Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 9, Problem 43P

(a)

To determine

The velocity of the third particle.

(a)

Expert Solution
Check Mark

Answer to Problem 43P

The velocity of third particle is (9.33×106i^8.33×106j^)m/s .

Explanation of Solution

Atomic nucleous splits into three particles hence their momentum remains conserved as no extrenal force is applied.

Write the expession for the conservation of the momentum.

  mnvn=m1v1+m2v2+m3v3                                                                            (I)

Here mn is the mass of the nucleus, vn is the velocity of nucleus, m1 is the mass of first paricle, v1 is the velocit of first particle, m2 is the mass of second particle, v2 is the velocity of second particle, m3 is the mass of the third particle and v3 is the velocity of the third particle.

As nucleus is at rest. Substitute 0m/s for vn and simplify equation (I) for  v3.

    v3=(m1v1+m2v2)m3                                                                                    (II)

Conclusion:

Substitute 5×1027kg for m1, (6×106m/s) j^ for v1, 8.4×1027kg for m2, (4×106m/s) i^ for v2 and 3.6×1027kg for m3 in equation (II).

    v3=[(5×1027kg)((6×106m/s) j^)+(8.4×1027kg)((4×106m/s) i^)]3.6×1027kg=(9.33i^+8.33j^)m/s

Thus, the velocity of third particle is (9.33×106i^8.33×106j^)m/s.

(b)

To determine

The total kinetic energy increase in the process.

(b)

Expert Solution
Check Mark

Answer to Problem 43P

The total kinetic energy increase in the process is 4.39×1013J .

Explanation of Solution

Initially nucleus is at rest hence increase in kinetic energy is the final kinetic energy.

    ΔK=KfKi

Here, ΔK is the increase in kinetic energy, Kf is the final kinetic energy and Ki is the initial kinetic energy.

Substitute 0J for Ki and solve for ΔK in above equation.

    ΔK=Kf                                                                                                    (III)

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

    K1=12m1v12                                                                                                (IV)

Here, K1 is the kinetic energy of first particle.

Write the expression for the final kinetic energyof second particle .

    K2=12m2v22                                                                                                 (V)

Here, K2 is the kinetic energy of second particle.

Write the expression for the final kinetic energyof third particle .

    K3=12m3v32                                                                                                (VI)

Here, K3 is the kinetic energy of third particle.

    Kf=K1+K2+K3                                                                                    (VII)

Substitute 12m1v12 for K1 , 12m2v22 for K2 and 12m3v32 for K3 in above equation.

    Kf=12m1v12+12m2v22+12m3v32

Substitute 12m1v12+12m2v22+12m3v32 for Kf in equation (III) and solve for ΔK.

    ΔK=12m1v12+12m2v22+12m3v32                                                                (VIII)

Conclusion:

Substitute 5×1027kg for m1, (6×106m/s) j^ for v1, 8.4×1027kg for m2, (4×106m/s) i^ for v2, 3.6×1027kg for m3 and  (9.33×106i^8.33×106j^)m/s for v3 in equation (VIII).

    ΔK=[12(5×1027kg)((6×106m/s) j^)2+12(8.4×1027kg)((4×106m/s) i^)2+12(3.6×1027kg)((9.33×106i^)2+(8.33×106j^m/s)2)]=(9×1014 J)+(6.72×1014 J)+(2.816×1013)=4.39×1013J

Thus, the total kinetic energy increase in the process is 4.39×1013J .

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Chapter 9 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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