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course phy 201
9/19 10I hope that you still accept this assignment. I wanted to talk to you about not getting it submitted by this morning but didn't have the change. I also apologize for leaving during class today, I had an emergency.
@& No problem. If figured there was a good reason for you leaving. I'll see you in the morning.*@
`q001. If you roll a ball down a 30-cm ramp, from rest, and it requires 3 seconds to travel the length of the ramp, what are its average velocity, final velocity and acceleration?
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vave=('ds)/('dt)= 30cm/3s=10cm/s
@& Right
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a=(vf-v0)/'dt = (30cm/s^2-0cm/s^2)/3s=10cm/s
vf=v0+a 'dt = 0+10cm/s*3s=30cm/s^2
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@& Right equation, but vf is not 30 cm/s^2.*@
For this interval which three of the quantities v0, vf, `dt, `ds and a are you given?
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v0 'dt 'ds are given
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These three quantities all appear in one of the two equations above. Which is it?
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'ds=((vf+v0)/2)*'dt
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There are four quantities in that equation. What is the fourth?
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vf=final velocity
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Solve that equation for the fourth variable, in terms of the three known variables. Do this symbolically. Don't substitute and numbers until you have a symbolic solution. For example, if you were to solve the first equation for `dt (not something you would do with the present example), you would get `dt = (vf - v0) / a. Include a brief explanation of the algebra steps you used to solve the equation. Don't worry at this point if the algebra gives you a little trouble; the algebra in the General College Physics course isn't that bad, and we can remedy it if necessary. (University Physics students won't have any trouble with the algebra).
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v0=initial velocity 'ds=change in position change in position/change in time= 30cm/3s=10cm/s
'dt=change in time vf=final velocity which is unknown
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@& The equation
`ds = (vf + v0) / 2 * `dt
is solved for vf by first multiplying both sides by 2 to get
2 `ds = (vf + v0) * `dt
then dividing both sides by `dt to get
2 `ds / `dt = vf + v0
and finally subtracting v0 from both sides to get
vf = 2 `ds / `dt - v0.
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Now substitute the values of the three known quantities into your rearranged equation, and simplify. Include units. Again, most likely not everyone at this point will be able to do this with complete success, but with practice this won't be difficult.
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'ds=((vf+v0)/2)*'dt = vf=v0+a*'dt vf=(0cm/s+10cm/s)*3s=30cm/s^2
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@& cm/s^2 is not a unit of velocity.
When you multiply cm/s by s you get cm, which is also not a unit of velocity.
Using the solution for vf from the preceding:
vf = 2 `ds / `dt - v0
Substituting known values we have
vf = 2 * 30 cm / (3 s) - 0 = 20 cm/s.
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Now suppose that the ball rolls off the end of the first ramp, right onto a second ramp of length 100 cm. If the ball requires 2 seconds to roll down this ramp, then for this new interval, which of the quantities v0, vf, `dt, a and `ds do you know? Note in particular that the initial velocity on this ramp is not zero, so for this interval v0 is not zero.
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v0=30cm/s^2 'dt=2s 'ds=100cm vf=unknown a=unknown
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Apply the definitions of velocity and acceleration to figure out the other two quantities for this interval.
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Vave=('ds)/('dt)=100cm/2s=50cm/s
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@& Good start, but you still need to find vf and a.*@
Now figure out which of the two equations can be applied to your new information. Solve that equation for that quantity, in the manner used previously, substitute your known values, and see what you get.
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vf=vo+a*'dt = (30cm/s^2+100cm/s^2)*2s = 130cm/s62*2s = 260cm/s^2
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@& None of these quantities is equal to 100 cm/s^2, or to 100 cm/s.
a * `dt is done before adding to v0. The expression is v0 + a `dt, not (v0 + a) `dt.
The three quantities you know are v0, `ds and `dt. The given information doesn't include vf for this ramp.*@
@& The reasoning:
ave velocity = ave roc of position wrt clock time = change in position / change in clock time = 100 cm / (2 sec) = 50 cm/s.
ave accel = ave roc of velocity wrt clock time = change in velocity / change in clock time.
To get the change in velocity, knowing the initial velocity, we have to find the final velocity. Initial velocity is 20 cm/s, average velocity is 50 cm/s. What do we average with 20 to get 50? The answer is 80, since (80 + 20) / 2 = 50. So the final velocity is 80 cm/s.
Now we can find the change in velocity, which is vf - v0 = 80 cm/s - 20 cm/s = 60 cm/s.
Knowing this we can find the average acceleration
aAve = `dv / `dt = 60 cm/s / (2 s) = 30 cm/s^2.
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`q002. When you coasted the toy car to rest along the tabletop, how far did it travel after your finger lost contact with it, and how long did it take to come to rest?
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60cm is the length it traveled 3 sec is how long it took
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Using the definitions of velocity and acceleration, find the car's initial velocity and acceleration, assuming that acceleration to be constant.
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Vave=60cm/3s=20cm/s v0=2*(Vave+vf)= 2*(20cm/s+0cm/s)=40cm/s
@& You didn't find the acceleration.
Change in velocity is -40 cm/s, so acceleration is (change in vel) / (change in clock time) = -40 cm/s / (3 s) = -13.3 cm/s^2.*@
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Assuming the mass of the car to be 10 grams, how much force was required to produce the acceleration you observed?
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Fnet=m*a =10g*((0+40cm/s)/3s)= 10g*(40cm/s/3s)= 10g*13.33cm=133.33g cm
@& acceleration is 13.3 cm/s^2, not 13.3 cm.
So the units would be g cm^2 / s^2.*@
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How much work did this force do as the ball coasted to rest?
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'dw=F*'ds work=133.33g cm *60cm = 7999.8g cm^2
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What was the initial kinetic energy of the car, i.e., the kinetic energy at the instant it lost contact with your finger?
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KE=((1/2)+10g)*20cm/s= 5g*20cm/s=100g cm/s
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@& KE = 1/2 m v^2, not 1/2 m v.
You need to use the initial velocity of the car during the interval, which is 40 cm/s.
We almost never use average velocity as a basic for calculating kinetic energy.*@
`q003. When you set up the system with the four rubber band chains, two toy cars, the strap and the axel:
As seen by someone facing the strap from the side to which the cars were attached, was the more massive car on the left or the right side of the strap?
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more massive car wa on the right
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While the strap was being held stationary, on which side were the rubber band tensions greater?
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more tension on the left rubber band
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@& The rubber band whose length matched its length when supporting the more massive car had the greater tension. So the greater tension was on the side of the more massive car.*@
Just after release, on which side did the car move away from the strap?
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car on the right
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Just after release, on which side(s) was the tension in the chain connecting the car to the strap less than the tension in the chain pulling 'down' on the car? Explain your thinking.
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more tension was on the car on the left when released. less tension was on the more massive car so when all tension was released it caused the more massive car to move away from the strap and the less massive car with more tension to move toward the strap.
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@& The side on which the tension pulling 'down' on the car is greater than the tension between car and strap will be the side on which the car initially accelerates 'down' and away from the strap. The car's initial motion will follow the greater tension.*@
`q004. Assume that the mass of the car and magnet was 12 grams.
How close did you get the magnet to the car, and how far did the car then travel before coming to rest?
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magnet was 1cm from car. the car traveled 2cm
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Assuming that the car's acceleration was the same as when you coasted it across the table, how fast was it going when it started to slow down? (This is actually a complicated situation, since you don't know where the car was when the magnet's force became negligible, so just assume that this occurred about a centimeter from the car's initial position).
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Vave=60cm/3s=20cm/s 20cm/s=(0+x)/2=40cm/s 'ds=((vf+v0)/2)*'dt= (0+40cm/s)*3s=120cm/s^2
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What was the force on the coasting car?
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Fnet=12g*20cm/s=240g cm/s
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@& You would multiply the mass by the acceleration, not by the average velocity.*@
How much work did this force do on the car?
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240g cm/s*120cm/s^2=28800(cm/s)^2
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@& Here you would multiply the force obtained in the previous step by the distance the car moved.*@
How much kinetic energy do you conclude the car gained from the interaction of the two magnets?
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KE=work= 28800(cm/s)^2
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What was the momentum of the car just before it started to slow down?
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12g*20cm/s=240g cm/s
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@& You had some errors, but you're doing OK here.
Be sure to see my notes, and let me know if there's anything you don't understand.*@