Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 25, Problem 55AP

(a)

To determine

The velocity of the particles at the instant of closest approach.

(a)

Expert Solution
Check Mark

Answer to Problem 55AP

The velocity of the particles at the instant of the closest approach is 6.00ˆim/sec.

Explanation of Solution

Write the expression for the velocity of the closest approach using the law of conservation of energy.

    vc=m1v1+m2v2m1+m2                              (I)

Here, vc is the velocity at the closest approach, m1 is the mass of the particle, m2 is the mass of the second particle, v1 is the velocity of the first particle AND v2 is the velocity of the second particle.

Conclusion:

Substitute 2.00gm for m1, 5.00gm for m2, 21.0ˆim/sec for v1 and 0.00m/sec for v2 in equation (I) to calculate the value of vc.

    vc=(2.00gm×103kg1gm)(21.0ˆim/sec)+(5.00gm×103kg1gm)(0m/sec)(2.00gm×103kg1gm)+(5.00gm×103kg1gm)=(42.0ˆim/sec)7=6.00ˆim/sec

Therefore, the velocity of the particles at the instant of the closest approach is 6.00ˆim/sec.

(b)

To determine

The distance of the closest approach.

(b)

Expert Solution
Check Mark

Answer to Problem 55AP

The distance of the closest approach is 3.64m.

Explanation of Solution

Write the expression for the law of conservation of energy.

    (KEf+PEf)=(KEi+PEi)                          (II)

Here, KEf is the final kinetic energy, KEi is the initial kinetic energy, PEf is the final potential energy and PEi is the initial potential energy.

Substitute 12(m1+m2)vc2 for KEf, keq1q2r for PEf, 12(m1v12+m2v22) for KEi and 0 for PEi in equation (II) to get the expression for r.

    (12(m1+m2)vc2+keq1q2r)=(12(m1v12+m2v22)+0)12(m1+m2)vc2+keq1q2r=12(m1v12+m2v22)                           (III)

Here, q1 is the charge on the first particle, q2 is the charge on the second particle, ke is the Coulomb’s constant and r is the distance of closest approach.

Conclusion:

Substitute 2.00gm for m1, 5.00gm for m2, 21.0ˆim/sec for v1, 0.00m/sec for v2, 6.0ˆim/sec for vc, 15.0μC for q1, 8.50μC for q2 and 8.9876×109N for ke in equation (III) to calculate the value of r.

    [12((2.00gm×103kg1gm)+(5.00gm103kg1gm))(6.0ˆim/sec)2+((8.9876×109N)(15.0μC×106C1μC)(8.50μC×106C1μC)r)]=[12(((2.00gm103kg1gm)(21.0ˆim/sec)2)+((5.00gm103kg1gm)(0.00m/sec)2))]0.126kg-m2/sec2+1.13475N-C2r=0.441kg-m2/sec21.145919N-C2r=0.441kg-m2/sec20.126kg-m2/sec2r=1.145919N-C20.315kg-m2/sec2

Solving further,

    r=3.64m

Therefore, the distance of the closest approach is 3.64m.

(c)

To determine

The velocity of 2.00gm particle.

(c)

Expert Solution
Check Mark

Answer to Problem 55AP

The velocity of 2.00gm particle is 9.00ˆim/s.

Explanation of Solution

Since the collision between the charged particle is elastic. Hence, the expression for the final velocity of the first particle is,

    v1f=[(m1m2m1+m2)v1+(2m2m1+m2)v2]                                 (IV)

Here, v1f is the final velocity of the initial particle.

Conclusion:

Substitute 2.00gm for m1, 5.00gm for m2, 21.0ˆim/sec for v1 and 0.00m/sec for v2 in equation (IV) to calculate the value of r.

    v1f=[((2.00gm×103kg1gm)(5.00gm×103kg1gm)(2.00gm×103kg1gm)+(5.00gm×103kg1gm))21.0ˆim/sec+(2(5.00gm×103kg1gm)(2.00gm×103kg1gm)+(5.00gm×103kg1gm))0.0m/sec]=37(21.0ˆim/sec)=9.00ˆim/s

Therefore, the velocity of the 2.00gm particle is 9.00ˆim/s.

(d)

To determine

The velocity of 5.00gm particle.

(d)

Expert Solution
Check Mark

Answer to Problem 55AP

The velocity of 5.00gm particle is 12.0ˆim/sec.

Explanation of Solution

Since the collision between the charged particle is elastic. Hence, the expression for the final velocity of the second particle is,

    v2f=[(m1m2m1+m2)v2+(2m1m1+m2)v1]                                 (V)

Here, v2f is the final velocity of the second particle.

Conclusion:

Substitute 2.00gm for m1, 5.00gm for m2, 21.0ˆim/sec for v1 and 0.00m/sec for v2 in equation (V) to calculate the value of r.

    v2f=[((2.00gm×103kg1gm)(5.00gm×103kg1gm)(2.00gm×103kg1gm)+(5.00gm×103kg1gm))0.0m/sec+(2(2.00gm×103kg1gm)(2.00gm×103kg1gm)+(5.00gm×103kg1gm))21.0ˆim/sec]=47(21.0ˆim/sec)=12.0ˆim/sec

Therefore, the velocity of the 5.00gm particle is 12.0ˆim/sec.

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Chapter 25 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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