Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 39, Problem 39.62P

An unstable particle with mass m = 3.34 × 10−27 kg is initially at rest. The particle decays into two fragments that fly off along the x axis with velocity components u1 = 0.987c and u2 = −0.868c. From this information, we wish to determine the masses of fragments 1 and 2. (a) Is the initial system of the unstable particle, which becomes the system of the two fragments, isolated or nonisolated? (b) Based on your answer to part (a), what two analysis models are appropriate for this situation? (c) Find the values of γ for the two fragments after the decay. (d) Using one of the analysis models in part (b), find a relationship between the masses m1 and m2 of the fragments. (e) Using the second analysis model in part (b). find a second relationship between the masses m1 and m2. (f) Solve the relationships in parts (d) and (c) simultaneously for the masses m1 and m2.

(a)

Expert Solution
Check Mark
To determine
The initial system of the unstable particle.

Answer to Problem 39.62P

The initial system of the unstable particle is isolated.

Explanation of Solution

Isolated system is system which does not allow the flow of mass energy in our out.

Here, the unstable particle system does not allow any disturbance form the interference of mass-energy from outside the system or didn’t allowed any mass–energy to flow out of the system initially. The fragments of smaller mass of the unstable nuclei are also a part of initial system which further carries out the mass and energy with themselves after the decay but there is no involvement of the mass and energy from outside the system.

Thus the initial system is isolated.

Conclusion:

Therefore, the initial system before the decay is isolated.

(b)

Expert Solution
Check Mark
To determine
The analysis model appropriate for analysis of the given isolated system in part (a).

Answer to Problem 39.62P

The two appropriate analysis models are the conservation of momentum in an isolated system and conservation of energy is an isolated system.

Explanation of Solution

Explanation

In an isolated system both the total momentum as well as total energy is conserved.

Determine the mass of the fragmented particles using the concept of conservation of total momentum and conservation of total energy.

The single unstable nuclei decayed into two fragments and there is no interference of any type of energy from outside and neither there is any flow of energy from outside the system. Therefore the momentum and the energy will be conserved during and after the fragmentation.

Thus the analysis which can be used to determine the mass of the two fragmented particles conservation of momentum in an isolated system and conservation of energy in an isolated system.

Conclusion:

Therefore, the two appropriate analysis models are the conservation of momentum in an isolated system and conservation of energy is an isolated system.

(c)

Expert Solution
Check Mark
To determine
The relativistic factor for the two particles after decay.

Answer to Problem 39.62P

The values for γ for the two particles are 6.22 and 2.01 .

Explanation of Solution

Explanation

Given info: The mass of the unstable particle is 3.34×1027kg . The mass of the fragmented particle one is m1 , the mass of fragmented particle m2 . the velocity of particle with mass m1 is 0.987c and the velocity of the particle with mass m2 is 0.868c .

The speed of light is 3×108ms-1 .

The formula to calculate the relativistic value is,

γ=11u2c2 (1)

Here,

γ is the relativistic factor.

u is the velocity for the fragment.

c is the speed of the light.

For first fragment,

Substitute 0.987c for u in equation (1).

γ1=11(0.987c)2c2=6.22

Thus the value of γ for fragmented particle on is 6.22 .

For second fragment

Substitute 0.868c for u in equation (1).

γ2=11(0.868c)2c2=2.01

Thus the value of γ for fragmented particle second is 2.01 .

Conclusion:

Therefore, value of relativistic factor for fragment one is 6.22 and for the fragment second is 2.01 .

(d)

Expert Solution
Check Mark
To determine
The relation between the masses of two fragments using the conservation of energy analysis.

Answer to Problem 39.62P

The relation between the mass of fragment one and fragment second is m2+3.09m1=1.66×1027kg .

Explanation of Solution

Explanation

From the conservation of energy the sum of the energy of the fragments is equal to the sum of the energy of the unstable nuclei.

E1+E2=Etotal (2)

Here,

E1 is the energy of the m1 fragment.

E2 is the energy of the m2 fragment.

Etotal is the total energy.

The formula to calculate the total relativistic energy is,

Etotal=mtotalc2

Here,

mtotal is the total mass of the unstable particle.

The formula to calculate relativistic energy of a particle m1 is,

E1=γ1m1c2

The formula to calculate the energy for particle m2 is,

E2=γ2m2c2

Substitute γ1m1c2 for E1 , γmtotalc2 for Etotal and γ2m2c2 for E2 in equation (2).

γ1m1c2+γ2m2c2=mtotalc2

Substitute 6.22 for γ1 and 2.01 for γ2 and 3.34×1027kg for mtotal in the above equation.

(6.22)m1c2+(2.01)m2c2=(3.34×1027kg)c2m2+3.09m1=1.66×1027kg

Thus the relation between the mass of fragment one and fragment second is,

m2+3.09m1=1.66×1027kg

Conclusion:

Therefore, the relation between the mass of fragment one and fragment second is m2+3.09m1=1.66×1027kg .

(e)

Expert Solution
Check Mark
To determine
The relation between the masses of two fragments.

Answer to Problem 39.62P

The relation between the masses m1 and m2 using the conservation of momentum analysis is m2=3.52m1 .

Explanation of Solution

Explanation

Given info: The mass of the unstable nuclei is 3.34×1027kg . The mass of the fragmented particles is m1 and m2 . the velocity of particle m1 is 0.987c and the velocity of the particle m2 is 0.868c .

From the conservation of momentum the final momentum is zero,

p1=p2 (3)

Here,

p1 is the momentum of first fragment.

p2 is the momentum of the second fragment.

The momentum of first fragment is,

p1=γ1m1u1

The momentum of the second fragment is,

p2=γ2m2u2

Substitute γ2m2u2 for p2 and γ1m1u1 for p1 in equation (1).

γ1m1u1=γ2m2u2

Substitute 6.22 for γ1 , 0.987c for u1 , 2.01 for γ2 and 0.868c for u2 in the above equation.

γ1m1u1=γ2m2u2(6.22)(0.987c)m1=(2.01)(0.868c)m2m2=3.52m1

The relation between the two fragment is m2=3.52m1 .

Conclusion:

Therefore, from the analysis model of conservation of momentum the relation between the two masses is m2=3.52m1 .

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Chapter 39 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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