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course phy241
09/22/10 1:30am
Assuming that the ball fell to the floor in .4 seconds, after leaving the end of the ramp, and that after leaving the ramp its horizontal velocity remains constant: How fast was it traveling in the horizontal direction when the domino was lying flat on its side?
7/.4=17.5cm/s
It's not clear where the 7 came from, or why you divided it by .4.
The calculation should have units with every quantity.
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How fast was it traveling in the horizontal direction when the domino was lying on its long edge?
16.5/.4=41.25cm/s
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How fast was it traveling in the horizontal direction when the domino was lying on its short edge?
26/.4=65cm/s
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Pendulum count:
What was the length of the pendulum you counted, and how many counts did you get in 30 seconds?
10.5cm, 41 sways in 30 sec
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What therefore is the period of motion of that pendulum?
.73 sec per cycle
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How does your result compare with the formula given on the board, T = .2 sqrt(L) where T is period of oscillation in seconds and L the length in centimeters?
mine was .12 sec longer than the formula calculated
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How well did the freely oscillating pendulum synchronize with the bouncing pendulum of the same length? Which was 'quicker'?
ok,it was very hard to tell, the bouncing was quicker
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Ball drop
From what height did the drop of the ball synchronize with the second 'hit' of the pendulum, and what was the length of the pendulum?
approx 1 m, 10.5 cm
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How long should it have taken the pendulum between release and the second 'hit'? On what do you base this answer?
?
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Given you answer to the preceding, you know the time required for the ball to fall from rest to the floor, and you know how far it fell. What therefore was its acceleration??
You should be able to figure out how long the pendulum took from release to second 'hit', based on its period. If you don't get it right, you shouldn't be too far off. You would then be able to find the acceleration.
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Ball down long ramp
How would you design an experiment to measure the velocities v0, v_mid_x, v_mid_t and v_f for different values of v0?
use the timer program to measure the velocities by finding the time intervals for midx and midt
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How would you design an experiment to measure v0 and `dv for different values of v0?
set a ramp up in front of the ramp you want to use for your vo and figure out what the final of that ramp is for the initial of the next ramp(assuming a smooth transition) you can the ajust the first ramp to create different v0 for the experiment
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Rotating strap
For the strap rotating about the threaded rod, give your data indicating through how many degrees it rotated, how long it took and the average number of degrees per second. Report one trial per line, with a line containing three numbers, the number of degrees, the number of seconds, and the average number of degrees per second, separated by commas.
1080,4 sec, 270 degrees/sec
2430,7 sec,approx 347degress/sec
4320,15sec,288degrees/sec
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To do with the materials you took home:
Using the TIMER program (link to Java version: Timer_b) with the materials you took home:
Bracket pendulum:
Shim the bracket pendulum until the 'strikes' appear to occur with a constant interval. Click when you release the bead, then click for alternate 'strikes' of the ball on the bracket pendulum (that is, click on release, on the second 'strike', on the fourth 'strike', etc., until the pendulum stops striking the bracket). Practice until you think you think your clicks are synchronized with the 'strikes'. Report the length of the pendulum in the first line, then in the second line report the corresponding time intervals below, separated by commas:
10cm
.6166992,.6459961,.7041016,.6821289,.6889618,.698711
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Using the same length, set the pendulum so it swings freely back and forth. Click each time the bead passes through the equilibrium position. Continue until you have recorded 11 'clicks'. Report the corresponding time intervals below in one line, separated by commas.
.3729297,.3862505,.3540039,.3779297,.3359375,.3393555,.3359795,.362793,.3520508,.3647461,.340332
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For both sets of trials, how do your results compare with the prediction of the formula T = .2 sqrt(L)?
t=.2sqrt10=.63. my clicks were pretty close but not my swings
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Ball down ramp:
Do your best to take measurements you can use to find vf, v_mid_x, v_mid_t and `dv using your ramp and ball, releasing the ball from rest. (You could use the TIMER to get decent data. If you wish you can use the fact that a ball falling off a typical table or countertop will reach the floor in about .4 seconds. Note: Don't let the ball fall on a tile or vinyl-covered floor. You don't want broken tile, and you don't want dents in your vinyl. You could put your book on a carpeted or otherwise protected floor and land the ball on the book.)
Briefly describe what you did and what your results were:
I used the timer program, measured and marked where my midpoint was on the ramp with a pen,then clicked on initial release, midpoint position and final.
vmidx=15cm,vmidt=1.26sec,`dv=2.51
you don't give `dt for the entire ramp
it's unclear what your `dv is based on or for what interval you calculated it
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Rotating strap:
Let the strap rotate on the threaded rod, as before. Click the TIMER at the start, and then at 180 degree and/or 360 degree intervals (the latter if it's moving too fast to do the former). Copy the output of the TIMER program below:
1 5732.583 5732.583
2 5733.503 .9199219
3 5734.521 1.018555
4 5735.833 1.311523
5 5737.691 1.858398
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On the average through how many degrees per second was the strap rotating during each interval? Report in a single line, giving the numbers separated by commas. Starting in the second line explain how you did your calculations.
391.337, 353.44, 274.49, 193.72
divided 360 by the interval
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The second column of the TIMER output shows the clock times. For a given interval the 'midpoint clock time' is the clock time in the middle of the interval. Report clock times at the beginning, middle and end of your first interval in the first line below. Do the same for your second interval, in the second line. Starting in the third line explain how you got your results.
?
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midpoint clock time was (hopefully) clarified in class the other day
Notes
Equations of motion
From analyzing the v0, vf, `dt trapezoid we find that whenever the v vs. t trapezoid is an accurate representation of the behavior of the object, the acceleration is uniform and we have
`ds = (vf + v0) / 2 * `dt.
and
vf = v0 + a `dt.
These equations apply as long as the straight line segment accurately represents the motion of the object during the interval.
These equations involve five quantities: v0, vf, a, `ds and `dt.
If we know three of these quantities, then only two remain unknown.
For example if we know a, v0 and `dt:
The first equation includes the quantities `ds, vf, v0 and `dt. Knowing a, v0 and `dt, we would have the values of three of the four quantities in that equation, and would therefore be able to easily substitute the known values to find `ds.
The second equation includes the quantities vf, v0, a and `dt. Knowing a, v0 and `dt we could easily substitute to find the value of vf.
Of course we don't really need the equations to solve this situation. Knowing a, v0 and `dt, we can use a and `dt to find the change in velocity, which we can then add to v0 to obtain the final velocity. Having found the final velocity we can average it with the initial velocity to obtain the average velocity (this works because the v vs. t graph is a straight line, i.e., acceleration is uniform), which we then multiply by the known `dt to find `ds.
As another example, if we know a, v0 and vf:
The first equation includes our known quantities v0 and vf, but also includes the unknown quantities `dt and `ds. We can't solve a single equation for two different unknowns, so the first equation isn't helpful.
The second equation does contain all three of our known quantities, plus the quantity `dt. So we will be able to use this equation to find `dt:
vf = v0 + a `dt. Subtracting v0 from both sides we get
a `dt = vf - v0. Dividing both sides by a we get
`dt = (vf - v0) / a.
If we substitute our values for v0, vf and a into this last form of the equation, we easily find `dt.
As another example, suppose we know v0, a and `ds.
The first equation contains v0 and `ds, but it also contains the unknown quantities vf and `dt.
The second equation contains v0 and a, but it also contains the unknown quantities vf and `dt.
Neither equation can be solved by itself; you can't solve a single equation with two unknowns.
However we do have two equations in the two unknowns vf and `dt. Given two equations in the same two unknowns, we have a system of simultaneous equations, and we can solve the system.
To save time, we go ahead and solve the system symbolically for the variables vf and `dt:
... identifying quantities, units
... problems involving various combinations, using eqns and/or using definitions
These equations should not prevent us from directly reasoning out as much of the solution as possible, using the seven quantities vf, v0, vAve, `dv, a, `dt and `ds and the definitions of average velocity and acceleration.
Most combinations of three of the five variables can be reasoned out without resorting to either of these equations. Only two combinations (namely, (v0, a `ds) and (vf, a, `ds)) cannot be directly reasoned out using the definitions. In those cases, we would require several steps to solve the two simultaneous equations. We can take care of that complication now so we don't have to worry about it later. We will use the process of eliminating one variable from the system, in the first case eliminating vf, in the second case eliminating `dt:
We can pretty easily eliminate vf from the two equations:
The equation vf = v0 + a `dt gives us an expression to substitute for vf in the equation `ds = (vf + v0) / 2 * `dt. We get
`ds = ( (v0 + a `dt) + v0 ) / 2 * `dt. Simplifying the numerator (v0 + a `dt) + v0 we get
`ds = ( 2 v0 + a `dt) / 2 * `dt. Dividing the numerator by 2, being sure to use the distributive law, we obtain
`ds = ( (2 v0) / 2 + a `dt / 2) * `dt = (v0 + 1/2 a `dt) * `dt. Multiplying through by `dt we get
`ds = v0 `dt + 1/2 a `dt^2. We will refer to this as the third equation of uniformly accelerated motion.
We can eliminate `dt from the two equations.
We solve the first for `dt:
a = (vf - v0) / `dt so
`dt = (vf - v0) / a.
We now substitute this expression for `dt in the second equation `ds = (v0 + vf) / 2 * `dt. We obtain
`ds = (v0 + vf) / 2 * (vf - v0) / a.
The right-hand side can be written (vf + v0) ( vf - v0) / (2 a). Since (vf - v0) ( vf + v0) = vf^2 - v0^2, our equation becomes
`ds = (vf^2 - v0^2) / (2 a).
We could leave the equation as it is, but you will see later that there is good reason to rearrange it as follows:
Multiply both sides by 2 a to get
2 a `ds = vf^2 - v0^2.
Then solve for vf^2, obtaining
vf^2 = v0^2 - 2 a `ds.
We will refer to this as the fourth equation of uniformly accelerated motion.
Our v vs. t trapezoid therefore gives us four equations that apply to uniformly accelerated motion on an interval:
`ds = (v0 + vf) / 2 * `dt (we will call this our first equation)
vf = v0 + a `dt (we will call this our second equation)
`ds = v0 `dt + 1/2 a `dt^2 (we will call this our third equation)
vf^2 = v0^2 + 2 a `ds. (we will call this our fourth equation)
University Physics students:
We can do something similar with the equations we derived previously using integration. Integrating a(t) = a = constant, assuming velocity v0 and position x0 at clock time t = 0, we get the two equations
v(t) = v0 + a t
x(t) = x0 + v0 t + 1/2 a t^2.
We can easily eliminate a between these equations to obtain
x(t) - x0 = (v(t) - v0) / 2 * t
We can also eliminate t between the equations, obtaining
v(t) = v0^2 + 2 a ( x(t) - x0).
Our four equations are therefore
x(t) - x0 = (v(t) - v0) / 2 * t
v(t) = v0 + a t
x(t) = x0 + v0 t + 1/2 a t^2.
v^2(t) = v0^2 + 2 a * (x(t) - x0).
Our second equation can be rearranged to the form x(t) - x0 = v0 t + 1/2 a t^2, so using s(t) = x(t) - x0 this equation becomes s(t) = v0 t + 1/2 a t^2. If we use s(t) for the expression x(t) - x0, our four equations become our four equations can therefore be written
s(t) = (v(t) - v0) / 2 * t
v(t) = v0 + a t
s(t) = v0 t + 1/2 a t^2
v^2(t) = v0^2 + 2 a s(t).
The expression s(t) is understood to represent the displacement x(t) - x0 since initial clock time t = 0. s(t) can always be replaced by x(t) - x0, should it be convenient to do so. If x0 = 0, then we can replace s(t) with x(t).
If we are talking about motion on an interval, then s(t) is just the displacement `ds from the beginning of the interval to the end, and v(t) is just the final velocity on the interval. The clock time t becomes the time duration of the interval, which we call `dt. Thus, using the symbols `ds, vf and `dt, our equations become
`ds = (v0 + vf) / 2 * `dt
vf = v0 + a `dt
`ds = v0 `dt + 1/2 a `dt^2
vf^2 = v0^2 + 2 a `ds.
These are the same as the equations obtained previously from the v vs. t trapezoid.
The three forms of the four equations as given above are all equivalent. One form can be obtained from and of the other two.
You should be able to quickly derive the third set of equations from the v0, vf, `dt trapezoid.
You should be able to derive the first set using integration.
You should understand how the two are connected, so that given one form you can always write down the other.
Calculating Projections (university physics only)
The dot product of vectors `A and `B is `A dot `B = || `A || * || `B || * cos(theta), where theta is the angle between the vectors.
The magnitude of the projection of `A on `B is easily seen to be equal to || `A || * | cos(theta) |, which by the above is equal to | `A dot `B | / || `B ||.
The expression `A dot `B / || `B || is equal either to the magnitude of the projection, or the negative of the magnitude of the projection, depending on whether the projection is in the direction of `B or opposite the direction of `B. It this quantity is multiplied by a unit vector in the direction of `B, the result will be the projection vector. We might refer to this quantity as the 'signed magnitude'.
A unit vector in the direction of `B is the vector `B / || `B ||.
It follows that the projection of `A on `B is
projection = ( `A dot `B / || `B || ) * `B / || `B ||.
This form shows how the projection is constructed and a 'signed magnitude' multiplied by a unit vector. This results in the formula
projection of `A on `B = `A dot `B / || `B ||^2 * `B.
It is recommended that you use the previous expression ( `A dot `B / || `B || ) * `B / || `B || for now, in order to remind yourself of the construction. You should also sketch the corresponding diagram, for the same reason.
Calculating cross product
Use the determinant form to represent the cross product.
Alternatively, note that `i X `j = `k, `i X `k = - `j, and `j X `k = `i. The distributive law works for cross products. Thus
(a1 `i + a2 `j + a3 `k) X (b1 `i + b2 `j + b3 `k)
= a1 `i X (b1 `i + b2 `j + b3 `k) + a2 `i X (b1 `i + b2 `j + b3 `k) + a3 `i X (b1 `i + b2 `j + b3 `k)
= a1 b1 `i X `i + a1 b2 `i X `j + a1 b3 `i X `k +
a2 b1 `j X `i + a2 b2 `j X `j + a2 b3 `j X `k +
a3 b1 `k X `i + a3 b2 `k X `j + a3 b3 `k X `k
= (a1 b2 - a2 b1) `i X `j + (a1 b3 - a3 b1) `i X `k + (a2 b3 - a3 b2) `j X `k
= (a1 b2 - a2 b1) `k + (-a1 b3 + a3 b1) `j + (a2 b3 - a3 b2) `i
In the above we have made use of the fact that for any two vectors `A and `B we have `A X `B = - `B X `A (the reason should be obvious by the right-hand rule); so for example since `i X `k = -`j, `k X `i = `j.
The last form is identical to the form you get by evaluating the determinant form.
Applications of dot and cross product
This is one answer to a question about how you integrate in 3-dimensional space, the answer most immediately relevant to the dynamics class being taken by several students:
If `F and ``dx represent a force vector and a displacement through which the force vector acts, then the expression
`F dot ``dx
represents the product of || `F || cos(theta) with ``dx, where theta is the angle between ``dx and `F.
|| `F || cos(theta) is the component of `F in the direction of ``dx, i.e., the component of the force vector in the direction of the displacement.
It is this component that determines the work done by the force. The more of the force component in the direction of motion, the more work the force does, and the greater the displacement ``dx the greater the work.
The expression `F dot ``dx is the work done by the force `F if exerted through displacement ``dx.
Now if you have a force field `F and a curved path through that force field, you can divide that path up into a large number of nearly-straight segments of form ``dx. At the location of a typical segment you have a force `F, which does work `dW = `F ``dx over that part of the path. If we add all these `dW contributions up, we get a good approximation to the total work done over the entire path. The smaller the ``dx segments, i.e., the more finely we partition the curved path, the more accurate our approximation. In the limit as the lengths of the segments approach zero, we get the exact work done. This is clearly an integration process. The details depend on the force function `F, which is usually a function of position, e.g., `F(x, y, z), and the parameterization of the curve (for example a set of three continuous functions x(t), y(t), z(t), over some t interval, are said to parameterize the path in terms of the parameter t; that is, if you evaluate the three functions at some value of t, you get the corresponding position on the curve). So starting with force as a function of position, and a parameterization that gives us position (x, y, z) as a function of some parameter like t, we can integrate `F over the corresponding path. This requires more details and some practice, but this is an outline of how we might integrate over a curve in 3-dimensional space. Glad you asked? Maybe soon...
An example of cross product:
The torque of a force exerted perpendicular to a rod constrained to rotate about some axis (think of a wrench turning a bolt, with the axis along the bolt) is equal to the distance from the axis at which the force is exerted, and the force. There is a big advantage to regarding the torque as a cross product of the `r vector (from the axis to the point of applications) and the force vector. The direction of the torque is therefore perpendicular to the `r vector and the force vector, by the right-hand rule. The direction of the torque is therefore the direction in which it would advance a right-hand screw.
If the force vector is parallel to the `r vector no torque is exerted (think of trying to turn a nut by pushing on the end of the wrench, pushing it toward the nut; you will have no turning effect. Your force will be in the direction of the `r vector, the angle between the two being 180 degrees, and the cross product will be zero since the sine of 180 degrees is zero).
If the force is neither perpendicular nor parallel to the `r vector, then the force vector will have a component parallel to the `r vector, and a component perpendicular to the `r vector. It is the perpendicular component that turns the nut or bolt. The perpendicular component is the force is || `F || sin(theta), where theta is the angle between the `r vector and the force vector.
Sine and cosine functions in analysis of motion
If x(t) = A sin(omega * t), then x(t) describes a motion of amplitude A and period 2 pi / omega. An ideal pendulum or a mass on an ideal spring will move in such a manner. Much more about that later in the course.
If x(t) = A sin(omega * t) then
v(t) = x ' (t) = omega A cos(omega * t)
describes the corresponding velocity as a function of clock time, and
a(t) = v ' (t) = -omega^2 A sin(omega * t)
describes the corresponding acceleration.
For future reference: This motion would be called simple harmonic motion, and it results whenever an object experiences a net force toward some equilibrium position, with the net force being directly proportional to the distance from that position.
Related note: One student asked whether optics is a second-semester topic because it is difficult. Any topic can be difficult to just about any degree, depending on how deeply we go into it. However second-semester topics aren't delayed until second semester because they are difficult. They are delayed because they rely on very basic first-semester topics like motion, force, energy and momentum, as well as on other first-semester topics such as simple harmonic motion, and/or gravitation, and/or angular dynamics and/or others. The course gets progressively more difficult if early topics aren't mastered; with good mastery the topics get progressively easier.
Note on questions (accessible to University Physics students)
Two vectors are parallel if their cross product is zero, or if their dot product is equal to plus or minus the product of their magnitudes, or if one is a scalar multiple of the other (which leads to three simultaneous equations in a single unknown; the value of the unknown must be the same for all three of the equations). The simplest test for orthogonality is whether the dot product is zero.
An object thrown up in the air will return to its original vertical position with a speed equal to its original speed, provided only gravity acts on it. If other forces such as air resistance act on it, this is no longer the case.
To find the final velocity without actually measuring it, you could integrate the acceleration function to get the change in velocity. If you know the original velocity you can then add the change to the original velocity to get the final velocity. If it's a uniform acceleration situation, then if you know three of the quantities v0, a, `dt and `ds, you can either reason out the final velocity or use the equations to find it.
The x and y components of the length of the rubber band are both displacement; neither is a force. The x and y components of the force vector are both forces; neither is a displacement. The tension of a rubber band has a magnitude (e.g., how many Newtons of force) and a direction. The direction of the tension force will be the same as that of the displacement vector from one end of the rubber band (the one at which you wish to find the tension vector) to the other. To find a unit vector in the direction of the displacement vector, divide the components of the displacement vector by its magnitude (equal to the length of the rubber band). To get the force vector, multiply this unit vector by the magnitude of the force.
Questions and Problems
These should be submitted using the Submit Work Form. You can submit the entire document at once, or you can submit the document in parts.
Very Short Preliminary Activity with TIMER (should take 5 minutes or less once you get the TIMER loaded)
This exercise can be put off until you are near a computer. However it is best done before some of the problems that follow. If you can't do it before starting the problems, at least imagine doing it, actually doing the 8-counts and clicking an imaginary mouse, and making your best estimate of the time intervals.
Click the mouse as you start an 8-count, doing your best to count at the same rate you used in class. Complete four 8-counts and click the mouse again. Note the time interval required to complete your set of four 8-counts.
Repeat four more times.
Report your five time intervals in the first line below, separated by commas:
.218,.181,.221,.186,.175
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Based on your results, how long does your typical 8-count last?
1.57 sec
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Based on your result, what is the time interval of each of your counts?
.196 sec
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If you counted the motion of a ball down the ramp, completing two 8-counts and 1-2-3-4-5 of a third, how long would you conclude the ball spend moving down the ramp? Based on the TIMER data you reported above, what do you think is the percent uncertainty in your result?
4.12 sec, 30%
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Preliminary problems:
a. A ball travels down a ramp in 2 seconds, accelerating uniformly. Its initial velocity on the ramp is 20 cm/s and its final velocity is 40 cm/s.
Reasoning from the definitions of velocity and acceleration, and assuming a linear v vs. t graph, how long is the ramp, and what is the ball's acceleration (i.e., rate of change of velocity with respect to clock time)?
60 cm, 10m/s^2. Average velocity is 30cm/s so in 2sec it travels 60cm. Acceleration was simply based on it accelerated 20cm/s in 2 sec so 10m/s^2
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In the first line below list the quantities v0, vf, `ds, `dt and a for this motion and give the value of each (or as many as you were able to identify or reason out). In the second line identify which of the quantities were given, and which were reasoned out. In the reasoning process you would have found vAve and `dv; identify these quantities also and give their values.
v0=20c,/s,vf=40cm,/s, `ds= 60cm, `dt=2sec,a=10m/s^2
v0,vf,and`dt were given and everything else is reasoned out. Vave=30cm/s `dv=20cm/s
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b. A ball travels down a ramp for 3 seconds, starting with velocity 20 cm/s and with its velocity changing with respect to clock time at 10 cm/s^2.
Reasoning from the definitions of velocity and acceleration, and assuming a linear v vs. t graph, how far did the ball travel along the ramp, and what is the ball's velocity at the end of the 3 seconds?
105cm,50cm/s
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In the first line below list the quantities v0, vf, `ds, `dt and a for this motion and give the value of each (or as many as you were able to identify or reason out). In the second line identify which of the quantities were given, and which were reasoned out. In the reasoning process you would have found vAve and `dv; identify these quantities also and give their values.
v0=20cm/s,vf=50cm/s,`ds=105cm, `dt=3sec, a=35cm/sec
v0,a,and`dt were given all others reasoned. vave= 35cm/s, dv=30cm/s
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c. A ball travels 30 cm down a ramp in 5 seconds, ending with a velocity of 20 cm/s.
Identify, by giving the value of each, which of the quantities v0, vf, a, `ds and `dt are given.
`ds,`dt,vf
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Identify which of the four equations of uniformly accelerated motion contain the three given quantities (identify all the equations that apply; there will be at least one such equation, and no more than two).
`ds=(v0+vf)/2*`dt
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For each of the equations you identified, identify the quantity that was not given, and do your best to solve that equation for that quantity.
v0=-8cm/s
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d. A ball travels 30 cm down a ramp, accelerating at 10 cm/s^2 and ending with a velocity of 20 cm/s.
Identify, by giving the value of each, which of the quantities v0, vf, a, `ds and `dt are given.
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Identify which of the four equations of uniformly accelerated motion contain the three given quantities (identify all the equations that apply; there will be at least one such equation, and no more than two).
vf^2=vo^2-2a`ds
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For each of the equations you identified, identify the quantity that was not given, and do your best to solve that equation for that quantity.
v0=31.62cm/s
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All students should be able to make a good attempt at the questions in the Preliminary Questions, though it is expected that there will be questions on some of the details.
Problems
1. Each ramp used in constructing the series of ramps used in today's lab was 24 inches long. The 5-ramp series had a length of 10 feet, or about about 300 cm. Assume that the ball takes 10 seconds to travel the length of the ramp when released from rest. If this time interval is accurate, then what is the value of each of the following:
The average velocity of the ball on the ramp.
30cm/s
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The final velocity of the ball on the ramp.
60cm/s
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The change in the velocity of the ball from start to finish.
60cm/s
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The average rate of change of the velocity with respect to clock time.
6cm/s^2
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The velocity of the ball at the midpoint of the ramp.
42.42cm/s
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The velocity of the ball at the clock time halfway between the start and the ball reaching the end of the ramp.
30cm/s
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Hint: sketch a trapezoid that you think represents the v vs. t behavior of the ball on the ramp (... time on each ... need equal-area divisions ... etc.)
It is expected that some phy 201 students will be able to make a good attempt on all the above questions, and all should be able to answer the first two and make a good attempt on the next two. University Physics students should be able to make a good attempt on all questions.
2. Based on your in-class counts and your timing of your counts, estimate as accurately as you can the time required for the ball to travel the length of this series of ramps, starting from rest.
Find the average velocity of the ball, and based on this result find its final velocity.
vave=31.56cm/s,vf=63.12cm/s
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Using your results, find its acceleration.
6.64cm/s^2
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Find the velocities v_mid_x and v_mid_t.
midx=44.63cm/s,midt=31.56cm/s
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Everyone should be able to do the first. University Physics students should certainly be able to do the second, and General College Physics students should be able to make a good attempt.
3. If the ball was given an initial velocity of 20 cm/s, then given the acceleration you found in the preceding problem:
How long would it take the ball to travel the length of the ramp, and what would be its final velocity?
6.96s,66.21cm/s
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Where would it be at the halfway clock time?
109.81cm
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How fast would it be moving at the midpoint between the two ends of the ramp?
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What would be its velocity at the halfway clock time?
43.11cm/s
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What would be the change in its velocity from one end of the ramp to the other?
46.21cm/s
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Everyone should be able to make a good attempt at some of these questions. University physics students should be able to make a good attempt at all.
4. A ball requires a count of 24 to accelerate from rest down a 60 cm ramp. It rolls from that ramp onto an identical ramp with an identical slope, and requires 13 counts from one end of the ramp to the other. Does it lose any speed in making the transition? If you simply answer 'yes' or 'no' without supporting your answer in detail, you haven't answered the question.
yes,because the vf of the first ramp should be the v0 of the next and it wasnt. vf of the first was 5cm/s and v0 of the second was 4.23cm/s
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This question is somewhat challenging. You should begin by figuring out everything you can from the given information. Then see how your information might be used to answer the question.
5. I just timed myself for five sets of counts, with four fast 8-counts in each set (imilar to the preliminary exercise I asked you to do at the beginning of these problems). My times for the sets were all between 4.4 and 4.6 seconds. Starting with 1 at release and counting until the ball reached the end of the last ramp, I counted two sets of four 8-counts, plus a count of 1-2-3 at the end. On three additional repetitions I always got two sets of four 8-counts, and the counts at the end were always 1-2 or 1-2-3. Based on these figures:
What is the best estimate of the time required for the ball to travel the entire distance?
9.42s
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What is the percent uncertainty in the time required for the ball to travel down the ramp, based on the given information and without making any extraneous assumptions?
?
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How long should the ball have spent on the first ramp, if the acceleration was indeed constant?
4.71s
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If I got to the 8 of the third set of 8-counts by the time the ball reached the end of the first ramp, what was the acceleration on that ramp?
17.78cm/s^2
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Everyone should be able to answer the first questions and make a good attempt on the others.
6. The ball is at the end of the first ramp when I reach the count 1 of the fourth set of 8-counts. Where will it be when I get to the 1 of the seventh set of 8-counts, assuming a constant acceleration throughout?
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7. I give the ball a quick flick, starting from the low end of the ramp, starting my count at 1 at the instant the ball leaves the end of my finger. It traveled up the ramp through one count of 8, and came to rest for an instant as I counted 5 during the next count of 8. I continued my count as it rolled back down, getting to the end of my third count of 8 and reaching 1-2 of the next set of 8 before the ball reached its original point.
Was the magnitude of the ball's acceleration the same going up as coming down?
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If not, what was the approximate percent difference in the accelerations?
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8. If acceleration down a ramp is constant, then where will an object released from rest reach its average velocity?
If the initial velocity is not zero, how will this affect the position at which the object reaches its average velocity?
It should still be in the same spot"
Like most, you appear to have lost the thread on the last few questions. Everything looks pretty good. See my notes.
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