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Chapter 28, Problem 40P

(a)

To determine

The minimum average distance by which the target was missed.

(a)

Expert Solution
Check Mark

Answer to Problem 40P

The minimum average miss distance is (2m)1/2(2Hg)1/4.

Explanation of Solution

The initial height of the pellet is H, its mass is m and let the target be the origin

Write the given expression for deviation from the target

  Δxf=(Δxi)+(Δvxt)                                                                                                    (I)

Here, Δxf is the uncertainty in position, Δxf is the uncertainty in initial position, Δvx is the uncertainty in velocity and t is the time taken.

Write the uncertainty relation.

  Δxi(mΔvx)=2

Here, Δxf is the uncertainty in position, Δvx is the uncertainty in velocity, m is the mass and is the reduced Planck’s constant.

Substitute (I) and rearrange

  Δvx=2(mΔxi)                                                                                                          (II)

Write the expression for change in velocity using equation of motion

    s=ut+12at2                                                                                                      (III)

Here, u is initial velocity, a is the acceleration, s is the distance travelled.

Substitute 0 for u, g for a and H for s in (III) and rearrange for t

Write the expression for time taken to reach distance H

  t=2Hg                                                                                                             (IV)

Substitute (IV) and (II) in (I)

  Δxf=(Δxi)+2mΔxi2Hg                                                                                       (V)

Find the uncertainty in initial position by differentiating the above equation with respect to Δxi

  d(Δxf)d(Δxi)=012m(Δxi)22Hg=0

Since the second derivative of equation (V) is positive. The above found Δxi is the minimum.

Rearrange for Δxi

  Δxi=(2m)1/2(2Hg)1/4                                                                                            (V)

Conclusion

Substitute Δxi in (V) to find Δxf

  Δxf=(2m)1/2(2Hg)1/4+2m2Hg((2m)1/2(2Hg)1/4)1=2((2m)1/2(2Hg)1/4)=(2m)1/2(2Hg)1/4

Thus, the minimum average miss distance is (2m)1/2(2Hg)1/4.

(b)

To determine

Find the average miss distance given the mass and height.

(b)

Expert Solution
Check Mark

Answer to Problem 40P

The average miss distance is 5.19×1016 m.

Explanation of Solution

The mass of the pellet is 0.500g and the height from which it is dropped is 2.00m.

Substitute 0.500g for m2.00m for H, 9.8ms2 for g and 1.054×1034 Js for in (V) to find Δxf

    Δxf=(2×1.054×1034 kgm2s10.500g)1/2(2×2.00m9.8ms2)1/4=(2×1.054×1034 kgm2s10.500×103kg)1/2(2×2.00m9.8ms2)1/4=5.19×1016 m

Thus, the average miss distance is 5.19×1016 m.

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Chapter 28 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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