Concept explainers
(a)
The velocity of the particles at the instant of closest approach.
(a)
Answer to Problem 55AP
The velocity of the particles at the instant of the closest approach is 6.00ˆi m/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 2.00 gm for m1, 5.00 gm for m2, 21.0ˆi m/sec for v1 and 0.00 m/sec for v2 in equation (I) to calculate the value of vc.
vc=(2.00 gm×10−3 kg1 gm)(21.0ˆi m/sec)+(5.00 gm×10−3 kg1 gm)(0 m/sec)(2.00 gm×10−3 kg1 gm)+(5.00 gm×10−3 kg1 gm)=(42.0ˆi m/sec)7=6.00ˆi m/sec
Therefore, the velocity of the particles at the instant of the closest approach is 6.00ˆi m/sec.
(b)
The distance of the closest approach.
(b)
Answer to Problem 55AP
The distance of the closest approach is 3.64 m.
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 2.00 gm for m1, 5.00 gm for m2, 21.0ˆi m/sec for v1, 0.00 m/sec for v2, 6.0ˆi m/sec for vc, 15.0 μC for q1, 8.50 μC for q2 and 8.9876×109 N for ke in equation (III) to calculate the value of r.
[12((2.00 gm×10−3 kg1 gm)+(5.00 gm10−3 kg1 gm))(6.0ˆi m/sec)2+((8.9876×109 N)(15.0 μC×10−6 C1 μC)(8.50 μC×10−6 C1 μC)r)]=[12(((2.00 gm10−3 kg1 gm)(21.0ˆi m/sec)2)+((5.00 gm10−3 kg1 gm)(0.00 m/sec)2))]0.126 kg-m2/sec2+1.13475 N-C2r= 0.441 kg-m2/sec21.145919 N-C2r=0.441 kg-m2/sec2−0.126 kg-m2/sec2r=1.145919 N-C20.315 kg-m2/sec2
Solving further,
r=3.64 m
Therefore, the distance of the closest approach is 3.64 m.
(c)
The velocity of 2.00 gm particle.
(c)
Answer to Problem 55AP
The velocity of 2.00 gm particle is −9.00ˆi m/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=[(m1−m2m1+m2)v1+(2m2m1+m2)v2] (IV)
Here, v1f is the final velocity of the initial particle.
Conclusion:
Substitute 2.00 gm for m1, 5.00 gm for m2, 21.0ˆi m/sec for v1 and 0.00 m/sec for v2 in equation (IV) to calculate the value of r.
v1f=[((2.00 gm×10−3 kg1 gm)−(5.00 gm×10−3 kg1 gm)(2.00 gm×10−3 kg1 gm)+(5.00 gm×10−3 kg1 gm))21.0ˆi m/sec+(2(5.00 gm×10−3 kg1 gm)(2.00 gm×10−3 kg1 gm)+(5.00 gm×10−3 kg1 gm))0.0 m/sec]=−37(21.0ˆi m/sec)=−9.00ˆi m/s
Therefore, the velocity of the 2.00 gm particle is −9.00ˆi m/s.
(d)
The velocity of 5.00 gm particle.
(d)
Answer to Problem 55AP
The velocity of 5.00 gm particle is 12.0ˆi m/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=[(m1−m2m1+m2)v2+(2m1m1+m2)v1] (V)
Here, v2f is the final velocity of the second particle.
Conclusion:
Substitute 2.00 gm for m1, 5.00 gm for m2, 21.0ˆi m/sec for v1 and 0.00 m/sec for v2 in equation (V) to calculate the value of r.
v2f=[((2.00 gm×10−3 kg1 gm)−(5.00 gm×10−3 kg1 gm)(2.00 gm×10−3 kg1 gm)+(5.00 gm×10−3 kg1 gm))0.0 m/sec+(2(2.00 gm×10−3 kg1 gm)(2.00 gm×10−3 kg1 gm)+(5.00 gm×10−3 kg1 gm))21.0ˆi m/sec]=47(21.0ˆi m/sec)=12.0ˆi m/sec
Therefore, the velocity of the 5.00 gm particle is 12.0ˆi m/sec.
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Chapter 25 Solutions
Physics for Scientists and Engineers with Modern Physics, Technology Update
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