Unformatted text preview: ˆ’ mOd sin 60.0o + mO d sin 60.0o + mS ( 0 )
=
= 0 nm
mO + mO + mS This result makes sense, because the sulfur atomâ€™s x coordinate is x3 = 0 nm, and the oxygen
atoms, which have equal masses, have x coordinates of equal magnitude but opposite signs:
x1 = âˆ’x2.
b. The y coordinate of the sulfur atom is zero (y3 = 0 nm), but the y coordinates of the two
oxygen atoms are both positive, so the center of mass of the sulfur dioxide molecule has a
positive y coordinate:
ycm = = + mO d cos 60.0o + mO d cos 60.0o + mS ( 0 nm )
mO + mO + mS
+2 mO d cos 60.0o
4 mO = = +2mO d cos 60.0o
2mO + 2mO o
+ d cos 60.0o + ( 0.143 nm ) cos 60.0
=
= +0.0358 nm
2
2 53. SSM WWW REASONING The system consists of the lumberjack and the log. For
this system, the sum of the external forces is zero. This is because the weight of the system
is balanced by the corresponding normal force (provided by the buoyant force of the water) 382 IMPULSE AND MOMENTUM and the water is assumed to be frictionless. The lumberjack and the log, then, constitute an
isolated system, and the principle of conservation of linear momentum holds.
SOLUTION
a. The total linear momentum of the system before the lumberjack begins to move is zero,
since all parts of the system are at rest. Momentum conservation requires that the total
momentum remains zero during the motion of the lumberjack. m1vf1 + m2vf2 = 1 24
40 3
14 244
4
3
Total momentum
just before the jump Initial momentum Here the subscripts "1" and "2" refer to the first log and lumberjack, respectively. Let the
direction of motion of the lumberjack be the positive direction. Then, solving for v1f gives
v f1 = â€“ m2 v f2
m1 =â€“ (98 kg)(+3.6 m / s)
= â€“1.5 m / s
230 kg The minus sign indicates that the first log recoils as the lumberjack jumps off.
b. Now the system is composed of the lumberjack, just before he lands on the second log,
and the second log. Gravity acts on the system, but for the short time under consideration
while the lumberjack lands, the effects of gravity in changing the linear momentum of the
system are negligible. Therefore, to a very good approximation, we can say that the linear
momentum of the system is very nearly conserved. In this case, the initial momentum is not
zero as it was in part (a); rather the initial momentum of the system is the momentum of the
lumberjack just before he lands on the second log. Therefore,
m v f1 + m2 v f2 = m1 v 01 + m2 v 02
11 4 44
4 2 3 14 244
4
3
Total momentum
just after lumberjack lands Initial momentum In this expression, the subscripts "1" and "2" now represent the second log and lumberjack,
respectively. Since the second log is initially at rest, v 01 = 0 . Furthermore, since the
lumberjack and the second log move with a common velocity, v f1 = v f2 = v f . The statement
of momentum conservation then becomes
m1 v f + m2 v f = m2 v 02 Solving for vf, we have
vf = m2 v 02
m1 + m2 = (98 kg)(+3.6 m / s)
= + 1.1 m / s
230 kg + 98 kg The positive sign indicates that the system moves in the same direction as the original
direction of the lumberjack's motion. Chapter 7 Problems 383 54. REASONING Since all of the collisions are elastic, the total mechanical energy of the ball
is conserved. However, since gravity affects its vertical motion, its linear momentum is not
conserved. If hf is the maximum height of the ball on its final bounce, conservation of
energy gives
1
mvf2 + mghf
24
14 244
3 1 2
= mv0 + mgh0
24
14 244
3 Ef E0 Solving this equation for hf gives
hf = 2
v0 âˆ’ vf2 2g + h0 SOLUTION In order to use this expression, we must obtain the values for the velocities v0
and vf. The initial velocity has only a horizontal component, v0 = v0 x .The final velocity also
has only a horizontal component since the ball is at the top of its trajectory, vf = vf x . No
forces act in the horizontal direction so the momentum of the ball in this direction is
conserved, hence v0 = vf . Therefore, hf = h0 = 3.00 m 55. REASONING a. During the collision between the bullet and the wooden block, linear
momentum is conserved, since no net external force acts on the bullet and the block. The
weight of each is balanced by the tension in the suspension wire, and the forces that the
bullet and block exert on each other are internal forces. This conservation law will allow us
to find the speed of the bullet/block system immediately after the collision.
b. Just after the collision, the bullet/block rise up, ultimately reaching a final height hf before
coming to a momentary rest. During this phase, the tension in the wire (a nonconservative
force) does no work, since it acts perpendicular to the motion. Thus, the work done by
nonconservative forces is zero, and the total mechanical energy of the system is conserved.
An application of this conservation law will enable us to determine the height hf.
SOLUTION a. The principle of conservation of linear momentum states that the total
momentum after the collision is equal to that before the...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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