query 102

course phy201

010. `query 10.2

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Question: `qQuery introductory problem set 3 #'s 7-12

Describe two ways to find the KE gain of an object of known mass under the influence of a known force acting for a given time, one way

based on finding the distance the object moves and the other on the change in the velocity of the object, and explain why both approaches

reach the same conclusion.

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Your solution:

I we have

mass

Force

'dt

we can first find the acceleration of the object by f= m * a or a=f/m.. Then we can use the acceleration the time and the given velocity

to find the ds, or given 'ds acceleration and ime to find the velocity. The kE of the obkect can be found either by finding the work done

'dw = f * ds using the distance or by KE1= .5m V0^2 using the velocity.

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Given Solution:

First way: KE change is equal to the work done by the net force, which is net force * displacement, or Fnet * `ds.

Second way: KE change is also equal to Kef - KE0 = .5 m vf^2 - .5 m v0^2. **

STUDENT QUESTION:

I wasn’t sure what equation to use to find KE the second way. What does Kef stand for?

INSTRUCTOR RESPONSE:

In general f stands for 'final' and 0 for 'initial'. Just as v0 and vf stand for initial and final velocities, we'll use KEf and KE0 to

stand for initial and final kinetic energies.

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Self-critique (if necessary):

Ok, so we can use kef- keo whic is .5m Vf^2 - .5 m V0^2 using the mass and the final velocity minus the initial velocity.

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Self-critique rating #$&*:3

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Question: `q (This question applies primarily to General College Physics students and University Physics students, though Principles of

Physics students are encouraged, if they wish, to answer the question).

In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that

the change in v^2 is proportional to Fnet `ds?

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Your solution:

Velocity final squared is equal to the veolcoity intitial squared plus 2 time the acceleration time the displacement or V^2=V0^2 + 2a 'ds.

You can obtain your acceleration by the change in velocity once all velocities have been found. Force is equal to mass time acceleration,

or f=m *a, and work done is equal to net force * displacement, or 'dw= fnet * 'ds.

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Given Solution:

In a nutshell:

since vf^2 = v0^2 + 2 a `ds, a `ds = 1/2 (vf^2 - v0^2), so a `ds is proportional to the change in v^2

since F_net = m a, F_net * `ds = m a * `ds so F_net * `ds is proportional to a * `ds

Thus F_net `ds is proportional to a * `ds, which in turn is proportional to the change in v^2.

Thus F_net `ds is proportional to the change in v^2.

More detail:

It's very important in physics to be able to think in terms of proportionality.

To say that y is proportional to x is to say that for some k, y = k x.

That is, y is a constant multiple of x.

To say that a * `ds is proportional to the change in v^2 is to say that

for some k, a * `ds = k * ( change in v^2)--i.e., that

a * `ds is a constant multiple of the change in v^2.

In terms of the equations of motion, we know that

vf^2 = v0^2 + 2 a `ds so

a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2.

So a `ds is a constant multiple (1/2) of the change in v^2.

Formally we have

a `ds = k ( change in v^2) for the specific k value k = 1/2.

Now since Fnet = m a, we conclude that

Fnet * `ds = m a * `ds

and since a `ds = k * ( change in v^2) for the specific k value k = 1/2, we substitute for a * `ds to get

Fnet `ds = m * k * (change in v^2), for k = 1/2.

Now m and k are constants, so m * k is constant. We can therefore revise our value of k, so that it becomes m * 1/2 or m / 2

With this revised value of k we have

Fnet * `ds = k * (change in v^2), where now k has the value m / 2.

That is, we don't expect Fnet * `ds to be proportional to the change in velocity v, but to the change in the square v^2 of the velocity.

STUDENT COMMENT: I am still a bit confused. Going through the entire process I see how these

values correlate but on my own I am not coming up with the correct solution. I am getting lost after we discover the a `ds

is a constant multiple of (1/2) the change in v^2. Is it that I should simply substitute the k into the equation? Or am I

missing something else?

INSTRUCTOR RESPONSE: The short answer is that by the fourth equation of uniformly accelerated motion, a `ds = 1/2 (vf^2 - v0^2), which is

half the change in v^2, so that a `ds is proportional to the change in v^2. (The proportionality constant between a `ds and change in v^2

is the constant number 1/2).

F_net = m a, where m is the mass of the object. So F_net is proportional to a. (The proportionality constant between F_net and a is the

constant mass m).

Thus F_net `ds is proportional to a `ds, which we have seen is proportional to the change in v^2.

The conclusion is the F_net `ds is proportional to the change in v^2.

(The proportionality constant between F_net `ds and change in v^2 is 1/2 m.)

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Self-critique (if necessary):

Ok, I think I understand most of this. It is alot. BUt i think I get the just of it.

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self-critique rating #$&*:2

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Question: How do our experimental results confirm or cause us to reject this hypothesis?

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Your solution:

Our experiments confirm this hypothesis.

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Given Solution:

The explanation for this result:

On a ramp with fixed slope the acceleration is constant so

a `ds is simply proportional to `ds

specifically a `ds = k * `ds for k = a.

In the preceding question we saw why

a * `ds = k * (change in v^2), with k = 1/2.

In our experiment the object always accelerated from rest. So

the change in v^2 for each trial would be from 0 to vf^2.

the change would therefore be just

change in v^2 = vf^2 - 0^2 = vf^2.

Thus if a `ds is proportional to the change in vf^2, our graph of vf^2 vs. a `ds should be linear.

The slope of this graph would just be our value of k in the proportionality a * `ds = k * (change in v^2), where as we have seen k = 1/2

We wouldn't even need to determine the actual value of the acceleration a. To confirm the hypothesis all we need is a linear graph of

vf^2 vs. `ds.

(we could of course use that slope with our proportionality to determine a, if desired)

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Self-critique (if necessary):

ok I think I understand this

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Self-critique rating #$&*:2

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Question: `qGeneral College Physics and Principles of Physics: convert 35 mi/hr to km/hr, m/s and ft/s.

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Your solution:

35 mi/hr* 1.6093 km/hr= 56.32km/hr

56.32km/hr * 1000m/hr = 56320m/hr

56320m/hr / 60 seconds=

938.67m/s * 3.281ft/s=

3078.83ft/s

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Given Solution:

`aWe need a conversions between miles and meters, km and ft, and we also need conversions between hours and seconds.

We know that 1 mile is 5280 ft, and 1 hour is 3600 seconds. We also know that 1 inch is 2.54 cm, and of course 1 foot is 12 inches.

1 mile is therefore 1 mile * 5280 ft / mile = 5280 ft,

5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm, which is the same as 160934 cm * 1 m / (100 cm) = 1609.34 m, which in turn is the

same as 1609.34 m * 1 km / (1000 m) = 1.60934 km.

Thus

35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr.

We can in turn convert this result to m / s: 56 km/hr * (1000 m / km) * (1 hr / 3600 sec) = 15.6 (km * m * hr) / (hr * km * sec) = 15.6

(km / km) * (hr / hr) * (m / s) = 15.6 m/s.

The original 35 mi/hr can be converted directly to ft / sec: 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec.

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Self-critique (if necessary):

Ok, I should have divided 56320meters/3600seconds =15.64m/s

15.64m/s * 3.28 feet/s = 51.3ft/s approximately

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Self-critique rating #$&*:3

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Question: `qGen phy and prin phy

prob 2.16: sports car rest to 95 km/h in 6.2 s; find acceleration

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Your solution:

V0= 0

Vf= 95km/h

'dt=6.2 seconds

'dt 6.2 seconds in 1 hour there are 60 seconds = .1033 seconds in an hour

'dv= 95km/h - 0km/h

'dv95km/h

a= 'dv/'dt

a= 95km/h/ .1033

a=919.65km/h^2

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Given Solution:

`a** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s.

So change in velocity is `dv = 26.3 m/s = 0 m/s = 26.3 m/s.

Average acceleration is aAve = `dv / `dt = 26.3 m/s / (6.2 s) = 4.2 m/s^2.

Extension: One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is

-4.2m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.43 'g'.

STUDENT QUESTION:

How did we know that the final velocity was 0?

INSTRUCTOR RESPONSE:

The final velocity was 0 because the car came to rest.

Summary of what we were given:

Initial velocity is 95 km/hr, or 26.3 m/s.

Final velocity is 0, since the car came to rest.

The velocity makes this change in a time interval of 6.2 seconds.

We can easily reason out the result using the definition of acceleration:

The acceleration is the rate at which velocity changes with respect to clock time, which by the definition of rate is (change in velocity)

/ (change in clock time)

The change in velocity from the initial 0 m/s to the final 26.3 m/s is 26.3 m/s, so

acceleration = change in velocity / change in clock time = 26.3 m/s / (6.2 s) = 4.2 m/s^2.

We could also have used the equations of uniformly accelerated motion, with vf = 26.3 m/s, v0 = 0 and `dt = 6.2 seconds. However in this

case it is important to understand that the definition of acceleration can be applied directly, with no need of the equations. (solution

using equations: 2d equation is vf = v0 + a `dt, which includes our three known quantities; solving for a we get a = (vf - v0) / `dt =

(26.3 m/s - 0 m/s) / (6.2 s) = 4.2 m/s^2.)

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Self-critique (if necessary):

ok I should have convert to m/s first and then solved the problem.....so

V0= 0

Vf= 95km/h

'dt=6.2 seconds

95km/h * 1000m/3600seconds= 26.89 m/s

a='dv/'dt

a=26.89m/s/6.2seconds

a= 4.26m/s^2

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Question: univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe

your graph.

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Your solution:

1st train

v0=25m/s

'ds=200m

a= -.1m/s^2

Vf^2= (25m/s)^2 + 2(-.1m/s^2) '200m

v^2= 625m^2/s^2 - 40m^2/s^2

vf^2= 585m^2/s^2

vf=24.19m/s

Vf=v0 + a *'dt

24.19m/s= 25m/s + -.1m/s^2 'dt

-.81m/s= -.1m/s^2 'dt

'dt= 8.1 seconds

2nd train

v0=15m/s

200m

a= -.1m/s^2

vf^2=v02 + 2 * 'ds

Vf^2=(15m/s)^2 + 2(-.1m/s^2) * 200m

vf^2= 225m^2/s^2 - 40m^2/s^2

vf^2= 185m^2/s^2

vf= 13.60m/s

Vf= v0 + a * 'dt

13.60m/s= 15m/s * -.1m/s^2 'dt

-1.4m/s = -.1m/s^2 'dt

dt= 14seconds

No the first trian will get there first so they will just miss each other.

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Given Solution:

If we assume the passenger train is at position x = 0 at clock time t = 0 we conclude that the position function is x(t) = x0 + v0 t + .5

a t^2; in this case a = -.1 m/s&2 and x0 was chosen to be 0 so we have x(t) = 25 m/s * t + .5 * (-.1m/s^2) * t^2 = 25 m/s * t - .05 m/s^2

* t^2. To distinguish the two trains we'll rename this function x1(t) so that

x1(t) = 25 m/s * t - .05 m/s^2 * t^2.

At t = 0 the freight train, which does not change speed so has acceleration 0 and constant velocity 15 m/s, is 200 m ahead of the

passenger train, so the position function for the freight train is

x2(t) = 200 m + 15 m/s * t .

The positions will be equal if x1 = x2, which will occur at any clock time t which solves the equation

25 t - .05 t^2 = 200 + 15 t(units are suppressed here but we see from the units of the original functions that solutions t will be in

seconds).

Rearranging the equation we have

-.05 t^2 + 10 t - 200 = 0.

The quadratic formula tells us that solutions are

t = [ - 10 +- sqrt( 10^2 - 4 * (-.05) * (-200) ) ] / ( 2 * .05 )

Simplifying we get solutions t = 22.54 and t = 177.46.

At t = 22.54 seconds the trains will collide.

Had the trains been traveling on parallel tracks this would be the instant at which the first train overtakes the second. t = 177.46 sec

would be the instant at which the second train again pulled ahead of the slowing first train. However since the trains are on the same

track, the accelerations of both trains will presumably change at the instant of collision and the t = 177.46 sec solution will not apply.

GOOD STUDENT SOLUTION:

for the two trains to colide, the 25 m/s train must have a greater velocity than the 15 m/s train. So I can use Vf = V0 + a('dt). 15 =

25 + (-.1)('dt)

-10 = -.('dt)

'dt = 100

so unless the displacement of the 25 m/s train is greater than the 15 m/s train in 100 s, their will be no colision.

'ds = 15 m/s(100) + 200 m

'ds = 1700 m

'ds = 25 m/s(100) + .5(-.1)(100^2) = 2000 m.

The trains collide. **

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Self-critique (if necessary):

ok, I gave it shot!!!!

good try

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Self-critique rating #$&*:1

&#Good responses. See my notes and let me know if you have questions. &#