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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
018. `query 18
Question: `qQuery intro problem sets
Explain how we determine the horizontal range of a projectile given its initial horizontal and vertical velocities and the vertical displacement.
Your solution:
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Given Solution:
`a** We treat the vertical and horizontal quantities independently.
We are given vertical displacement and initial velocity and we know that the vertical acceleration is the acceleration of gravity. So we solve the vertical motion first, which will give us a `dt with which to solve the horizontal motion.
We first determine the final vertical velocity using the equation vf^2 = v0^2 + 2a'ds, then average the result with the initial vertical velocity. We divide this into the vertical displacement to find the elapsed time.
We are given the initial horizontal velocity, and the fact that for an ideal projectile the only force acting on it is vertical tells us that the acceleration in the horizontal direction is zero. Knowing `dt from the analysis of the vertical motion we can now solve the horizontal motion for `ds. This comes down to multiplying the constant horizontal velocity by the time interval `dt. **
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Question: `qQuery class notes #17
Why do we expect that in an isolated collision of two objects the momentum change of each object must be equal and opposite to that of the other?
Your solution:
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Given Solution:
`a Briefly, the force exerted on each object on the other is equal and opposite to the force exerted on it by the other, by Newton's Third Law.
By assumption the collision is isolated (i.e., this is a closed system); the two objects interact only with one another. So the net force on each object is the force exerted on it by the other.
So the impulse F_net `dt on one object is equal and opposite the impulse experienced by the other.
By the impulse-momentum theorem, F_net `dt = `d ( m v). The impulse on each object is equal to its change in momentum.
Since the impulses are equal and opposite, the momentum changes are equal and opposite.
**COMMON ERROR AND INSTRUCTION CORRECTION: This is because the KE change is going to be equal to the PE change.
Momentum has nothing directly to do with energy.
Two colliding objects exert equal and opposite forces on one another, resulting in equal and opposite impulses. i.e., F1 `dt = - F2 `dt. The result is that the change in momentum are equal and opposite: `dp1 = -`dp2. So the net momentum change is `dp1 + `dp2 = `dp1 +(-`dp1) = 0. **
STUDENT QUESTION
Are impulses the same as momentum changes?
INSTRUCTOR RESPONSE
impulse is F * `dt
momentum is m v, and as long as mass is constant momentum change will be m `dv
by the impulse-momentum theorem impulse is equal to change in momentum (subject, of course, to the conditions of the theorem)
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Question: `qWhat are the six quantities in terms of which we analyze the momentum involved in a collision of two objects which, before and after collision, are both moving along a common straight line? How are these quantities related?
Your solution:
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Given Solution:
`a** We analyze the momentum for such a collision in terms of the masses m1 and m2, the before-collision velocities v1 and v2 and the after-collision velocities v1' and v2'.
Total momentum before collision is m1 v1 + m2 v2.
Total momentum after collision is m1 v1' + m2 v2'.
Conservation of momentum, which follows from the impulse-momentum theorem, gives us
m1 v1 + m2 v2 = m1 v1' + m2 v2'. **
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Question: `1prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much thermal energy is produced in the collision?
Your solution:
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Given Solution:
`aThere is no change in PE. All the initial KE of the cars will be lost to nonconservative forces, with nearly all of this energy converted to thermal energy.
The initial speed are 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s, so each car has initial KE of .5 m v^2 = .5 * 7650 kg * (26.4 m/s)^2 = 2,650,000 Joules, so that their total KE is 2 * 2,650,000 J = 5,300,000 J.
This KE is practically all converted to thermal energy.
STUDENT QUESTIONS
Why is the kinetic energy multiplied by two?
And why is all of the kinetic energy practically converted to thermal energy?
Is thermal energy simply two times the kinetic energy?
Is this what happens to all kinetic energy
in real life?
INSTRUCTOR RESPONSE
You've calculated the KE of one of the
cars. There are two cars, which is why we multiply that result by 2.
Some of the KE does go into producing sound, but loud as the crash might be only
a small fraction of the energy goes into the sound. Practically all the rest
goes into thermal energy. A lot of the metal in the cars is going to
twist, buckle and otherwise deform, and warm up some in the process. They
probably won't become hot to the touch, but it takes a lot more thermal energy
that that involved in this collision to achieve an overall temperature change we
would be likely to notice.
If two cars of unequal mass and equal speeds collide they don't come to rest, so
they have some KE after the collision.
It the cars were perfectly elastic they would rebound with their original
relative speed. A perfectly elastic collision is one in which kinetic energy is
conserved. No energy would go into thermal energy and there would be no sound.
This is an idean and cannot actually be achieved with railroad cars (nor with
steel balls, or marbles, or pool balls, etc.). However the collisions of
molecules in a gas are perfectly elastic, and analyzing the statistics of those
collisions allows us to explain a lot of what we observe about gases.
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Question `gen, `prin (Openstax #32):
A person in good physical condition can put out 100 W of useful power for
several hours at a stretch, perhaps by pedaling a mechanism that drives an
electric generator. Neglecting any problems of generator efficiency and
practical considerations such as resting time:
(a) How many people would it take to run a 4.00-kW electric clothes dryer?
(b) How many people would it take to replace a large electric power plant that
generates 800 MW?
Your solution:
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Given Solution:
4 kW is 4000 watts. At 100 watts per person, it would require
4000 watts / (100 watts / person) =
40 (watts / (watts / person)) =
40 watts * persons / watt =
40 persons
to power the dryer.
To produce 800 MW = 800 (1 000 000 watts) = 800 000 000 watts, or 8 * 10^8 watts, would require
800 000 000 watts / (100 watts / person) = 8 000 000 people.
This is a little more than the average population per State in the U.S.
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Question:
Query* 7.53. gen phy roller coaster 1.7 m/s at point 1, ave frict 1/5 wt, reaches poin 28 m below at what vel (`ds = 45 m along the track)?
Your solution:
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Given Solution:
`a**GOOD STUDENT SOLUTION WITH ERROR IN ONE DETAIL, WITH INSTRUCTOR CORRECTION:
Until just now I did not think I could work that one, because I did not know the mass, but I retried it.
Conservation of energy tells us that `dKE + `dPE + `dWnoncons = 0.
PE is all gravitational so that `dPE = (y2 - y1).
The only other force acting in the direction of motion is friction.
Thus .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * `ds = 0 and
.5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m)
It looks like the M's cancel so I don't need to know mass.
.5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 so
v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s.
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Question: Very large forces are produced in joints when a person jumps from some height to the ground.
(a) Calculate the force produced if an 80.0-kg person jumps from a 0.600–m-high ledge and lands stiffly, compressing joint material 1.50 cm as a result. (Be certain to include the weight of the person.)
(b) In practice the knees bend almost involuntarily to help extend the distance over which you stop. Calculate the force produced if the stopping distance is 0.300 m.
(c) Compare both forces with the weight of the person.
Your solution:
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Given Solution:
Intuitively, the person's potential energy decreases during the .6 meter fall, increasing his or her kinetic energy, which is then dissipated over a 1.5 cm distance by the compressive forces in the joint. So we could calculate the kinetic energy gained during the fall and figure out what force exerted through a 1.5 cm distance would be required to dissipate this energy. To do so would be to neglect the additional potential energy lost during the 1.5 cm compression, so this approach wouldn't be quite right, but that additional energy is in this case small so, at least for part (a), the error would be fairly insignificant.
A more general and complete approach is to apply the work-energy theorem.
Recall that the work-energy theorem says that
`dW_noncons_ON = `dKE + `dPE
(the work done by nonconservative forces on the system is equal to the sum of the changes in its kinetic and potential energies).
We will apply the work-energy theorem to the interval between leaving the ledge and coming to rest. We consider the changes in the person's kinetic and potential energy, and the work done by nonconservative forces on the person.
The person has negligible kinetic energy when stepping off the ledge, and zero kinetic energy when coming to rest, so between the beginning and the end of the interval there is no change in kinetic energy (there are changes in kinetic energy during the interval, but the kinetic energy at the end is the same as at the beginning of the interval). So for this interval `dKE = 0.
The potential energy change is equal and opposite to the work done by conservative forces. The only conservative force acting in this situation is the gravitational force, which is 80 kg * 9.8 m/s^2 = 780 N, approx.. This force is downward, in the same direction as the displacement, so the work done by the conservative force is positive. The downward displacement between the initial and final events is the 6 meter height of the ledge, plus the 1.5 cm through which the person's mass (or nearly all of it) descends after reaching the ground. The work is therefore equal to 780 N * 0.615 m = 470 Joules, approx.. The potential energy change is equal and opposite to this work and is therefore negative, equal to
`dPE = -470 Joules.
The nonconservative compressive force exerted by the joint material acts upward, in the direction opposite the compression. This force is exerted as (most of) the person continues to descend 1.5 cm = .015 m, so the force and the displacement have opposite directions.
If F_compression_ave represents the average compressive force, then, we have
`dW_noncons_ON = - (F_compression_ave * .015 cm).
Intuitively, the work done by the compressive force as it acts through the .015 cm distance of compression has to do 470 Joules of work to dissipate the energy the person gained from the gravitational force. However to be sure we have everything right, we apply the work-energy theorem:
`dW_noncons_ON = `dKE + `dPE
so
-(F_compression_ave * .015 cm) = 0 Joules + (-470 Joules).
Solving for the compressive force
F_compression_ave = 470 Joules / (.015 cm) = 30 000 Newtons, approximately.
MORE RIGOROUS SOLUTION (choose positive direction, use symbols)
We can be more rigorous by choosing a positive direction.
We will use `ds_y for the vertical displacement, `ds_compression for the displacement during compression, F_grav for the force exerted by gravity on the person, and other symbols as previously defined.
If we choose our positive direction as upward then
`dPE = - (F_grav * (`ds_y + `ds_compression) ) = - (-780 N * (-0.6 m + (-.015 m)) ) = 470 Joules
and
`dW_noncons_ON = F_compression_ave * `ds_compression = F_compression_ave * (-.015 m)
`dKE is still 0.
In symbols, then, we get
F_compression_ave * `ds_compression = - (F_grav * (`ds_y + `ds_compression) )
Solving for the force of compression
F_compression_ave = - (F_grav * (`ds_y + `ds_compression) ) / `ds_compression =
- (-780 N * (-0.6 m + (-.015 m)) ) / (-.015 m) = -470 Joules / (-.015 m) = 30 000 N, approx..
This is about 40 times the weight of the person.
(b) If the knees bend so that the person descends 0.3 meters, then the -.015 m in the preceding becomes -0.3 m and the average force becomes about 1500 N, roughly double the weight of the person. Note that the forces will no longer be concentrated only in the joint, but will be spread as well over the surrounding musculature.
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Question: `
Your solution:
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Given Solution:
`a** The forces acting on the package while sliding down the incline, include gravitiational force, normal force and friction; and after encountering the spring the tension force of the spring also acts on the package.
The normal force is Fnormal = 2 kg * 9.8 m/s^2 * cos(53.1 deg) = 11.7 N, approx.. This force is equal and opposite to the component of the weight which is perpendicular to the incline.
The frictional force is f = .2 * normal force = .2 * 11.7 N = 2.3 N, approx..
The component of the gravitational force down the incline is Fparallel = 2 kg * 9.8 m/s^2 * sin(53.1 deg) = 15.7 N, approx..
Friction acts in the direction opposite motion, up the incline in this case.
If we choose the downward direction as positive the net force on the package is therefore 15.7 N - 2.3 N = 13.4 N. So in traveling 4 meters down the incline the work done on the system by the net force is
13.4 N * 4 m = 54 Joules approx.
Just before hitting the spring we therefore have
.5 m v^2 = KE so v = +-sqrt(2 * KE / m) = +-sqrt(2 * 54 J / (2 kg) ) = +- 7.4 m/s.
If we ignore the gravitational and frictional forces on the object while the spring is compressed, which we really don't want to do, we would conclude the spring would be compressed until its elastic PE was equal to the 54 J of KE which is lost when the object comes to rest. The result we would get here by setting .5 k x^2 equal to the KE loss is x = sqrt(2 * KE / k) = .9 meters, approx..
However we need to include gravitational and frictional forces. So we let x stand for the distance the spring is compressed.
As the object moves the distance x its KE decreases by 54 Joules, its gravitational PE decreases by Fparallel * x, the work done against friction is f * x (where f is the force of friction), and the PE gained by the spring is .5 k x^2. So we have
`dKE + `dPE + `dWnoncons = 0 so
-54 J - 15.7N * x + .5 * 120 N/m * x^2 + 2.3 N * x = 0 which gives us the quadratic equation
60 N/m * x^2 - 13.4 N * x - 54 N m = 0. (note that if x is in meters every term has units N * m). Suppressing the units and solving for x using the quadratic formula we have
x = ( - (-13.4) +- sqrt(13.4^2 - 4 * 60 (-54) ) / ( 2 * 60) = 1.03 or -.8
meaning 1.07 m or -.8 m (see previous note on units).
We reject the negative result since the object will continue to move in the direction down the incline, and conclude that the spring would compress over 1 m as opposed to the .9 m obtained if gravitational and frictional forces are not accounted for during the compression. This makes sense because we expect the weight of the object (more precisely the weight component parallel to the incline) to compress the spring more than it would otherwise compress. Another way of seeing this is that the additional gravitational PE loss as well as the KE loss converts into elastic PE.
If the object then rebounds the spring PE will be lost to gravitational PE and to work against friction. If the object goes distance xMax back up the incline from the spring's compressed position we will have`dPE = -.5 k x^2 + Fparallel * xMax, `dKE = 0 (velocity is zero at max compression as well as as max displacement up the incline) and `dWnoncons = f * xMax. We obtain
`dPE + `dKE + `dWnoncons = 0 so
-.5 k x^2 + Fparallel * xMax + 0 + 2.3 N * xMax = 0 or
-.5 * 120 N/m * (1.07 m)^2 + 15.7 N * xMax + 2.3 N * xMax = 0
We obtain
18 N * xMax = 72 N m, approx., so that
xMax = 72 N m / (18 N) = 4 meters, approx..
This is only 2.93 meters beyond the position of the object when the spring was compressed. Note that the object started out 4 meters beyond this position. **
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