College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Chapter 26, Problem 34P

An unstable particle with a mass equal to 3.34 × 10−27 kg is initially at rest. The particle decay's into two fragments that fly off with velocities of 0.987c and −0.868c, respectively. Find the masses of the fragments. Hint: Conserve both mass-energy and momentum.

Expert Solution & Answer
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To determine
The masses of the fragments.

Answer to Problem 34P

  The masses of the fragments are 2.51×108kg and 8.84×1028kg .

Explanation of Solution

Given info:

The speed of first segment is 0.868c .

The speed of second segment is 0.987c .

The formula used to calculate the rest energy of the particle is,

ER=mc2

  • ER is energy of the particle
  • c is the speed of light
  • m is the mass of the particle

The formula used to calculate the rest energy of the first fragment is,

ER1=m1c2

  • ER1 is energy of the first fragment
  • m1 is the mass of the first fragment

The formula used to calculate the rest energy of the second fragment is,

ER2=m2c2

  • ER2 is energy of the second fragment
  • m2 is the mass of the second fragment

The formula used to calculate the gamma factor for first fragment is,

γ1=11(v1c)2

  • γ1 is the gamma factor of the first fragment
  • v1 is the speed of the first fragment

The formula used to calculate the gamma factor for second fragment is,

γ2=11(v2c)2

  • γ2 is the gamma factor of the second fragment
  • v2 is the speed of the second fragment

The formula used to calculate the total energy of the first fragment is,

E1=γ1ER1=(11(v1c)2)(m1c2)

The formula used to calculate the total energy of the second fragment is,

E2=γ2ER1=(11(v2c)2)(m2c2)

The formula define the conservation of energy is,

ER=E1+E2mc2=(11(v1c)2)(m1c2)+(11(v2c)2)(m2c2)m=m11(v1c)2+m21(v2c)2

Substitute 3.34×1027kg for m , 0.987c for v1 and 0.868c for v2 to find the masses.

3.34×1027kg=m11(0.987cc)2+m21(0.868cc)23.34×1027kg=6.22m1+2.01m2

Substitute 3.34×1027kg for m , 0.987c for v1 and 0.868c for v2 to find the masses.

3.34×1027kg=m11(0.987cc)2+m21(0.868cc)23.34×1027kg=6.22m1+2.01m2 I

The formula used to calculate the momentum of the first fragment is,

P1=γ1m1v1=(11(v1c)2)m1v1

The formula used to calculate the momentum of the second fragment is,

P2=γ2m2v2=(11(v2c)2)m2v2

The formula define the conservation of momentum is,

P1=P2γ1m1v1=γ2m2v2(11(0.987cc)2)(m1)(0.987c)=(11(0.868cc)2)(m2)(0.868c)m2=3.52m1 II

Substitute 3.52m1 for m2 in equation I to find m1 .

3.34×1027kg=6.22m1+2.01(3.52m1)m1=2.51×1028kg

Substitute 2.51×1028kg for m1 in equation II to find m2 .

m2=3.52(2.51×1028kg)=8.84×1028kg

Thus, the masses of the fragments are 2.51×108kg and 8.84×1028kg .

Conclusion:

The masses of the fragments are 2.51×108kg and 8.84×1028kg

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Chapter 26 Solutions

College Physics

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