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course Phy 201
`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?#$&*
For this interval which three of the quantities v0, vf, `dt, `ds and a are you given?
You are given the quantities ’dt, v0 and ‘ds.
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These three quantities all appear in one of the two equations above. Which is it?
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@& The equations are
`ds = (vf + v0) / 2 * `dt
and
vf = v0 + a `dt.
v0, `ds and `dt all appear in the first of these equations.
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There are four quantities in that equation. What is the fourth?
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@& The first equation includes vf, v0, `ds and `dt. The one that isn't given is vf.*@
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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@& 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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@& 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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@& The ball ends the first ramp at 20 cm/s, so this will be the initial velocity on the second. So for this inteval v0 = 20 cm/s.
The ramp is 100 cm long so `ds = 100 cm.
The time down the ramp is 2 seconds so `dt = 2 sec.
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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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@& 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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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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@& Our new information includes `ds, `dt and v0.
The first equation includes `ds, v0, vf and `dt. This include all our information and leaves only vf undetermined.
Solving the first equation for vf we start with the equation
`ds = (vf + v0) / 2 * `dt.
We multiply both sides by 2, divide by both by `dt then subtract v0 from both sides to get
vf = 2 * `ds / `dt - v0 = 2 * 100 cm / (2 s) - 20 cm/s = 100 cm/s - 20 cm/s = 80 cm/s
This is the same result we reasoned out earlier.
We now know v0, vf, `ds and `dt. We can easily find the acceleration, either by direct reasoning or by use of the second of our equations.
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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?
The car traveled a total distance of 60cm, and it took it 3 seconds to roll that far after my finger lost contact with it.
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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.
The cars initial velocity was 0 cm/ second and it acceleration was 13.3 cm / s^2. I found this by applying the definition (40 cm / s - 0 cm / s) / (3s) = 13.3 cm/ s.
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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?
Applying the definition F=ma I found the force produced by my finger to be 133N.
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How much work did this force do as the ball coasted to rest?
The work that the car did as it coasted to rest was 7980 N*cm. I found this by multiplying 133N and 60 cm together to get 7980 N*cm.
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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?
The initial kinetic energy was 0g*cm, applying the definition I got (1/2)(10)(0)^2= 0 g*cm.
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@& When it left your finger the car had velocity 40 cm/s. *@
`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?
The more massive car was on the right side of the strap.
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While the strap was being held stationary, on which side were the rubber band tensions greater?
The rubber band straps had the most tension on the left side because it had less weight and therefore had to be pulled back further to equal the distance that the right side rubber band was pulled back.
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@& The straps should have been pulled back to the same length they had when supporting the cars. So the side with the less weight would have had less tension.*@
Just after release, on which side did the car move away from the strap?
The right side car moved away from the strap because its mass was greater.
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@& The greater mass had something to do with this, since that car would have been pulled away from the strap with the greater force.*@
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.
The left side or the side with the less massive car would have less tension in the strap than the right side because the right side is heavier and pulls down on the rubber band causing it to have more tension.
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@& If the tension in the strap was less then the tension pulling the car 'down' would have been greater and the car would have accelerated 'down' and away from the original position of the strap.*@
`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?
I got the magnet about 9cm away from the car before it started rolling and the car rolled 20cm.
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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).
The car was going about 17.73cm/s^2. I found this by applying the acceleration formula Vf-Vo/Change in clock time.
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What was the force on the coasting car?
The force on coasting car was 212.76N. I found this by applying the formula F=ma.
@& The force might have been 213 g cm / sec^2, but it wasn't 213 Newtons.*@
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How much work did this force do on the car?
The car did 4255.2 N*cm. I found this answer by applying the formula work=F*d.
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How much kinetic energy do you conclude the car gained from the interaction of the two magnets?
The car gained about 1061.34 g*cm/s kinetic I found this by applying the formula for KE. I used the cars average velocity for (v) in the formula I’m not sure if that was correct to use that please correct me if I am in the wrong here.
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@& You would use the car's maximum velocity, which since the car came to rest would be double its average velocity. This would give you four times the KE you would get from the average velocity.
We almost never use average velocity in the KE formula.*@
What was the momentum of the car just before it started to slow down?
The momentum of the car was 159.6g * cm/s. I used the formula momentum= mass * velocity to find this answer.
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9/18 around 7:45" ""
@& You didn't put much information into the first series of questions, so be sure to see my notes.
You did well with the rest of the questions, but check my notes there as well regarding some of the details.*@