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Concept explainers
(a)
The velocity of the particles.
(a)
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Answer to Problem 70P
The velocity of the particles is vf=(m1v1i/m1+m2).
Explanation of Solution
Write the expression from conservation of momentum.
→Pi=→Pf (I)
Here, →Pi is the initial momentum and →Pf is the final momentum.
Write the equation for initial momentum.
→Pi=(m1→v1i)+(m2→v2i) (II)
Here, m1,m2 are the masses, →v1i,→v2i are the initial velocities of the objects.
Write the equation for final momentum.
→Pf=(m1+m2)→vf (III)
Here, →vf are the final velocities of the objects.
Conclusion:
Substitute, (m1+m2)→vf for →Pf, (m1→v1i)+(m2→v2i) for →Pi, 0 m/s for →v2i in Equation (I) to find →vf.
(m1→v1i)+(m2→v2i)=(m1+m2)→vf→vf=(m1→v1i)+(m2(0))(m1+m2)=(m1→v1i)(m1+m2)
Thus, the velocity of the particles is vf=(m1v1i/m1+m2).
(b)
The closest distance.
(b)
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Answer to Problem 70P
The closest distance is r(2kq1q2(m1+m2)/m1m2v21i)_.
Explanation of Solution
Here, initial potential energy is zero.
Write the expression from conservation of energy.
Ki=Kf+Uf (IV)
Here, Ki,Kf are the initial and final kinetic energy and Uf is the final potential energy.
Write the equation for initial kinetic energy.
Ei=12(m1(v1i)2)+12(m2(v2i)2) (V)
Write the equation for final kinetic energy.
Ef=12(m1+m2)(vf)2 (VI)
Write the equation for final potential energy.
Uf=kq1q2r (VII)
Here, q1,q2 are the charges, k is Coulomb’s constant, r is distance.
Conclusion:
Substitute, 12(m1(v1i)2)+12(m2(v2i)2) for Ei, 12(m1+m2)(vf)2 for Ef, kq1q2r for Ufq2, ,0 m/s for v2i in Equation (VII) to find r.
12(m1(v1i)2)+12(m2(0)2)=12(m1+m2)(vf)2+[kq1q2r]m1m2v21i=2kq1q2(m1+m2)rr=2kq1q2(m1+m2)m1m2v21i
Thus, the closest distance is r(2kq1q2(m1+m2)/m1m2v21i)_.
(c)
The velocity of first particle.
(c)
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Answer to Problem 70P
The velocity of first particle is v1f=((m1−m2)v1i/m1+m2)_.
Explanation of Solution
The initial velocity of second particle is zero.
Write the expression from relative velocity equation.
v1i=v2f−v1f (X)
Conclusion:
Substitute, v1i+v1f for v2f, (m1+m2)→vf for →Pf, (m1→v1i)+(m2→v2i) for →Pi, 0 m/s for →v2i in Equation (I) to find v1f.
m1v1i=m1v1f+m2(v1i+v1f)v1f=(m1−m2m1+m2)v1i
Thus, the velocity of first particle is v1f=((m1−m2)v1i/m1+m2)_.
(d)
The velocity of second particle.
(d)
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Answer to Problem 70P
The velocity of second particle is (2m1v1i/m1+m2)_.
Explanation of Solution
Write the expression from relative velocity equation.
v1i=v2f−v1f (XI)
Conclusion:
Substitute, ((m1−m2)v1i/m1+m2) for v1f in Equation (XI) to find v2f.
v1i=v2f−[((m1−m2)v1i/m1+m2)]v2f=v1i+[((m1−m2)v1i/m1+m2)]=2m1v1im1+m2
Thus, the velocity of second particle is (2m1v1i/m1+m2)_.
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Chapter 20 Solutions
Principles of Physics: A Calculus-Based Text
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