Unformatted text preview: s 2 ) 46. REASONING AND SOLUTION During the elastic collision both the total kinetic energy
and the total momentum of the balls are conserved. The conservation of momentum gives
mvf1 + mvf2 = mv01 + mv02
where ball 1 has v01 = + 7.0 m/s and ball 2 has v02 = â€“ 4.0 m/s. Hence,
vf1 + vf2 = v01 + v02 = 3.0 m/s
The conservation of kinetic energy gives
(1/2)mvf12 + (1/2)mvf22 = (1/2)mv012 + (1/2)mv022
or
vf12 + vf22 = v012 + v022 = 65 m2/s2
Solving the first equation for vf1 and substituting into the second gives
2vf22 â€“ (6.0 m/s)vf2 â€“ 56 m2/s2 = 0
The quadratic formula yields two solutions, vf2 = 7.0 m/s and vf2 = â€“ 4.0 m/s.
The first equation now gives vf1 = â€“ 4.0 m/s and vf1 = + 7.0 m/s. Hence, the 7.0m/sball
has a final velocity of âˆ’4.0 m/s , opposite its original direction. 378 IMPULSE AND MOMENTUM The 4.0m/sball has a final velocity of +7.0 m/s , opposite its original direction. 47. REASONING AND SOLUTION Initially, the ball has a total mechanical energy given by
E0 = mgh0. After one bounce it reaches the top of its trajectory with an energy of
E1 = 0.900 E0 = 0.900 mgh0
After two bounces it has energy
E2 = (0.900)2mgh0
After N bounces it has a remaining energy of
EN = (0.900)Nmgh0
In order to just reach the sill the ball must have EN = mgh where h = 2.44 m. Hence,
N (0.900) mgh0 = mgh or N (0.900) = h/h0 Taking the log of both sides gives
N log(0.900) = log(h/h0)
Then 2.44 m log 6.10 m = 8.7
N=
log (0.900)
The ball can make 8 bounces and still reach the sill. 48. REASONING
We will use the relation
m1 x1 + m2 x2
xcm =
(Equation 7.10) to determine
m1 + m2
the mass of the smaller star. Let the line joining
the centers of the two stars be the x axis, and let
the position of the larger starâ€™s center be x1 = 0 m
(see the drawing). Then the position of the
smaller starâ€™s center is x2 = x1 + d = d, where d is
the distance separating the centers of the stars. d x
xcm
x1 = 0 m x2 SOLUTION With the substitutions x1 = 0 m and x2 = d, Equation 7.10 simplifies to Chapter 7 Problems xcm = 379 0 + m2 d
m2 d
=
m1 + m2 m1 + m2 Solving this relation for the mass m2 of the smaller star, we obtain m1 + m2 = m2d
xcm or m2 = m1
d
âˆ’1
xcm Thus, the mass of the smaller star is
m2 = m1
3.70 Ã—1030 kg
=
= 1.51Ã—1030 kg
d
7.17 Ã— 1011 m
âˆ’1
âˆ’1
xcm
2.08 Ã—1011 m 49. SSM REASONING AND SOLUTION The velocity of the center of mass of a system is
given by Equation 7.11. Using the data and the results obtained in Example 5, we obtain the
following:
a. The velocity of the center of mass of the twocar system before the collision is ( vcm )before = m1v01 + m2v02 = m1 + m2 (65 Ã—103 kg)( + 0.80 m/s) + (92 Ã— 103 kg)( +1.2 m/s)
65 Ã— 103 kg + 92 Ã—103 kg = +1.0 m/s b. The velocity of the center of mass of the twocar system after the collision is ( vcm )after = m1vf + m2vf
m1 + m2 = vf = +1.0 m/s c. The answer in part (b) should be the same as the common velocity vf. Since the cars are
coupled together, every point of the twocar system, including the center of mass, must
move with the same velocity. 380 IMPULSE AND MOMENTUM 50. REASONING Using Equation 7.10, we can calculate the location of the center of mass of
John and Barbara:
m x + mB xB
xcm = J J
mJ + mB
By calculating John and Barbaraâ€™s center of mass before and after they change positions, we
can determine how far and in what direction their center of mass move as a result of the
switch.
SOLUTION Before xcm = mJ xJ + mB xB (86 kg ) ( 9.0 m ) + ( 55 kg ) ( 2.0 m )
=
= 6.3 m
mJ + mB
86 kg + 55 kg After xcm = mJ xJ + mB xB (86 kg )( 2.0 m ) + ( 55 kg )( 9.0 m )
=
= 4.7 m
mJ + mB
86 kg + 55 kg The center of mass moves by an amount 6.3 m âˆ’ 4.7 m = 1.6 m . Since it moves from the
6.3m point to the 4.7m point, the center of mass moves toward the origin. 51. REASONING AND SOLUTION The velocity of the center of mass is given by
Equation 7.11,
m v + m2v2 ( m1 / m2 ) v1 + v2
vcm = 1 1
=
m1 + m2
( m1 / m2 ) + 1
a. When the masses are equal, m1 / m2 = 1 , and we have vcm = v1 + v2
2 = 9.70 m/s + (â€“11.8 m/s)
= â€“1.05 m/s
2 b. When the mass of the ball moving at 9.70 m/s is twice the mass of the other ball, we
have m1/m2 = 2, and the velocity of the center of mass is
vcm = 2v1 + v2
2 +1 = 2(9.70 m/s) + ( â€“11.8 m/s)
= +2.53 m/s
3 Chapter 7 Problems 381 52. REASONING The coordinates xcm and ycm of the moleculeâ€™s center of mass are given by
m x + m2 x2 + m3 x3
m y + m2 y2 + m3 y3
xcm = 1 1
and ycm = 1 1
(Equation 7.10), where we make
m1 + m2 + m3
m1 + m2 + m3
the following identifications:
Atom
Oxygen (left side) Mass m1 = mO x2 = +d sin 60.0Â° y2 = +d cos 60.0Â° m3 = mS = 2mO Sulfur ycoordinate
y1 = +d cos 60.0Â° m2 = mO Oxygen (right side) xcoordinate
x1 = âˆ’d sin 60.0Â° x3 = 0 nm y3 = 0 nm Here, we have used d = 0.143 nm to represent the distance between the center of the sulfur
atom and the center of either oxygen atom.
SOLUTION
a. Making substitutions into xcm = m1 x1 + m2 x2 + m3 x3 (Equation 7.10) from the data table
m1 + m2 + m3
above, we obtain the x coordinate of the moleculeâ€™s center of mass:
xcm â...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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