General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 22, Problem 69E

(a)

To determine

The fractional transmitted energy.

(a)

Expert Solution
Check Mark

Answer to Problem 69E

The fractional transmitted energy is nt=4m1m2(m1+m2)2.

Explanation of Solution

Write the expression for conservation of momentum.

  m1v1i+m2v2i=m1v1f+m2v2f

Here, m1 is the mass of first body, v1i is the initial speed of first body, v1f is the final speed of first body, m2 is the mass of second body, v2i is the initial speed of second body and v2f is the final speed of second body.

Substitute 0 for v2i and rearrange the above equation for value v2f.

  v2f=m1m2(v1iv1f)

Write the expression for conservation of Kinetic energy.

  12m1v1i2+12m2v2i2=12m1v1f2+12m2v2f2        (I)

Substitute 0 for v2i in above expression and divide by 12m1v1i2.

  1=v1f2v1i2+12m2v2f212m1v1i2

Substitute nt for  12m2v2f212m1v1i2 in above expression.

  1=v1f2v1i2+nt

Rearrange the above expression in terms of nt.

  nt=1v1f2v1i2        (II)

Here, nt is fractional transmitted energy.

Substitute 0 for v2i and m1m2(v1iv1f) for v2f in equation (I).

  12m1v1i2+12m2(0)2=12m1v1f2+12m2[m1m2(v1iv1f)]2m1v1i2=m1v1f2+m12m2(v1iv1f)2

Simplify the above expression.

  v1i2v1f2=m1m2(v1iv1f)2(v1i+v1f)(v1iv1f)=m1m2(v1iv1f)(v1iv1f)v1i+v1f=m1m2(v1iv1f)

Rearrange the above expression in terms of v1f/v1i.

  v1fv1i=m1m2m1+m2

Square both sides of above expression.

  (v1fv1i)2=(m1m2m1+m2)2

Substitute (m1m2m1+m2)2 for (v1fv1i)2 in equation (II).

  nt=1(m1m2m1+m2)2=(m1+m2)2(m1m2)2(m1+m2)2

Simplify the above expression.

  nt=4m1m2(m1+m2)2        (III)

Conclusion:

Thus, the fractional transmitted energy is nt=4m1m2(m1+m2)2.

(b)

To determine

The fractional reflected energy.

(b)

Expert Solution
Check Mark

Answer to Problem 69E

The fractional reflected energy is nr=(m1m2m1+m2)2.

Explanation of Solution

Write the expression for fractional reflected energy.

  nr=v1f2v1i2

Substitute (m1m2m1+m2)2 for (v1fv1i)2 in above equation.

  nr=(m1m2m1+m2)2        (IV)

Here, nr is fractional reflected energy.

Conclusion:

Thus, the fractional reflected energy is nr=(m1m2m1+m2)2.

(c)

To determine

The fraction of energy transmitted and reflected.

(c)

Expert Solution
Check Mark

Answer to Problem 69E

The fractional transmitted energy is 1 and fractional reflected energy is 0 for m1=m2; the fractional transmitted energy is 40/121 and fractional reflected energy is 81/121 for m1=10m2; the fractional transmitted energy is 0.04 and fractional reflected energy is 0.96 for m1=100m2.

Explanation of Solution

Conclusion:

For m1=m2:

Substitute m2 for m1 in equation (III).

  nt=4m2m2(m2+m2)2=4m224m22=1

Substitute m2 for m1 in equation (IV).

  nr=(m2m2m2+m2)2=(02m2)2=0

For m1=10m2:

Substitute 10m2 for m1 in equation (III).

  nt=4(10m2)m2(10m2+m2)2=40m22121m22=40121

Substitute 10m2 for m1 in equation (IV).

  nr=(10m2m210m2+m2)2=(9m211m2)2=81121

For m1=100m2:

Substitute 100m2 for m1 in equation (III).

  nt=4(100m2)m2(100m2+m2)2=400m2210201m22=0.04

Substitute 100m2 for m1 in equation (IV).

  nr=(100m2m210m2+m2)2=(99m211m2)2=980110201=0.96

Thus, the fractional transmitted energy is 1 and fractional reflected energy is 0 for m1=m2; the fractional transmitted energy is 40/121 and fractional reflected energy is 81/121 for m1=10m2; the fractional transmitted energy is 0.04 and fractional reflected energy is 0.96 for m1=100m2.

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