s)=53(1 m0.72 m)=0.47 m

Substitute t2=411 ns=411×109 s, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation x2=γ(x2vt2) to get x2. Thus,

x2=53(121 m0.8×3×108 m / s×411×109 s)=53(121 m98.64 m)=37.3 m

Conclusion:

Therefore, the coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

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