Unformatted text preview: 0 m, and we have
xcm = mc xc + mo xo
mc + mo = xo ( mc / mo ) + 1 = 1.13 ×10−10 m
= 6.46 × 10−11 m
0.750 mo / mo ) + 1
( 61. SSM REASONING AND SOLUTION The comet piece and Jupiter constitute an isolated
system, since no external forces act on them. Therefore, the headon collision obeys the
conservation of linear momentum:
( mcomet + mJupiter ) v f =
144 2444
4
3
Total momentum
after collision mcomet v comet + mJupiter v Jupiter
14444 4444
2
3
Total momentum
before collision where vf is the common velocity of the comet piece and Jupiter after the collision. We
assume initially that Jupiter is moving in the +x direction so v Jupiter = +1.3 × 10 4 m / s . The
comet piece must be moving in the opposite direction so v comet = –6.0 × 10 4 m / s . The final Chapter 7 Problems 387 velocity vf of Jupiter and the comet piece can be written as v f = v Jupiter + ∆v where ∆v is the
change in velocity of Jupiter due to the collision. Substituting this expression into the
conservation of momentum equation gives
( mcomet + mJupiter )( v Jupiter + ∆v ) = mcomet v comet + mJupiter v Jupiter Multiplying out the left side of this equation, algebraically canceling the term mJupiter v Jupiter
from both sides of the equation, and solving for ∆v yields ∆v =
= mcomet ( v comet – v Jupiter)
mcomet + mJupiter
(4.0 × 10 12 kg) (–6.0 × 10 4 m / s – 1.3 × 10 4 m / s)
= –1.5 × 10 –10 m / s
12
27
4.0 × 10 kg + 1.9 × 10 kg The change in Jupiter's speed is 1.5 × 10 –10 m / s . 62. REASONING We consider the boy and the skateboard as a single system. Friction between
the skateboard and the sidewalk is minimal, so there is no net horizontal force acting on the
system. Therefore, the horizontal component of the system’s linear momentum remains
constant while the boy pushes off from the skateboard: m1vf 1x + m2 vf2x = m1v01x + m2 v02x
(Equation 7.9a). We will solve this equation to find the final velocity of the skateboard.
SOLUTION The data given in the problem may be summarized as follows: Horizontal velocity component
Mass Initial Final Boy m1 = 40.0 kg v01x = v0 = +5.30 m/s vf1x = +(6.00 m/s)(cos 9.50°)
= +5.92 m/s Skateboard m2 = 2.50 kg v02x = v0 = +5.30 m/s vf2x = ? Solving equation 7.9a for the final horizontal component of the skateboard’s velocity, we
obtain
m ( v − v ) + m2v0
m2vf2x = m1v0 + m2v0 − m1vf 1x
or
vf2x = 1 0 f 1x
m2
Thus,
vf2x = m1 ( v0 − vf 1x )
m2 + v0 = ( 40.0 kg ) ( +5.30 m/s − 5.92 m/s ) + 5.30 m/s =
2.50 kg −4.6 m/s 388 IMPULSE AND MOMENTUM 63. REASONING For the system consisting of the female character, the gun and the bullet, the
sum of the external forces is zero, because the weight of each object is balanced by a
corresponding upward (normal) force, and we are ignoring friction. The female character,
the gun and the bullet, then, constitute an isolated system, and the principle of conservation
of linear momentum applies.
SOLUTION a. The total momentum of the system before the gun is fired is zero, since all
parts of the system are at rest. Momentum conservation requires that the total momentum
remains zero after the gun has been fired. m1 v f1 + m2 v f2 = 14 244
40
3
14 244 Total momentum before
4
3
Total momentum after
gun is fired gun is fired where the subscripts 1 and 2 refer to the woman (plus gun) and the bullet, respectively.
Solving for vf1, the recoil velocity of the woman (plus gun), gives
v f1 = – m2 v f2
m1 = – (0.010 kg)(720 m / s)
= – 0.14 m / s
51 kg b. Repeating the calculation for the situation in which the woman shoots a blank cartridge,
we have
mv
– (5.0 × 10 –4 kg)(720 m / s)
v f1 = – 2 f2 =
= – 7.1 × 10 –3 m / s
m1
51 kg
In both cases, the minus sign means that the bullet and the woman move in opposite
directions when the gun is fired. The total momentum of the system remains zero, because
momentum is a vector quantity, and the momenta of the bullet and the woman have equal
magnitudes, but opposite directions. 64. REASONING The mass of each part of the
seated human figure will be treated as if it
were all located at the corresponding center
of mass point. See the drawing given in the
text. In effect, then, the problem deals with
the threeparticle system shown in the
drawing at the right. To determine the x and
y coordinates of the center of mass of this
system, we will employ equations analogous
to Equation 7.10. The values for the masses
are m1 = 41 kg, m2 = 17 kg, and m3 = 9.9 kg. +y
m1 0.39 m
0.17 m
m2 +x
0.26 m 0.43 m m3 Chapter 7 Problems 389 SOLUTION The x and y coordinates of the center of mass for the threeparticle system in
the drawing are
xcm =
ycm = m1x1 + m2 x2 + m3 x3
m1 + m2 + m3 m1 y1 + m2 y2 + m3 y3
m1 + m2 + m3 = ( 41 kg ) ( 0 m ) + (17 kg ) ( 0.17 m ) + ( 9.9 kg ) ( 0.43 m ) = = ( 41 kg )( 0.39 m ) + (17 kg )( 0 m ) + ( 9.9 kg )( −0.26 m ) = 41 kg + 17 kg + 9.9 kg 0.11 m 41 kg + 17 kg + 9.9 kg 0.20 m 65. SSM REASONING We will define the system to be the platform, the two people and
the ball. Since the ball travels nearly ho...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details