Preparation problems
10-19. An object initally at rest at the origin bursts
into two pieces with masses 560 g and 390 g. The 560-g piece moves off
at 2.1 m/s in the positive x-direction. Describe the motion of the
other piece.
Solution:
Since we are dealing with an explosion, there are no external
forces present, and we can apply the principle of conservation of
momentum. Since the initial momentum is zero, the final momentum must
be zero. Thus the momenta of the two pieces must be equal but opposite,
or
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0.56 kg
0.39 kg
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10-37. An Arianne rocket ejects 1.0 ×105 kg of
fuel in the 90 s after launch. (a) How much thrust is developed
if the fuel is ejected at 3.0 km/s with respect to the rocket?
(b) What is the maximum total mass of th rocket if it is to get
off the ground?
Solution:
(a) We need to use our thrust equation to determine how
much thrust that rate of change in mass will provide.
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T = - vex |
dM
dt
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DM
Dt
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Plugging in our numbers, we get
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- (3000 m/s) |
-1 ×105 kg
90 s
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(b) For the rocket to just get off the ground, its
thrust must be at least equal to the weight of the full rocket. Then
10-38. Ninety percent of a rocket's initial mass is
fuel. If the fuel is exhausted at 3.5 km/s relative to the rocket, what
is the terminal speed of the rocket?
Solution:
Since ninety percent of the rocket's mass is fuel, when all
the fuel is expended the final mass will be only 10 percent of the
initial mass. Thus
We can now use the rocket equation from our constants and equation sheet to
solve the problem.
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vi + vex ln |
é
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mi
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Follow-up problems
8-52. A 1.5-kg block is launched up a 30°
incline with an initial speed of 6.4 m/s. It comes to a halt after
moving 3.4 m along the incline, as shown in Fig. 8-34 (p. 199). Find
(a) the change in the block's kinetic energy; (b) the
change in the block's potential energy; (c) the work done by
friction; (d) the coefficient of kinetic friction.
Solution:
(a) The change in the kinetic energy of the block is
given by
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m vf2 - |
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m vi2 = - |
1
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m vi2 |
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1
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(1.5 kg) (6.4 m/s)2 |
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(b) The change in the potential energy of the block as
it moves a length l = 3.4 m along the incline is given by
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(1.5 kg) (9.8 m/s2) (3.4 m) sin30° |
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(c) The work done by friction in stopping the block is
given by
(d) The coefficient of kinetic friction of the surface
is found remembering that Wnc = -Ff ·l and Ff = mN. We
must further remember that the normal force N = m g cos30°.
Putting these together, we see
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-5.73 J
(1.5 kg) (9.8 m/s2) (cos30°) (3.4 m)
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8-61. A 190-g block is launched by compressing a spring
of constant k = 200 N/m a distance of 15 cm. The spring is mounted
horizontally, and the surface directly under it is frictionless. But
beyond the equilibrium position of the spring end, the surface has
coefficient of friction m = 0.27. This frictional surface exteds
85 cm, followed by a frictionless curved rise, as shown in Fig. 8-38 (p.
200). After launch, where does the block finally come to rest? Measure
from the left end of the frictional zone.
Solution:
Our problem obviously deals with the conservation of energy in
the presence of non-conservative forces. As such, we know
At this point, there are several ways we can approach the
problem. I will find the amount of energy lost to non-conservative work
during each pass through the frictional zone. By determining what my
initial energy was, I can then determine the number of passes through
the frictional zone required to stop the block. From this, I can
determin the stopping position of the block.
In passing through the zone, friction does an amount of work
on the block given by
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(0.27) (0.19 kg) (9.8 m/s2) (0.85 m) |
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Now for the block to stop, all of the energy imparted by the spring must
be dissipated by friction, so the total amount of work necessary is
So the number of passes must be given by the total amount of work done
divided by the amount of work done on one pass, or
Now we can find the position by remembering that on the first
trip, the block is traveling to the right, on the second to the left,
right on the third, etc. So the block should end up 0.265 of the way
from the right hand edge, or
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0.624 m from the left edge of the zone |
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Solutions translated from TEX by TTH, version 1.57.
