Modern Physics
Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Chapter 2, Problem 29P

(a)

To determine

The mass of the original object.

(a)

Expert Solution
Check Mark

Answer to Problem 29P

The mass of the original object is 3.65MeVc2

Explanation of Solution

The formula to calculate energy of the first fragment is,

    E12=p12c2+(m1c2)2        (I)

Here, m1 is the mass of the fragment, p1 is the linear momentum of the first fragment, and c is the speed of light.

The formula to calculate energy of the second fragment is,

  E22=p22c2+(m2c2)2

Here, m2 is the mass of the fragment, p2 is the linear momentum of the second fragment, and c is the speed of light.

Conclusion:

Substitute 1.75MeVc for p2, and 1.00MeVc2 for m2 in expression (I) to calculate the value for E2,

    E22=(1.75MeVc)2c2+((1.00MeVc2)c2)2E2=(1.75MeV)2+(1.00MeV)2=2.02MeV

Thus the total energy of the first fragment is 2.02MeV.

Substitute 2.00MeVc for p2, and 1.50MeVc2 for m2 in expression (I) to calculate the value for E2,

    E22=(2.00MeVc)2c2+((1.50MeVc2)c2)2E2=(2.00MeV)2+(1.50MeV)2=2.50MeV

Thus the total energy of the first fragment is 2.50MeV.

The total energy of the original object is, by substituting 2.50MeV and 2.02MeV,

  E=E1+E2=2.02MeV+2.50MeV=4.52MeV 

The total linear momentum of the original object is,

  p2=p12+p22=(2.00MeVc)2+(1.75MeVc)2=1c2((2.00MeV)2+(1.75MeV)2)

The equation for energy of the object is,

  E2=p2c2+(mc2)2     

Here, m is the mass of the object.

Substitute 4.52MeV for E and 1c2((2.00MeV)2+(1.75MeV)2) for p2 in above equation E2=p2c2+(mc2)2

  (4.52MeV)2=1c2((2.00MeV)2+(1.75MeV)2)c2+(mc2)2=(1.75MeV)2+(2.00MeV)2+(mc2)2=3.65MeVc2

Therefore the mass of the original object is 3.65MeVc2

(b)

To determine

The speed of the original object.

(b)

Expert Solution
Check Mark

Answer to Problem 29P

The speed of the original object is 0.589c

Explanation of Solution

Write the expression for the total energy of the object is,

  E=mc21u2c2        (I)

Here, u is the speed of the original object, m is the mass of the object, and c is the speed of light.

Conclusion:

Substitute 4.52MeV for E, and 3.65MeVc2 for m expression (I) and solve for u

  4.52MeV=(3.65MeVc2)c21u2c21u2c2=0.652u2c2=0.348c2=0.589c

Therefore the speed of the original object is 0.589c

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