The velocity of the particles.

Question

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Answer 1

Question

Chapter 20, Problem 69P

(a)

To determine

The velocity of the particles.

(a)

Expert Solution

Answer to Problem 69P

The velocity of the particles is $6.00i^ m/s$ .

Explanation of Solution

Write the expression from conservation of momentum.

$P→i=P→f$ (I)

Here, $P→i$ is the initial momentum and $P→f$ is the final momentum.

Write the equation for initial momentum.

$P→i=(m1v→1i)+(m2v→2i)$ (II)

Here, $m1,m2$ are the masses, $v→1i,v→2i$ are the initial velocities of the objects.

Write the equation for final momentum at the distance of closest approach.

$P→f=(m1+m2)v→f$ (III)

Here, $v→f$ are the final velocities of the objects.

Rewrite the expression from equation (I) by using (II), (III) in terms of final velocity.

$(m1v→1i)+(m2v→2i)=(m1+m2)v→fv→f=(m1v→1i)+(m2v→2i)(m1+m2)$ (IV)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0i^ m/s$ for $v→1i$ , $0 m/s$ for $v→2i$ in Equation (IV) to find $v→f$ .

Thus, the velocity of the particles is $6.00i^ m/s$ .

(b)

To determine

The closest distance.

(b)

Expert Solution

Answer to Problem 69P

The closest distance is $3.64 m_$ .

Explanation of Solution

Here, initial potential energy is zero.

Write the expression from conservation of energy.

$Ki=Kf+Uf$ (V)

Here, $Ki,Kf$ are the initial and final kinetic energy and $Uf$ is the final potential energy.

Write the equation for initial kinetic energy.

$Ki=12(m1(v1i)2)+12(m2(v2i)2)$ (VI)

Write the equation for final kinetic energy.

$Kf=12(m1+m2)(vf)2$ (VII)

Write the equation for final potential energy.

$Uf=kq1q2r$ (VIII)

Here, $q1,q2$ are the charges, $k$ is Coulomb’s constant, $r$ is distance.

Rewrite the expression from equation (V) by using (VI), (VII) and (VIII) in terms of distance.

$r=2kq1q2(m1+m2)m1m2v1i2$ (IX)

Conclusion:

Substitute, $8.99×109 N-m2/C2$ for $k$ , $15.0 μC$ for $q1$ , $8.50 μC$ for $q2$ , $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (IX) to find $r$ .

$r=2(8.99×109 N-m2/C2)[(15.0 μC)(8.50 μC)(1×10−6 C1 μC)2][{(2.00 g)+(5.00 g)}(1×10−3kg1 g)][{(2.00 g)(5.00 g)}(1×10−3kg1 g)2](21.0 m/s)2=3.64 m$

Thus, the closest distance is $3.64 m_$ .

(c)

To determine

The velocity of first particle.

(c)

Expert Solution

Answer to Problem 69P

The velocity of first particle is $−9.00 i^m/s_$ .

Explanation of Solution

The initial velocity of second particle is zero.

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (X)

Rewrite the expression from equation (IV) by using (X) in terms of final velocity of first particle.

$v1f=(m1−m2m1+m2)v1i$ (XI)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (XI) to find $v1f$ .

$v1f=(2.00 g−5.00 g2.00 g+5.00 g)(21.0 m/s)=−9.00 m/s$

Thus, the velocity of first particle is $−9.00 i^m/s_$ .

(d)

To determine

The velocity of second particle.

(d)

Expert Solution

Answer to Problem 69P

The velocity of second particle is $12.0 i^m/s_$ .

Explanation of Solution

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (XII)

Conclusion:

Substitute, $−9.00 m/s$ for $v1f$ , $21.0 m/s$ for $v1i$ in Equation (XII) to find $v2f$ .

$(21.0 m/s)=v2f−(−9.00 m/s)v2f=12.0 m/s$

Thus, the velocity of second particle is $12.0 i^m/s_$ .

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Answer 2

Question

Chapter 20, Problem 69P

(a)

To determine

The velocity of the particles.

(a)

Expert Solution

Answer to Problem 69P

The velocity of the particles is $6.00i^ m/s$ .

Explanation of Solution

Write the expression from conservation of momentum.

$P→i=P→f$ (I)

Here, $P→i$ is the initial momentum and $P→f$ is the final momentum.

Write the equation for initial momentum.

$P→i=(m1v→1i)+(m2v→2i)$ (II)

Here, $m1,m2$ are the masses, $v→1i,v→2i$ are the initial velocities of the objects.

Write the equation for final momentum at the distance of closest approach.

$P→f=(m1+m2)v→f$ (III)

Here, $v→f$ are the final velocities of the objects.

Rewrite the expression from equation (I) by using (II), (III) in terms of final velocity.

$(m1v→1i)+(m2v→2i)=(m1+m2)v→fv→f=(m1v→1i)+(m2v→2i)(m1+m2)$ (IV)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0i^ m/s$ for $v→1i$ , $0 m/s$ for $v→2i$ in Equation (IV) to find $v→f$ .

Thus, the velocity of the particles is $6.00i^ m/s$ .

(b)

To determine

The closest distance.

(b)

Expert Solution

Answer to Problem 69P

The closest distance is $3.64 m_$ .

Explanation of Solution

Here, initial potential energy is zero.

Write the expression from conservation of energy.

$Ki=Kf+Uf$ (V)

Here, $Ki,Kf$ are the initial and final kinetic energy and $Uf$ is the final potential energy.

Write the equation for initial kinetic energy.

$Ki=12(m1(v1i)2)+12(m2(v2i)2)$ (VI)

Write the equation for final kinetic energy.

$Kf=12(m1+m2)(vf)2$ (VII)

Write the equation for final potential energy.

$Uf=kq1q2r$ (VIII)

Here, $q1,q2$ are the charges, $k$ is Coulomb’s constant, $r$ is distance.

Rewrite the expression from equation (V) by using (VI), (VII) and (VIII) in terms of distance.

$r=2kq1q2(m1+m2)m1m2v1i2$ (IX)

Conclusion:

Substitute, $8.99×109 N-m2/C2$ for $k$ , $15.0 μC$ for $q1$ , $8.50 μC$ for $q2$ , $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (IX) to find $r$ .

$r=2(8.99×109 N-m2/C2)[(15.0 μC)(8.50 μC)(1×10−6 C1 μC)2][{(2.00 g)+(5.00 g)}(1×10−3kg1 g)][{(2.00 g)(5.00 g)}(1×10−3kg1 g)2](21.0 m/s)2=3.64 m$

Thus, the closest distance is $3.64 m_$ .

(c)

To determine

The velocity of first particle.

(c)

Expert Solution

Answer to Problem 69P

The velocity of first particle is $−9.00 i^m/s_$ .

Explanation of Solution

The initial velocity of second particle is zero.

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (X)

Rewrite the expression from equation (IV) by using (X) in terms of final velocity of first particle.

$v1f=(m1−m2m1+m2)v1i$ (XI)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (XI) to find $v1f$ .

$v1f=(2.00 g−5.00 g2.00 g+5.00 g)(21.0 m/s)=−9.00 m/s$

Thus, the velocity of first particle is $−9.00 i^m/s_$ .

(d)

To determine

The velocity of second particle.

(d)

Expert Solution

Answer to Problem 69P

The velocity of second particle is $12.0 i^m/s_$ .

Explanation of Solution

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (XII)

Conclusion:

Substitute, $−9.00 m/s$ for $v1f$ , $21.0 m/s$ for $v1i$ in Equation (XII) to find $v2f$ .

$(21.0 m/s)=v2f−(−9.00 m/s)v2f=12.0 m/s$

Thus, the velocity of second particle is $12.0 i^m/s_$ .

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Answer 3

Question

Chapter 20, Problem 69P

(a)

To determine

The velocity of the particles.

(a)

Expert Solution

Answer to Problem 69P

The velocity of the particles is $6.00i^ m/s$ .

Explanation of Solution

Write the expression from conservation of momentum.

$P→i=P→f$ (I)

Here, $P→i$ is the initial momentum and $P→f$ is the final momentum.

Write the equation for initial momentum.

$P→i=(m1v→1i)+(m2v→2i)$ (II)

Here, $m1,m2$ are the masses, $v→1i,v→2i$ are the initial velocities of the objects.

Write the equation for final momentum at the distance of closest approach.

$P→f=(m1+m2)v→f$ (III)

Here, $v→f$ are the final velocities of the objects.

Rewrite the expression from equation (I) by using (II), (III) in terms of final velocity.

$(m1v→1i)+(m2v→2i)=(m1+m2)v→fv→f=(m1v→1i)+(m2v→2i)(m1+m2)$ (IV)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0i^ m/s$ for $v→1i$ , $0 m/s$ for $v→2i$ in Equation (IV) to find $v→f$ .

Thus, the velocity of the particles is $6.00i^ m/s$ .

(b)

To determine

The closest distance.

(b)

Expert Solution

Answer to Problem 69P

The closest distance is $3.64 m_$ .

Explanation of Solution

Here, initial potential energy is zero.

Write the expression from conservation of energy.

$Ki=Kf+Uf$ (V)

Here, $Ki,Kf$ are the initial and final kinetic energy and $Uf$ is the final potential energy.

Write the equation for initial kinetic energy.

$Ki=12(m1(v1i)2)+12(m2(v2i)2)$ (VI)

Write the equation for final kinetic energy.

$Kf=12(m1+m2)(vf)2$ (VII)

Write the equation for final potential energy.

$Uf=kq1q2r$ (VIII)

Here, $q1,q2$ are the charges, $k$ is Coulomb’s constant, $r$ is distance.

Rewrite the expression from equation (V) by using (VI), (VII) and (VIII) in terms of distance.

$r=2kq1q2(m1+m2)m1m2v1i2$ (IX)

Conclusion:

Substitute, $8.99×109 N-m2/C2$ for $k$ , $15.0 μC$ for $q1$ , $8.50 μC$ for $q2$ , $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (IX) to find $r$ .

$r=2(8.99×109 N-m2/C2)[(15.0 μC)(8.50 μC)(1×10−6 C1 μC)2][{(2.00 g)+(5.00 g)}(1×10−3kg1 g)][{(2.00 g)(5.00 g)}(1×10−3kg1 g)2](21.0 m/s)2=3.64 m$

Thus, the closest distance is $3.64 m_$ .

(c)

To determine

The velocity of first particle.

(c)

Expert Solution

Answer to Problem 69P

The velocity of first particle is $−9.00 i^m/s_$ .

Explanation of Solution

The initial velocity of second particle is zero.

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (X)

Rewrite the expression from equation (IV) by using (X) in terms of final velocity of first particle.

$v1f=(m1−m2m1+m2)v1i$ (XI)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (XI) to find $v1f$ .

$v1f=(2.00 g−5.00 g2.00 g+5.00 g)(21.0 m/s)=−9.00 m/s$

Thus, the velocity of first particle is $−9.00 i^m/s_$ .

(d)

To determine

The velocity of second particle.

(d)

Expert Solution

Answer to Problem 69P

The velocity of second particle is $12.0 i^m/s_$ .

Explanation of Solution

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (XII)

Conclusion:

Substitute, $−9.00 m/s$ for $v1f$ , $21.0 m/s$ for $v1i$ in Equation (XII) to find $v2f$ .

$(21.0 m/s)=v2f−(−9.00 m/s)v2f=12.0 m/s$

Thus, the velocity of second particle is $12.0 i^m/s_$ .

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Answer 4

Question

Chapter 20, Problem 69P

(a)

To determine

The velocity of the particles.

(a)

Expert Solution

Answer to Problem 69P

The velocity of the particles is $6.00i^ m/s$ .

Explanation of Solution

Write the expression from conservation of momentum.

$P→i=P→f$ (I)

Here, $P→i$ is the initial momentum and $P→f$ is the final momentum.

Write the equation for initial momentum.

$P→i=(m1v→1i)+(m2v→2i)$ (II)

Here, $m1,m2$ are the masses, $v→1i,v→2i$ are the initial velocities of the objects.

Write the equation for final momentum at the distance of closest approach.

$P→f=(m1+m2)v→f$ (III)

Here, $v→f$ are the final velocities of the objects.

Rewrite the expression from equation (I) by using (II), (III) in terms of final velocity.

$(m1v→1i)+(m2v→2i)=(m1+m2)v→fv→f=(m1v→1i)+(m2v→2i)(m1+m2)$ (IV)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0i^ m/s$ for $v→1i$ , $0 m/s$ for $v→2i$ in Equation (IV) to find $v→f$ .

Thus, the velocity of the particles is $6.00i^ m/s$ .

(b)

To determine

The closest distance.

(b)

Expert Solution

Answer to Problem 69P

The closest distance is $3.64 m_$ .

Explanation of Solution

Here, initial potential energy is zero.

Write the expression from conservation of energy.

$Ki=Kf+Uf$ (V)

Here, $Ki,Kf$ are the initial and final kinetic energy and $Uf$ is the final potential energy.

Write the equation for initial kinetic energy.

$Ki=12(m1(v1i)2)+12(m2(v2i)2)$ (VI)

Write the equation for final kinetic energy.

$Kf=12(m1+m2)(vf)2$ (VII)

Write the equation for final potential energy.

$Uf=kq1q2r$ (VIII)

Here, $q1,q2$ are the charges, $k$ is Coulomb’s constant, $r$ is distance.

Rewrite the expression from equation (V) by using (VI), (VII) and (VIII) in terms of distance.

$r=2kq1q2(m1+m2)m1m2v1i2$ (IX)

Conclusion:

Substitute, $8.99×109 N-m2/C2$ for $k$ , $15.0 μC$ for $q1$ , $8.50 μC$ for $q2$ , $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (IX) to find $r$ .

$r=2(8.99×109 N-m2/C2)[(15.0 μC)(8.50 μC)(1×10−6 C1 μC)2][{(2.00 g)+(5.00 g)}(1×10−3kg1 g)][{(2.00 g)(5.00 g)}(1×10−3kg1 g)2](21.0 m/s)2=3.64 m$

Thus, the closest distance is $3.64 m_$ .

(c)

To determine

The velocity of first particle.

(c)

Expert Solution

Answer to Problem 69P

The velocity of first particle is $−9.00 i^m/s_$ .

Explanation of Solution

The initial velocity of second particle is zero.

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (X)

Rewrite the expression from equation (IV) by using (X) in terms of final velocity of first particle.

$v1f=(m1−m2m1+m2)v1i$ (XI)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (XI) to find $v1f$ .

$v1f=(2.00 g−5.00 g2.00 g+5.00 g)(21.0 m/s)=−9.00 m/s$

Thus, the velocity of first particle is $−9.00 i^m/s_$ .

(d)

To determine

The velocity of second particle.

(d)

Expert Solution

Answer to Problem 69P

The velocity of second particle is $12.0 i^m/s_$ .

Explanation of Solution

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (XII)

Conclusion:

Substitute, $−9.00 m/s$ for $v1f$ , $21.0 m/s$ for $v1i$ in Equation (XII) to find $v2f$ .

$(21.0 m/s)=v2f−(−9.00 m/s)v2f=12.0 m/s$

Thus, the velocity of second particle is $12.0 i^m/s_$ .

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Answer 5

Chapter 20, Problem 69P

(a)

To determine

The velocity of the particles.

(a)

Expert Solution

Answer to Problem 69P

The velocity of the particles is $6.00i^ m/s$ .

Explanation of Solution

Write the expression from conservation of momentum.

$P→i=P→f$ (I)

Here, $P→i$ is the initial momentum and $P→f$ is the final momentum.

Write the equation for initial momentum.

$P→i=(m1v→1i)+(m2v→2i)$ (II)

Here, $m1,m2$ are the masses, $v→1i,v→2i$ are the initial velocities of the objects.

Write the equation for final momentum at the distance of closest approach.

$P→f=(m1+m2)v→f$ (III)

Here, $v→f$ are the final velocities of the objects.

Rewrite the expression from equation (I) by using (II), (III) in terms of final velocity.

$(m1v→1i)+(m2v→2i)=(m1+m2)v→fv→f=(m1v→1i)+(m2v→2i)(m1+m2)$ (IV)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0i^ m/s$ for $v→1i$ , $0 m/s$ for $v→2i$ in Equation (IV) to find $v→f$ .

Thus, the velocity of the particles is $6.00i^ m/s$ .

(b)

To determine

The closest distance.

(b)

Expert Solution

Answer to Problem 69P

The closest distance is $3.64 m_$ .

Explanation of Solution

Here, initial potential energy is zero.

Write the expression from conservation of energy.

$Ki=Kf+Uf$ (V)

Here, $Ki,Kf$ are the initial and final kinetic energy and $Uf$ is the final potential energy.

Write the equation for initial kinetic energy.

$Ki=12(m1(v1i)2)+12(m2(v2i)2)$ (VI)

Write the equation for final kinetic energy.

$Kf=12(m1+m2)(vf)2$ (VII)

Write the equation for final potential energy.

$Uf=kq1q2r$ (VIII)

Here, $q1,q2$ are the charges, $k$ is Coulomb’s constant, $r$ is distance.

Rewrite the expression from equation (V) by using (VI), (VII) and (VIII) in terms of distance.

$r=2kq1q2(m1+m2)m1m2v1i2$ (IX)

Conclusion:

Substitute, $8.99×109 N-m2/C2$ for $k$ , $15.0 μC$ for $q1$ , $8.50 μC$ for $q2$ , $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (IX) to find $r$ .

$r=2(8.99×109 N-m2/C2)[(15.0 μC)(8.50 μC)(1×10−6 C1 μC)2][{(2.00 g)+(5.00 g)}(1×10−3kg1 g)][{(2.00 g)(5.00 g)}(1×10−3kg1 g)2](21.0 m/s)2=3.64 m$

Thus, the closest distance is $3.64 m_$ .

(c)

To determine

The velocity of first particle.

(c)

Expert Solution

Answer to Problem 69P

The velocity of first particle is $−9.00 i^m/s_$ .

Explanation of Solution

The initial velocity of second particle is zero.

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (X)

Rewrite the expression from equation (IV) by using (X) in terms of final velocity of first particle.

$v1f=(m1−m2m1+m2)v1i$ (XI)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (XI) to find $v1f$ .

$v1f=(2.00 g−5.00 g2.00 g+5.00 g)(21.0 m/s)=−9.00 m/s$

Thus, the velocity of first particle is $−9.00 i^m/s_$ .

(d)

To determine

The velocity of second particle.

(d)

Expert Solution

Answer to Problem 69P

The velocity of second particle is $12.0 i^m/s_$ .

Explanation of Solution

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (XII)

Conclusion:

Substitute, $−9.00 m/s$ for $v1f$ , $21.0 m/s$ for $v1i$ in Equation (XII) to find $v2f$ .

$(21.0 m/s)=v2f−(−9.00 m/s)v2f=12.0 m/s$

Thus, the velocity of second particle is $12.0 i^m/s_$ .

Answer 6

Chapter 20, Problem 69P

(a)

To determine

The velocity of the particles.

(a)

Expert Solution

Answer to Problem 69P

The velocity of the particles is $6.00i^ m/s$ .

Explanation of Solution

Write the expression from conservation of momentum.

$P→i=P→f$ (I)

Here, $P→i$ is the initial momentum and $P→f$ is the final momentum.

Write the equation for initial momentum.

$P→i=(m1v→1i)+(m2v→2i)$ (II)

Here, $m1,m2$ are the masses, $v→1i,v→2i$ are the initial velocities of the objects.

Write the equation for final momentum at the distance of closest approach.

$P→f=(m1+m2)v→f$ (III)

Here, $v→f$ are the final velocities of the objects.

Rewrite the expression from equation (I) by using (II), (III) in terms of final velocity.

$(m1v→1i)+(m2v→2i)=(m1+m2)v→fv→f=(m1v→1i)+(m2v→2i)(m1+m2)$ (IV)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0i^ m/s$ for $v→1i$ , $0 m/s$ for $v→2i$ in Equation (IV) to find $v→f$ .

Thus, the velocity of the particles is $6.00i^ m/s$ .

Answer 7

Chapter 20, Problem 69P

(a)

To determine

The velocity of the particles.

Answer 8

(a)

Expert Solution

Answer to Problem 69P

The velocity of the particles is $6.00i^ m/s$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression from conservation of momentum.

$P→i=P→f$ (I)

Here, $P→i$ is the initial momentum and $P→f$ is the final momentum.

Write the equation for initial momentum.

$P→i=(m1v→1i)+(m2v→2i)$ (II)

Here, $m1,m2$ are the masses, $v→1i,v→2i$ are the initial velocities of the objects.

Write the equation for final momentum at the distance of closest approach.

$P→f=(m1+m2)v→f$ (III)

Here, $v→f$ are the final velocities of the objects.

Rewrite the expression from equation (I) by using (II), (III) in terms of final velocity.

$(m1v→1i)+(m2v→2i)=(m1+m2)v→fv→f=(m1v→1i)+(m2v→2i)(m1+m2)$ (IV)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0i^ m/s$ for $v→1i$ , $0 m/s$ for $v→2i$ in Equation (IV) to find $v→f$ .

Thus, the velocity of the particles is $6.00i^ m/s$ .

Answer 13

(b)

To determine

The closest distance.

(b)

Expert Solution

Answer to Problem 69P

The closest distance is $3.64 m_$ .

Explanation of Solution

Here, initial potential energy is zero.

Write the expression from conservation of energy.

$Ki=Kf+Uf$ (V)

Here, $Ki,Kf$ are the initial and final kinetic energy and $Uf$ is the final potential energy.

Write the equation for initial kinetic energy.

$Ki=12(m1(v1i)2)+12(m2(v2i)2)$ (VI)

Write the equation for final kinetic energy.

$Kf=12(m1+m2)(vf)2$ (VII)

Write the equation for final potential energy.

$Uf=kq1q2r$ (VIII)

Here, $q1,q2$ are the charges, $k$ is Coulomb’s constant, $r$ is distance.

Rewrite the expression from equation (V) by using (VI), (VII) and (VIII) in terms of distance.

$r=2kq1q2(m1+m2)m1m2v1i2$ (IX)

Conclusion:

Substitute, $8.99×109 N-m2/C2$ for $k$ , $15.0 μC$ for $q1$ , $8.50 μC$ for $q2$ , $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (IX) to find $r$ .

$r=2(8.99×109 N-m2/C2)[(15.0 μC)(8.50 μC)(1×10−6 C1 μC)2][{(2.00 g)+(5.00 g)}(1×10−3kg1 g)][{(2.00 g)(5.00 g)}(1×10−3kg1 g)2](21.0 m/s)2=3.64 m$

Thus, the closest distance is $3.64 m_$ .

Answer 14

(b)

To determine

The closest distance.

Answer 15

(b)

Expert Solution

Answer to Problem 69P

The closest distance is $3.64 m_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Here, initial potential energy is zero.

Write the expression from conservation of energy.

$Ki=Kf+Uf$ (V)

Here, $Ki,Kf$ are the initial and final kinetic energy and $Uf$ is the final potential energy.

Write the equation for initial kinetic energy.

$Ki=12(m1(v1i)2)+12(m2(v2i)2)$ (VI)

Write the equation for final kinetic energy.

$Kf=12(m1+m2)(vf)2$ (VII)

Write the equation for final potential energy.

$Uf=kq1q2r$ (VIII)

Here, $q1,q2$ are the charges, $k$ is Coulomb’s constant, $r$ is distance.

Rewrite the expression from equation (V) by using (VI), (VII) and (VIII) in terms of distance.

$r=2kq1q2(m1+m2)m1m2v1i2$ (IX)

Conclusion:

Substitute, $8.99×109 N-m2/C2$ for $k$ , $15.0 μC$ for $q1$ , $8.50 μC$ for $q2$ , $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (IX) to find $r$ .

$r=2(8.99×109 N-m2/C2)[(15.0 μC)(8.50 μC)(1×10−6 C1 μC)2][{(2.00 g)+(5.00 g)}(1×10−3kg1 g)][{(2.00 g)(5.00 g)}(1×10−3kg1 g)2](21.0 m/s)2=3.64 m$

Thus, the closest distance is $3.64 m_$ .

Answer 20

(c)

To determine

The velocity of first particle.

(c)

Expert Solution

Answer to Problem 69P

The velocity of first particle is $−9.00 i^m/s_$ .

Explanation of Solution

The initial velocity of second particle is zero.

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (X)

Rewrite the expression from equation (IV) by using (X) in terms of final velocity of first particle.

$v1f=(m1−m2m1+m2)v1i$ (XI)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (XI) to find $v1f$ .

$v1f=(2.00 g−5.00 g2.00 g+5.00 g)(21.0 m/s)=−9.00 m/s$

Thus, the velocity of first particle is $−9.00 i^m/s_$ .

Answer 21

(c)

To determine

The velocity of first particle.

Answer 22

(c)

Expert Solution

Answer to Problem 69P

The velocity of first particle is $−9.00 i^m/s_$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The initial velocity of second particle is zero.

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (X)

Rewrite the expression from equation (IV) by using (X) in terms of final velocity of first particle.

$v1f=(m1−m2m1+m2)v1i$ (XI)

Conclusion:

Substitute, $2.00 g$ for $m1$ , $5.00 g$ for $m2$ , $21.0 m/s$ for $v1i$ in Equation (XI) to find $v1f$ .

$v1f=(2.00 g−5.00 g2.00 g+5.00 g)(21.0 m/s)=−9.00 m/s$

Thus, the velocity of first particle is $−9.00 i^m/s_$ .

Answer 27

(d)

To determine

The velocity of second particle.

(d)

Expert Solution

Answer to Problem 69P

The velocity of second particle is $12.0 i^m/s_$ .

Explanation of Solution

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (XII)

Conclusion:

Substitute, $−9.00 m/s$ for $v1f$ , $21.0 m/s$ for $v1i$ in Equation (XII) to find $v2f$ .

$(21.0 m/s)=v2f−(−9.00 m/s)v2f=12.0 m/s$

Thus, the velocity of second particle is $12.0 i^m/s_$ .

Answer 28

(d)

To determine

The velocity of second particle.

Answer 29

(d)

Expert Solution

Answer to Problem 69P

The velocity of second particle is $12.0 i^m/s_$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write the expression from relative velocity equation.

$v1i=v2f−v1f$ (XII)

Conclusion:

Substitute, $−9.00 m/s$ for $v1f$ , $21.0 m/s$ for $v1i$ in Equation (XII) to find $v2f$ .

$(21.0 m/s)=v2f−(−9.00 m/s)v2f=12.0 m/s$

Thus, the velocity of second particle is $12.0 i^m/s_$ .

Answer 34

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The velocity of the particles.

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Explanation of Solution

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Explanation of Solution

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Answer to Problem 69P

Explanation of Solution

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Explanation of Solution

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