Prove that ξ lab = ( π − θ cm ) 2

Question

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Answer 1

Question

Chapter 14, Problem 14.28P

(a)

To determine

Prove that $ξlab=(π−θcm)2$

(a)

Expert Solution

Answer to Problem 14.28P

The angle is $ξlab=(π−θcm)2$ .

Explanation of Solution

Classical Mechanics, Chapter 14, Problem 14.28P

Consider particle 1 with initial momentum $plab 1$ and particle 2 with zero initial momentum. Final momentum of particle 1 is $p'lab 1$ and final momentum of the particle 2 is $p'lab 2$ .

According to law of conservation of momentum

$plab 1+0=p'lab 1+p'lab 2p'lab 2=plab 1−p'lab 1$

From the figure, $AC$ denotes $plab 1$ , $AD$ represents $p'lab 1$ and $p'lab 2$ is represented by $AC−AD$

From the triangular law of vectors, $AC−AD=DC$

$BD$ and $BC$ is the radius of the circle. Thus the angles opposite to these sides are equal.

From the triangle $ΔBCD$ ,

$∠BCD=∠BDC=ξlab$

And,

$∠DBC=θcm$

The sum of all three angles of the triangle is equal to $π rad$ .

Therefore for $ΔBCD$

$ξlab+ξlab+θcm=π2ξlab+θcm=πξlab=π−θcm2$

Conclusion:

The angle is $ξlab=(π−θcm)2$ .

(b)

To determine

Show that in the case of equal masses, the angle of these outgoing particles in elastic collision is $90°$ .

(b)

Expert Solution

Answer to Problem 14.28P

For the collision of equal masses, $ξlab+θlab=π2$

Explanation of Solution

Write down the expression connecting $θlab$ and $θcm$

$tanθlab=sinθcmλ+cosθcm$

$λ=1$ for $m1=m2$ .

Then,

$tanθlab=sinθcm1+cosθcm=2sinθcm2cosθcm22cos2θcm2=tanθcm2$

That is,

$θlab=θcm2θcm=2θlab$

Substitute $π−2ξlab$ for $θcm$

$ξlab=π−θcm2ξlab=π2−θlabπ2=ξlab+θlab$

Hence proved.

Conclusion:

For the collision of equal masses, $ξlab+θlab=π2$

(c)

To determine

Prove $ξlab+θlab=π2$ using law of conservation of momentum.

(c)

Expert Solution

Answer to Problem 14.28P

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

Explanation of Solution

Apply law of conservation of momentum in horizontal direction

$plab1=p'lab1cosθlab+p'lab2cosξlab$

Along vertical direction,

$0=p'lab1sinθlab−p'labsinξlabp'lab1sinθlab=−p'lab2sinξlab$

Apply the law of conservation of energy

$p2lab12m1=p'2lab12m1+p'2lab22m2$

Substitute $m1=m2$ ,

$p2lab1=p'2lab1+p'2lab2$

Then,

$p'2lab1+p'2lab2=(p'lab1cosθlab+p'lab2cosξlab)2p'2lab1+p'2lab2=p'2lab1cos2θlab+p'2lab2cos2ξlab+2p'lab1cosθlabp'lab2cosξlab0=p'2lab1(1−cos2θlab)+p'2lab2(1−cos2ξlab)−2p'lab1p'lab2cosθlabcosξlab0=p'2lab1sin2θlab+p'2lab2sin2cos2ξlab−p'lab1p'lab2cosθlabcosξlab$

On further simplification,

$p'2lab1sin2θlab+p'2lab2sin2cos2ξlab=p'lab1p'lab2cosθlabcosξlab(p'2lab1sin2θlab+p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)(p'2lab1sin2θlab−p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)0=2p'lab1p'lab2(cosξlab+cosθlab)$

Here $2p'lab1p'lab2≠0$ . Therefore $cosξlab+cosθlab=0$

Take cosine inverse on both sides

$ξlab+θlab=cos−10=π2$

Conclusion:

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

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Answer 2

Question

Chapter 14, Problem 14.28P

(a)

To determine

Prove that $ξlab=(π−θcm)2$

(a)

Expert Solution

Answer to Problem 14.28P

The angle is $ξlab=(π−θcm)2$ .

Explanation of Solution

Classical Mechanics, Chapter 14, Problem 14.28P

Consider particle 1 with initial momentum $plab 1$ and particle 2 with zero initial momentum. Final momentum of particle 1 is $p'lab 1$ and final momentum of the particle 2 is $p'lab 2$ .

According to law of conservation of momentum

$plab 1+0=p'lab 1+p'lab 2p'lab 2=plab 1−p'lab 1$

From the figure, $AC$ denotes $plab 1$ , $AD$ represents $p'lab 1$ and $p'lab 2$ is represented by $AC−AD$

From the triangular law of vectors, $AC−AD=DC$

$BD$ and $BC$ is the radius of the circle. Thus the angles opposite to these sides are equal.

From the triangle $ΔBCD$ ,

$∠BCD=∠BDC=ξlab$

And,

$∠DBC=θcm$

The sum of all three angles of the triangle is equal to $π rad$ .

Therefore for $ΔBCD$

$ξlab+ξlab+θcm=π2ξlab+θcm=πξlab=π−θcm2$

Conclusion:

The angle is $ξlab=(π−θcm)2$ .

(b)

To determine

Show that in the case of equal masses, the angle of these outgoing particles in elastic collision is $90°$ .

(b)

Expert Solution

Answer to Problem 14.28P

For the collision of equal masses, $ξlab+θlab=π2$

Explanation of Solution

Write down the expression connecting $θlab$ and $θcm$

$tanθlab=sinθcmλ+cosθcm$

$λ=1$ for $m1=m2$ .

Then,

$tanθlab=sinθcm1+cosθcm=2sinθcm2cosθcm22cos2θcm2=tanθcm2$

That is,

$θlab=θcm2θcm=2θlab$

Substitute $π−2ξlab$ for $θcm$

$ξlab=π−θcm2ξlab=π2−θlabπ2=ξlab+θlab$

Hence proved.

Conclusion:

For the collision of equal masses, $ξlab+θlab=π2$

(c)

To determine

Prove $ξlab+θlab=π2$ using law of conservation of momentum.

(c)

Expert Solution

Answer to Problem 14.28P

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

Explanation of Solution

Apply law of conservation of momentum in horizontal direction

$plab1=p'lab1cosθlab+p'lab2cosξlab$

Along vertical direction,

$0=p'lab1sinθlab−p'labsinξlabp'lab1sinθlab=−p'lab2sinξlab$

Apply the law of conservation of energy

$p2lab12m1=p'2lab12m1+p'2lab22m2$

Substitute $m1=m2$ ,

$p2lab1=p'2lab1+p'2lab2$

Then,

$p'2lab1+p'2lab2=(p'lab1cosθlab+p'lab2cosξlab)2p'2lab1+p'2lab2=p'2lab1cos2θlab+p'2lab2cos2ξlab+2p'lab1cosθlabp'lab2cosξlab0=p'2lab1(1−cos2θlab)+p'2lab2(1−cos2ξlab)−2p'lab1p'lab2cosθlabcosξlab0=p'2lab1sin2θlab+p'2lab2sin2cos2ξlab−p'lab1p'lab2cosθlabcosξlab$

On further simplification,

$p'2lab1sin2θlab+p'2lab2sin2cos2ξlab=p'lab1p'lab2cosθlabcosξlab(p'2lab1sin2θlab+p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)(p'2lab1sin2θlab−p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)0=2p'lab1p'lab2(cosξlab+cosθlab)$

Here $2p'lab1p'lab2≠0$ . Therefore $cosξlab+cosθlab=0$

Take cosine inverse on both sides

$ξlab+θlab=cos−10=π2$

Conclusion:

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

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Answer 3

Question

Chapter 14, Problem 14.28P

(a)

To determine

Prove that $ξlab=(π−θcm)2$

(a)

Expert Solution

Answer to Problem 14.28P

The angle is $ξlab=(π−θcm)2$ .

Explanation of Solution

Classical Mechanics, Chapter 14, Problem 14.28P

Consider particle 1 with initial momentum $plab 1$ and particle 2 with zero initial momentum. Final momentum of particle 1 is $p'lab 1$ and final momentum of the particle 2 is $p'lab 2$ .

According to law of conservation of momentum

$plab 1+0=p'lab 1+p'lab 2p'lab 2=plab 1−p'lab 1$

From the figure, $AC$ denotes $plab 1$ , $AD$ represents $p'lab 1$ and $p'lab 2$ is represented by $AC−AD$

From the triangular law of vectors, $AC−AD=DC$

$BD$ and $BC$ is the radius of the circle. Thus the angles opposite to these sides are equal.

From the triangle $ΔBCD$ ,

$∠BCD=∠BDC=ξlab$

And,

$∠DBC=θcm$

The sum of all three angles of the triangle is equal to $π rad$ .

Therefore for $ΔBCD$

$ξlab+ξlab+θcm=π2ξlab+θcm=πξlab=π−θcm2$

Conclusion:

The angle is $ξlab=(π−θcm)2$ .

(b)

To determine

Show that in the case of equal masses, the angle of these outgoing particles in elastic collision is $90°$ .

(b)

Expert Solution

Answer to Problem 14.28P

For the collision of equal masses, $ξlab+θlab=π2$

Explanation of Solution

Write down the expression connecting $θlab$ and $θcm$

$tanθlab=sinθcmλ+cosθcm$

$λ=1$ for $m1=m2$ .

Then,

$tanθlab=sinθcm1+cosθcm=2sinθcm2cosθcm22cos2θcm2=tanθcm2$

That is,

$θlab=θcm2θcm=2θlab$

Substitute $π−2ξlab$ for $θcm$

$ξlab=π−θcm2ξlab=π2−θlabπ2=ξlab+θlab$

Hence proved.

Conclusion:

For the collision of equal masses, $ξlab+θlab=π2$

(c)

To determine

Prove $ξlab+θlab=π2$ using law of conservation of momentum.

(c)

Expert Solution

Answer to Problem 14.28P

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

Explanation of Solution

Apply law of conservation of momentum in horizontal direction

$plab1=p'lab1cosθlab+p'lab2cosξlab$

Along vertical direction,

$0=p'lab1sinθlab−p'labsinξlabp'lab1sinθlab=−p'lab2sinξlab$

Apply the law of conservation of energy

$p2lab12m1=p'2lab12m1+p'2lab22m2$

Substitute $m1=m2$ ,

$p2lab1=p'2lab1+p'2lab2$

Then,

$p'2lab1+p'2lab2=(p'lab1cosθlab+p'lab2cosξlab)2p'2lab1+p'2lab2=p'2lab1cos2θlab+p'2lab2cos2ξlab+2p'lab1cosθlabp'lab2cosξlab0=p'2lab1(1−cos2θlab)+p'2lab2(1−cos2ξlab)−2p'lab1p'lab2cosθlabcosξlab0=p'2lab1sin2θlab+p'2lab2sin2cos2ξlab−p'lab1p'lab2cosθlabcosξlab$

On further simplification,

$p'2lab1sin2θlab+p'2lab2sin2cos2ξlab=p'lab1p'lab2cosθlabcosξlab(p'2lab1sin2θlab+p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)(p'2lab1sin2θlab−p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)0=2p'lab1p'lab2(cosξlab+cosθlab)$

Here $2p'lab1p'lab2≠0$ . Therefore $cosξlab+cosθlab=0$

Take cosine inverse on both sides

$ξlab+θlab=cos−10=π2$

Conclusion:

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

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Answer 4

Question

Chapter 14, Problem 14.28P

(a)

To determine

Prove that $ξlab=(π−θcm)2$

(a)

Expert Solution

Answer to Problem 14.28P

The angle is $ξlab=(π−θcm)2$ .

Explanation of Solution

Classical Mechanics, Chapter 14, Problem 14.28P

Consider particle 1 with initial momentum $plab 1$ and particle 2 with zero initial momentum. Final momentum of particle 1 is $p'lab 1$ and final momentum of the particle 2 is $p'lab 2$ .

According to law of conservation of momentum

$plab 1+0=p'lab 1+p'lab 2p'lab 2=plab 1−p'lab 1$

From the figure, $AC$ denotes $plab 1$ , $AD$ represents $p'lab 1$ and $p'lab 2$ is represented by $AC−AD$

From the triangular law of vectors, $AC−AD=DC$

$BD$ and $BC$ is the radius of the circle. Thus the angles opposite to these sides are equal.

From the triangle $ΔBCD$ ,

$∠BCD=∠BDC=ξlab$

And,

$∠DBC=θcm$

The sum of all three angles of the triangle is equal to $π rad$ .

Therefore for $ΔBCD$

$ξlab+ξlab+θcm=π2ξlab+θcm=πξlab=π−θcm2$

Conclusion:

The angle is $ξlab=(π−θcm)2$ .

(b)

To determine

Show that in the case of equal masses, the angle of these outgoing particles in elastic collision is $90°$ .

(b)

Expert Solution

Answer to Problem 14.28P

For the collision of equal masses, $ξlab+θlab=π2$

Explanation of Solution

Write down the expression connecting $θlab$ and $θcm$

$tanθlab=sinθcmλ+cosθcm$

$λ=1$ for $m1=m2$ .

Then,

$tanθlab=sinθcm1+cosθcm=2sinθcm2cosθcm22cos2θcm2=tanθcm2$

That is,

$θlab=θcm2θcm=2θlab$

Substitute $π−2ξlab$ for $θcm$

$ξlab=π−θcm2ξlab=π2−θlabπ2=ξlab+θlab$

Hence proved.

Conclusion:

For the collision of equal masses, $ξlab+θlab=π2$

(c)

To determine

Prove $ξlab+θlab=π2$ using law of conservation of momentum.

(c)

Expert Solution

Answer to Problem 14.28P

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

Explanation of Solution

Apply law of conservation of momentum in horizontal direction

$plab1=p'lab1cosθlab+p'lab2cosξlab$

Along vertical direction,

$0=p'lab1sinθlab−p'labsinξlabp'lab1sinθlab=−p'lab2sinξlab$

Apply the law of conservation of energy

$p2lab12m1=p'2lab12m1+p'2lab22m2$

Substitute $m1=m2$ ,

$p2lab1=p'2lab1+p'2lab2$

Then,

$p'2lab1+p'2lab2=(p'lab1cosθlab+p'lab2cosξlab)2p'2lab1+p'2lab2=p'2lab1cos2θlab+p'2lab2cos2ξlab+2p'lab1cosθlabp'lab2cosξlab0=p'2lab1(1−cos2θlab)+p'2lab2(1−cos2ξlab)−2p'lab1p'lab2cosθlabcosξlab0=p'2lab1sin2θlab+p'2lab2sin2cos2ξlab−p'lab1p'lab2cosθlabcosξlab$

On further simplification,

$p'2lab1sin2θlab+p'2lab2sin2cos2ξlab=p'lab1p'lab2cosθlabcosξlab(p'2lab1sin2θlab+p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)(p'2lab1sin2θlab−p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)0=2p'lab1p'lab2(cosξlab+cosθlab)$

Here $2p'lab1p'lab2≠0$ . Therefore $cosξlab+cosθlab=0$

Take cosine inverse on both sides

$ξlab+θlab=cos−10=π2$

Conclusion:

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

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Answer 5

Chapter 14, Problem 14.28P

(a)

To determine

Prove that $ξlab=(π−θcm)2$

(a)

Expert Solution

Answer to Problem 14.28P

The angle is $ξlab=(π−θcm)2$ .

Explanation of Solution

Classical Mechanics, Chapter 14, Problem 14.28P

Consider particle 1 with initial momentum $plab 1$ and particle 2 with zero initial momentum. Final momentum of particle 1 is $p'lab 1$ and final momentum of the particle 2 is $p'lab 2$ .

According to law of conservation of momentum

$plab 1+0=p'lab 1+p'lab 2p'lab 2=plab 1−p'lab 1$

From the figure, $AC$ denotes $plab 1$ , $AD$ represents $p'lab 1$ and $p'lab 2$ is represented by $AC−AD$

From the triangular law of vectors, $AC−AD=DC$

$BD$ and $BC$ is the radius of the circle. Thus the angles opposite to these sides are equal.

From the triangle $ΔBCD$ ,

$∠BCD=∠BDC=ξlab$

And,

$∠DBC=θcm$

The sum of all three angles of the triangle is equal to $π rad$ .

Therefore for $ΔBCD$

$ξlab+ξlab+θcm=π2ξlab+θcm=πξlab=π−θcm2$

Conclusion:

The angle is $ξlab=(π−θcm)2$ .

(b)

To determine

Show that in the case of equal masses, the angle of these outgoing particles in elastic collision is $90°$ .

(b)

Expert Solution

Answer to Problem 14.28P

For the collision of equal masses, $ξlab+θlab=π2$

Explanation of Solution

Write down the expression connecting $θlab$ and $θcm$

$tanθlab=sinθcmλ+cosθcm$

$λ=1$ for $m1=m2$ .

Then,

$tanθlab=sinθcm1+cosθcm=2sinθcm2cosθcm22cos2θcm2=tanθcm2$

That is,

$θlab=θcm2θcm=2θlab$

Substitute $π−2ξlab$ for $θcm$

$ξlab=π−θcm2ξlab=π2−θlabπ2=ξlab+θlab$

Hence proved.

Conclusion:

For the collision of equal masses, $ξlab+θlab=π2$

(c)

To determine

Prove $ξlab+θlab=π2$ using law of conservation of momentum.

(c)

Expert Solution

Answer to Problem 14.28P

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

Explanation of Solution

Apply law of conservation of momentum in horizontal direction

$plab1=p'lab1cosθlab+p'lab2cosξlab$

Along vertical direction,

$0=p'lab1sinθlab−p'labsinξlabp'lab1sinθlab=−p'lab2sinξlab$

Apply the law of conservation of energy

$p2lab12m1=p'2lab12m1+p'2lab22m2$

Substitute $m1=m2$ ,

$p2lab1=p'2lab1+p'2lab2$

Then,

$p'2lab1+p'2lab2=(p'lab1cosθlab+p'lab2cosξlab)2p'2lab1+p'2lab2=p'2lab1cos2θlab+p'2lab2cos2ξlab+2p'lab1cosθlabp'lab2cosξlab0=p'2lab1(1−cos2θlab)+p'2lab2(1−cos2ξlab)−2p'lab1p'lab2cosθlabcosξlab0=p'2lab1sin2θlab+p'2lab2sin2cos2ξlab−p'lab1p'lab2cosθlabcosξlab$

On further simplification,

$p'2lab1sin2θlab+p'2lab2sin2cos2ξlab=p'lab1p'lab2cosθlabcosξlab(p'2lab1sin2θlab+p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)(p'2lab1sin2θlab−p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)0=2p'lab1p'lab2(cosξlab+cosθlab)$

Here $2p'lab1p'lab2≠0$ . Therefore $cosξlab+cosθlab=0$

Take cosine inverse on both sides

$ξlab+θlab=cos−10=π2$

Conclusion:

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

Answer 6

Chapter 14, Problem 14.28P

(a)

To determine

Prove that $ξlab=(π−θcm)2$

(a)

Expert Solution

Answer to Problem 14.28P

The angle is $ξlab=(π−θcm)2$ .

Explanation of Solution

Classical Mechanics, Chapter 14, Problem 14.28P

Consider particle 1 with initial momentum $plab 1$ and particle 2 with zero initial momentum. Final momentum of particle 1 is $p'lab 1$ and final momentum of the particle 2 is $p'lab 2$ .

According to law of conservation of momentum

$plab 1+0=p'lab 1+p'lab 2p'lab 2=plab 1−p'lab 1$

From the figure, $AC$ denotes $plab 1$ , $AD$ represents $p'lab 1$ and $p'lab 2$ is represented by $AC−AD$

From the triangular law of vectors, $AC−AD=DC$

$BD$ and $BC$ is the radius of the circle. Thus the angles opposite to these sides are equal.

From the triangle $ΔBCD$ ,

$∠BCD=∠BDC=ξlab$

And,

$∠DBC=θcm$

The sum of all three angles of the triangle is equal to $π rad$ .

Therefore for $ΔBCD$

$ξlab+ξlab+θcm=π2ξlab+θcm=πξlab=π−θcm2$

Conclusion:

The angle is $ξlab=(π−θcm)2$ .

Answer 7

Chapter 14, Problem 14.28P

(a)

To determine

Prove that $ξlab=(π−θcm)2$

Answer 8

(a)

Expert Solution

Answer to Problem 14.28P

The angle is $ξlab=(π−θcm)2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Classical Mechanics, Chapter 14, Problem 14.28P

Consider particle 1 with initial momentum $plab 1$ and particle 2 with zero initial momentum. Final momentum of particle 1 is $p'lab 1$ and final momentum of the particle 2 is $p'lab 2$ .

According to law of conservation of momentum

$plab 1+0=p'lab 1+p'lab 2p'lab 2=plab 1−p'lab 1$

From the figure, $AC$ denotes $plab 1$ , $AD$ represents $p'lab 1$ and $p'lab 2$ is represented by $AC−AD$

From the triangular law of vectors, $AC−AD=DC$

$BD$ and $BC$ is the radius of the circle. Thus the angles opposite to these sides are equal.

From the triangle $ΔBCD$ ,

$∠BCD=∠BDC=ξlab$

And,

$∠DBC=θcm$

The sum of all three angles of the triangle is equal to $π rad$ .

Therefore for $ΔBCD$

$ξlab+ξlab+θcm=π2ξlab+θcm=πξlab=π−θcm2$

Conclusion:

The angle is $ξlab=(π−θcm)2$ .

Answer 13

(b)

To determine

Show that in the case of equal masses, the angle of these outgoing particles in elastic collision is $90°$ .

(b)

Expert Solution

Answer to Problem 14.28P

For the collision of equal masses, $ξlab+θlab=π2$

Explanation of Solution

Write down the expression connecting $θlab$ and $θcm$

$tanθlab=sinθcmλ+cosθcm$

$λ=1$ for $m1=m2$ .

Then,

$tanθlab=sinθcm1+cosθcm=2sinθcm2cosθcm22cos2θcm2=tanθcm2$

That is,

$θlab=θcm2θcm=2θlab$

Substitute $π−2ξlab$ for $θcm$

$ξlab=π−θcm2ξlab=π2−θlabπ2=ξlab+θlab$

Hence proved.

Conclusion:

For the collision of equal masses, $ξlab+θlab=π2$

Answer 14

(b)

To determine

Show that in the case of equal masses, the angle of these outgoing particles in elastic collision is $90°$ .

Answer 15

(b)

Expert Solution

Answer to Problem 14.28P

For the collision of equal masses, $ξlab+θlab=π2$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write down the expression connecting $θlab$ and $θcm$

$tanθlab=sinθcmλ+cosθcm$

$λ=1$ for $m1=m2$ .

Then,

$tanθlab=sinθcm1+cosθcm=2sinθcm2cosθcm22cos2θcm2=tanθcm2$

That is,

$θlab=θcm2θcm=2θlab$

Substitute $π−2ξlab$ for $θcm$

$ξlab=π−θcm2ξlab=π2−θlabπ2=ξlab+θlab$

Hence proved.

Conclusion:

For the collision of equal masses, $ξlab+θlab=π2$

Answer 20

(c)

To determine

Prove $ξlab+θlab=π2$ using law of conservation of momentum.

(c)

Expert Solution

Answer to Problem 14.28P

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

Explanation of Solution

Apply law of conservation of momentum in horizontal direction

$plab1=p'lab1cosθlab+p'lab2cosξlab$

Along vertical direction,

$0=p'lab1sinθlab−p'labsinξlabp'lab1sinθlab=−p'lab2sinξlab$

Apply the law of conservation of energy

$p2lab12m1=p'2lab12m1+p'2lab22m2$

Substitute $m1=m2$ ,

$p2lab1=p'2lab1+p'2lab2$

Then,

$p'2lab1+p'2lab2=(p'lab1cosθlab+p'lab2cosξlab)2p'2lab1+p'2lab2=p'2lab1cos2θlab+p'2lab2cos2ξlab+2p'lab1cosθlabp'lab2cosξlab0=p'2lab1(1−cos2θlab)+p'2lab2(1−cos2ξlab)−2p'lab1p'lab2cosθlabcosξlab0=p'2lab1sin2θlab+p'2lab2sin2cos2ξlab−p'lab1p'lab2cosθlabcosξlab$

On further simplification,

$p'2lab1sin2θlab+p'2lab2sin2cos2ξlab=p'lab1p'lab2cosθlabcosξlab(p'2lab1sin2θlab+p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)(p'2lab1sin2θlab−p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)0=2p'lab1p'lab2(cosξlab+cosθlab)$

Here $2p'lab1p'lab2≠0$ . Therefore $cosξlab+cosθlab=0$

Take cosine inverse on both sides

$ξlab+θlab=cos−10=π2$

Conclusion:

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

Answer 21

(c)

To determine

Prove $ξlab+θlab=π2$ using law of conservation of momentum.

Answer 22

(c)

Expert Solution

Answer to Problem 14.28P

The expression $ξlab+θlab=π2$ has been proven using the law of conservation of momentum.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Apply law of conservation of momentum in horizontal direction

$plab1=p'lab1cosθlab+p'lab2cosξlab$

Along vertical direction,

$0=p'lab1sinθlab−p'labsinξlabp'lab1sinθlab=−p'lab2sinξlab$

Apply the law of conservation of energy

$p2lab12m1=p'2lab12m1+p'2lab22m2$

Substitute $m1=m2$ ,

$p2lab1=p'2lab1+p'2lab2$

Then,

$p'2lab1+p'2lab2=(p'lab1cosθlab+p'lab2cosξlab)2p'2lab1+p'2lab2=p'2lab1cos2θlab+p'2lab2cos2ξlab+2p'lab1cosθlabp'lab2cosξlab0=p'2lab1(1−cos2θlab)+p'2lab2(1−cos2ξlab)−2p'lab1p'lab2cosθlabcosξlab0=p'2lab1sin2θlab+p'2lab2sin2cos2ξlab−p'lab1p'lab2cosθlabcosξlab$

On further simplification,

$p'2lab1sin2θlab+p'2lab2sin2cos2ξlab=p'lab1p'lab2cosθlabcosξlab(p'2lab1sin2θlab+p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)(p'2lab1sin2θlab−p'2lab2sin2cos2ξlab)2=2p'lab1p'lab2(cosξlabcosθlab+cosξlabcosθlab)0=2p'lab1p'lab2(cosξlab+cosθlab)$