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Physics Solution Manual for 1100 and 2101

# This amount equals the difference between the two

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Unformatted text preview: 0.88 c −12 1− = 1.3 × 10 m c 31. SSM REASONING The de Broglie wavelength λ is related to Planck’s constant h and the magnitude p of the particle’s momentum. The magnitude of the momentum can be related to the particle’s kinetic energy. Thus, using the given wavelength and the fact that the kinetic energy doubles, we will be able to obtain the new wavelength. SOLUTION The de Broglie wavelength is λ= h p (29.8) The kinetic energy and the magnitude of the momentum are 1 2 KE = mv 2 (6.2) p = mv (7.2) 1530 PARTICLES AND WAVES where m and v are the mass and speed of the particle. Substituting Equation 7.2 into Equation 6.2, we can relate the kinetic energy and momentum as follows: 1 2 KE = mv 2 = m2v 2 p 2 = 2m 2m p = 2m ( KE ) or Substituting this result for p into Equation 29.8 gives λ= h h = p 2m ( KE ) Applying this expression for the final and initial wavelengths λf and λi, we obtain λf = h 2m ( KE )f h λi = and 2m ( KE )i Dividing the two equations and rearranging reveals that h λf = λi 2m ( KE )f = h 2m ( KE )i ( KE )i ( KE )f λf = λi or ( KE )i ( KE )f Using the given value for λi and the fact that KE f = 2 ( KEi ) , we find λf = λi 32. REASONING ( KE )i ( KE )f ( = 2.7 × 10−10 m ) KEi ( 2 KEi ) = 1.9 ×10−10 m The de Broglie wavelength λ of the woman is found from p = h λ −34 (Equation 29.8), where p is the magnitude of her momentum and h = 6.63×10 J·s is Planck’s constant. We will use p = mv (Equation 7.2) to determine the magnitude p of the woman’s momentum from her mass m and her speed v at the instant she strikes the water. Once the woman jumps from the cliff, she is in free fall with an initial speed of v0 = 0 m/s, and an acceleration a = −9.8 m/s2. Since we have taken upward to be the positive direction, her displacement during the fall is H = −9.5 m. Her final speed v, then, is given by Chapter 29 Problems 2 2 v = v0 + 2aH Solving p = SOLUTION h λ 1531 (2.9) (Equation 29.8) for λ yields λ = h . Substituting p = mv p (Equation 7.2) into this result, we obtain λ= h h = p mv (1) Substituting v0 = 0 m/s into Equation (2.9) and taking the square root of both sides, we find that v 2 = ( 0 m/s ) + 2aH 2 or v 2 = 2aH or v = 2aH (2) Substituting Equation (2) into Equation (1), we find the de Broglie wavelength of the woman at the instant she strikes the water to be: λ= 6.63 ×10−34 J ⋅ s h h = = = 1.2 ×10−36 m mv m 2aH ( 41 kg ) 2 ( −9.8 m/s 2 ) ( −9.5 m ) 33. REASONING When the electron is at rest, it has electric potential energy, but no kinetic energy. The electric potential energy EPE is given by EPE = eV (Equation 19.3), where e is the magnitude of the charge on the electron and V is the potential difference. When the electron reaches its maximum speed, it has no potential energy, but its kinetic energy is 1 mv 2 . The conservation of energy states that the final total energy of the electron equals the 2 initial total energy: 1 mv 2 = eV 2 2 1 24 Initial 3 4 3 1 total Final total energy energy Solving this equation for the potential difference gives V = mv 2 / ( 2e ) . The speed of the electron can be expressed in terms of the magnitude p of its momentum by v = p/m (Equation 7.2). The magnitude of the electron’s momentum is related to its de Broglie wavelength λ by p = h/λ (Equation 29.8), where h is Planck’s constant. Thus, the speed can be written as v = h/(mλ). Substituting this expression for v into V = mv 2 / ( 2e ) gives V = h 2 / ( 2meλ 2 ) . 1532 PARTICLES AND WAVES SOLUTION The potential difference that accelerates the electron is ( 6.63 × 10−34 J ⋅ s ) h2 V= = = 1.86 ×104 V 2 2 −31 −19 )( −11 2meλ 2 ( 9.11×10 kg ) (1.60 ×10 C 0.900 ×10 m) 2 34. REASONING The linear momentum p of a particle is given by p = mv (Equation 7.2), where m and v are its mass and velocity. Since particle A is initially at rest, its momentum is zero. The initial momentum of particle B is p0B = mBv0B. This is also the total initial linear momentum of the two-particle system. After the collision the combined mass of the two particles is mA+ mB, and the common velocity is vf .Thus, the total linear momentum of the system after the collision is pf = (mA+ mB)vf . From Section 7.2, we know that the total linear momentum of an isolated system is conserved. An isolated system is one in which the vector sum of the external forces acting on the system is zero. Since there are no external forces acting on the particles, the two-particle system is an isolated system. Thus, the total linear momentum of the system after the collision equals the total linear momentum before the collision. The de Broglie wavelength λ is inversely related to the magnitude p of a particle’s momentum by λ = h/p (Equation 29.8), where h is Planck’s constant. SOLUTION The de Broglie wavelength λf of the object that moves off after the collision is given by λf = h/pf (Equation 29.8). Since momentum is mass times velocity, the magnitude of the momentum that the object has after the collision is pf = (...
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