Classical Mechanics
Classical Mechanics
5th Edition
ISBN: 9781891389221
Author: John R. Taylor
Publisher: University Science Books
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Chapter 15, Problem 15.74P

(a)

To determine

The speed of either b particle in the rest frame of particle a.

(a)

Expert Solution
Check Mark

Answer to Problem 15.74P

The speed of the newly formed particles in the rest frame of particle a are 0.8401c_ and 0.1122c_ or 0.3912c_ and 0.7122c_.

Explanation of Solution

Write the expression for the relativistic momentum of particle a.

    p=γmv        (I)

Here, p is the relativistic momentum, γ is the relativistic factor, m is the mass of the particle, v is the velocity of the particle.

Write the expression for γ.

    γ=11v2c2        (II)

Here, c is the speed of light.

Consider a particle a is moving along positive x axis of S frame. This particle decayed into two identical particle (band b). These two new formed particles are also moving along the x axis.

    ab+b        (III)

Write the expression for the conservation of momentum for the reaction in equation (III).

    pa=pb1+pb2        (IV)

Here, pb1 is the momentum of the first new particle b, pb2 is the momentum of the second new particle b.

Use equation (I) and (II) to solve for the relativistic momentum of particle a.

    pa=γamava=mava1va2c2        (V)

Substitute 2.5mb for ma, 0.5c for va in equation (V) to solve for pa.

    pa=(2.5mb)(0.5c)1(0.5c)2c2=1.44mbc        (VI)

Substitute 1.44mbc for pa in equation (IV) to solve for pb1.

    1.44mbc=pb1+pb2pb1=144mbcpb2        (V)

The total energy of the particle a is equal to the sum of the total energy of the newly formed particles.

Write the expression for the conservation of energy.

    mac2+Ta=mbc2+Tb1+mbc2+Tb2        (VI)

Here, Ti is the kinetic energy of ith particle.

Write the expression for the relativistic kinetic energy.

    T=12γmv2=12m1v2c2v2        (VII)

Use equation (VII) in (VI) to solve for the conservation of energy.

    mac2+12ma1va2c2va2=2mbc2+Tb1+Tb2        (VIII)

Substitute 2.5mb for ma, 0.5c for va in equation (VIII) to solve for the conservation of energy.

    2.5mbc2+122.5mb1(0.5c)2c2(0.5c)2=2mbc2+Tb1+Tb20.5mbc2+0.3608mbc2=Tb1+Tb20.8608mbc2=Tb1+Tb2        (IX)

Write the expression for the kinetic energy of first b particle.

    Tb1=(pb1c)2+(mbc2)2mbc2        (X)

Write the expression for the kinetic energy of second b particle.

    Tb2=(pb2c)2+(mbc2)2mbc2        (XI)

Substitute (pb1c)2+(mbc2)2mbc2 for Tb1, (pb2c)2+(mbc2)2mbc2 for Tb2 in equation (IX) and it becomes,

    0.8608mbc2=((pb1c)2+(mbc2)2mbc2)+((pb2c)2+(mbc2)2mbc2)(pb1c)2+(mbc2)2=2.8068mbc2(pb2c)2+(mbc2)2        (XII)

Use equation (V) in (XII) and squaring on both side to solve for pb2.

    (pb1c)2+(mbc2)2=8.1841mb2c4+[(pb2c)2+(mbc2)2]5.7216mbc2(pb2c)2+(mbc2)2[(1.44mbcpb2)c]2+(mbc2)2={8.1841mb2c4+[(pb2c)2+(mbc2)2]5.7216mbc2(pb2c)2+(mbc2)2}2.0736mb2c2+(pb2c)22.88mbc2pb2c={8.1841mb2c4+[(pb2c)2+(mbc2)2]5.7216mbc2(pb2c)2+(mbc2)2}6.1105mbc2+2.88pb2c=5.7216(pb2c)2+(mbc2)2        (XIII)

Squaring the equation (XIII) on both sides to solve for pb2.

    37.34mb2c4+8.2944(pb2c)2+17.59824mbc2(pb2c)=32.74[(pb2c)2+(mbc2)2]4.6mb2c424.4406(pb2c)2+17.59824mbc2(pb2c)=024.4406(pb2c)17.59824mbc2(pb2c)4.6mb2c4=0        (XIV)

Use the general solution of quadratic equation in (XIV) to solve for pb2.

    pb2=(17.5982mbc2)±(17.5982mbc2)24(24.4406)(4.6mb2c4)2(24.4406)=48.35mbc2±27.5572mbc22(24.4406)=1.533mbc2or 0.4253mbc2        (XV)

Write the expression for the relativistic momentum of the second newly formed particle.

    pb2=mbvb21(vb2)2c2        (XVI)

Substitute 1.533mbc2 for pb2 in equation (XVI) to solve for vb2.

    1.533mbc2=mbvb21(vb2)2c2(vb2)2=2.4c2(1(vb2)2c2)3.4(vb2)2=2.4c2vb2=(2.43.4)c=0.8401c        (XVII)

Substitute 0.4253mbc2 for pb2 in equation (XVI) to solve for vb2.

    0.4253mbc2=mbvb21(vb2)2c2(vb2)2=0.18088c2(1(vb2)2c2)1.18088(vb2)2=0.18088c2(vb2)2=0.180881.18088c2vb2=0.1531c=0.3912c        (XVIII)

Substitute 1.553mbc for pb2 in equation (V) to solve for pb1.

    pb1=1.44mbc1.553mbc=0.113mbc        (XIX)

Write the expression for the relativistic momentum of the first newly formed particle.

    pb1=mbvb11(vb1)2c2        (XX)

Substitute 0.113mbc for pb1 in equation (XX) to solve for vb1.

    0.113mbc=mbvb11(vb1)2c2(vb1)2=0.012769c2(1(vb1)2c2)vb1=(0.0127691.012769)c=0.1122c        (XXI)

Substitute 0.4253mbc for pb2 in equation (V) to solve for pb1.

    pb1=1.44mbc(0.4253mbc)=1.0147mbc        (XXII)

Substitute 1.0147mbc for pb1 in equation (XX) to solve for vb1.

    1.0147mbc=mbvb11(vb1)2c2(vb1)2=1.0296c2(1(vb1)2c2)vb1=(1.02962.0296)c2=0.7122c        (XXIII)

Conclusion:

Therefore, the speed of the newly formed particles in the rest frame of particle a are 0.8401c_ and 0.1122c_ or 0.3912c_ and 0.7122c_.

(b)

To determine

The velocities of the two b particles in the original frame S.

(b)

Expert Solution
Check Mark

Answer to Problem 15.74P

The velocities of the newly formed particles in the frame of reference S are 0.9436c_ and 0.5797c_ or 0.7454c_ and 0.8939c_ respectively.

Explanation of Solution

The relative velocity of the newly formed particles with respect to the frame of reference S.

Write the expression for the velocity of the newly formed particle in the S frame.

    v=v+V1+vVc2        (XXIV)

Here, v is the velocity of the particle in the original frame S, v is the speed of the moving frame, V is the speed of the newly formed particle.

Substitute 0.8401c for v, 0.5c for V in equation (XXIV) to solve for vb2.

    vb2=(0.8401c)+(0.5c)1+(0.8401c)(0.5c)c2=0.9436c        (XXV)

Here, vb2 is the speed of the second particle b in the frame of reference S.

Substitute 0.3912c for v, 0.5c for V in equation (XXIV) to solve for vb1.

    vb1=(0.3912c)+(0.5c)1+(0.3912c)(0.5c)c2=0.7454c        (XXVI)

Here, vb1 is the speed of the first particle b in the S frame.

Substitute 0.1122c for v, 0.5c for V in equation (XXIV) to solve for vb2.

    vb2=(0.1122c)+(0.5c)1+(0.1122c)(0.5c)c2=0.5797c        (XXVII)

Substitute 0.7122c for v, 0.5c for V in equation (XXIV) to solve for vb1.

    vb1=(0.7122c)+(0.5c)1+(0.7122c)(0.5c)c2=0.8939c        (XXVIII)

Conclusion:

Therefore, the velocities of the newly formed particles in the frame of reference S are 0.9436c_ and 0.5797c_ or 0.7454c_ and 0.8939c_ respectively.

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Chapter 15 Solutions

Classical Mechanics

Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42PCh. 15 - Prob. 15.43PCh. 15 - Prob. 15.44PCh. 15 - Prob. 15.45PCh. 15 - Prob. 15.46PCh. 15 - Prob. 15.47PCh. 15 - Prob. 15.48PCh. 15 - Prob. 15.49PCh. 15 - Prob. 15.50PCh. 15 - Prob. 15.51PCh. 15 - Prob. 15.52PCh. 15 - Prob. 15.53PCh. 15 - Prob. 15.54PCh. 15 - Prob. 15.55PCh. 15 - Prob. 15.56PCh. 15 - Prob. 15.57PCh. 15 - Prob. 15.58PCh. 15 - Prob. 15.59PCh. 15 - Prob. 15.60PCh. 15 - Prob. 15.61PCh. 15 - Prob. 15.62PCh. 15 - Prob. 15.63PCh. 15 - Prob. 15.64PCh. 15 - Prob. 15.65PCh. 15 - Prob. 15.66PCh. 15 - Prob. 15.67PCh. 15 - Prob. 15.68PCh. 15 - Prob. 15.69PCh. 15 - Prob. 15.70PCh. 15 - Prob. 15.71PCh. 15 - Prob. 15.72PCh. 15 - Prob. 15.73PCh. 15 - Prob. 15.74PCh. 15 - Prob. 15.75PCh. 15 - Prob. 15.76PCh. 15 - Prob. 15.79PCh. 15 - Prob. 15.80PCh. 15 - Prob. 15.81PCh. 15 - Prob. 15.82PCh. 15 - Prob. 15.83PCh. 15 - Prob. 15.84PCh. 15 - Prob. 15.85PCh. 15 - Prob. 15.88PCh. 15 - Prob. 15.89PCh. 15 - Prob. 15.90PCh. 15 - Prob. 15.91PCh. 15 - Prob. 15.94PCh. 15 - Prob. 15.95PCh. 15 - Prob. 15.96PCh. 15 - Prob. 15.97PCh. 15 - Prob. 15.98PCh. 15 - Prob. 15.101PCh. 15 - Prob. 15.102PCh. 15 - Prob. 15.103PCh. 15 - Prob. 15.104PCh. 15 - Prob. 15.105PCh. 15 - Prob. 15.106PCh. 15 - Prob. 15.107PCh. 15 - Prob. 15.109PCh. 15 - Prob. 15.110PCh. 15 - Prob. 15.111P
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