Physics HL/SL Help

[Peanut Butter Jelly]

Yes. “Heat engines and heat pumps” section removed. I am not sure about the carnot cycle, but since the carnot cycle is actually a heat engine like thing, I would assume it's not included.

[Guest Mohammed Rahman]

Question: A train of mass 1.5*10^5 kg is travelling at 40ms^-1 when the brakes are applied and it decelerates steadily. The train travels a distance of 250m before coming to a halt. Calculate the deceleration of the train.

My attempt:

s= 250m

u = 45

v = 0

a = ?

t = -

v^2 = u^2 + 2as, rearrange and I got -4.05. However, the book gives 3.2.

I'm assumed that since this in the mechanics section and the textbook has yet to introduce any other method of working this out, I use the suvat equations.

Also, question: A large helium balloon is attached to the ground by two fixing ropes. Each rope makes an angle of 50 degrees with the ground. There is a force F vertically upwards of 2.15*10^3 N. The total mass of the balloon and its basket is 1.95*10^2 kg. Calculate the tension in either of the fixing ropes.

This one, I have no idea. Since the book has only introduced free body diagrams with things like x=mgsin(theta) and y=mgcos(theta) in triangle quesion thingys, I'm assuming that the method lies somewhere there.

Thanks

[CkyBlue]

Wow, I don't believe I'm doing physics.

You're using the same equation.

v=0

u=40m/s

s=250m

a=?

0=40x40+2a(250)

-1600/500=a

-3.2=a

However, as it speaks of deceleration, the answer is 3.2m/s^2

I actually did the second question a few days ago myself...

First thing to do is to draw a free body diagram. You see that the forces acting on the balloon are upthrust, weight, and the tension of the two ropes. There are no relevant x-components, so you needn't worry about that. Find the net force on for the y axis.

F net y= upthrust-weight-2*tension of y

I'm sure you can do that on your own. When you've calculated the y tension, just use trig to find the hypotenuse.

Oh, and you will notice that the book uses g=10.

[Procrastination]

A bucket of mass 2.50kg is whirled in a vertical circle of radius1.40m. At the lowest point of its motion the tension in the rope supporting the bucket is35.0N

a)Find the speed of the bucket.

b)How fast must the bucket move at the top of the circle so that the rope does not go slack?

Edited February 27, 2012 by Procrastination

[Slovakov]

For circular motion F=mv²/r, where r is the radius, m is the mass and v is the speed.

So in point a) all you need to so is to put the data in the formula.

For point b), the rope doesn't go slack if the acting force ancting on it is greater than gravity. So you just need to solve for mg-=mv²/r.

[Grassroot]

The energy of molecultes of an ideal gas is

A.Thermal only

B. thermal and potential

C. Potential and kinetic

D. kinetic only

why D....

[Drake Glau]

"Thermal" energy is kinetic energy in theory so there goes A and B.

An ideal gas has the highest entropy/most favorable configuration which would be all the particles moving but never close enough to have potential energy due to the other particles repulsion.

Key word here was "ideal."

[Grassroot]

"Thermal" energy is kinetic energy in theory so there goes A and B.

An ideal gas has the highest entropy/most favorable configuration which would be all the particles moving but never close enough to have potential energy due to the other particles repulsion.

Key word here was "ideal."

Garcias

I have a question about doppler effect.

According to the book it says this is due to the change in frequency of the wave

However, in exam questions they appear to be change in wavelength

I am a bit confused about this part....

And also, just to make sure... in 2 slits interference and diffraction part of the core or HL syllabus????

I also encountered a topic called beats...I dont see it in the current syllabus.

Is that something we have to learn for the exam?

[timtamboy63]

First up, it's a percieved change in frequency/wavelength, not an actual change from the source - the observer just sees it as a different frequency/wavelength.

Also, if you keep in mind that c = fλ, you should be able to see that an increase in frequency is basically the same thing as a decrease in wavelength, and vice versa.

Off memory, double slit diffration is part of the EM waves topic, which is part of the Core HL Syllabus, but an option for SL. I'll need more answer regarding the beats topic to answer your last question.

[Award Winning Boss]

You walk up a 1000m high mountain, which takes you 5 hours. Walking on the flat consumes about 400W. You are hoping to lose some of your 70kg mass with all this exercise but on the way you eat 4 small chocolate bars, which according to the wrapper will each supply 1130KJ

A) Calculate the energy you expend during the 5 hour walk. While you're climbing your body only works at a 15% efficiency.

• I got 48 MJ as the total
  

B) Calculate the energy that must be obtained from your body after the energy of the chocolate has been used.

• I got 4520KJ (but i'm not sure whether to account for the efficiency)
  

C) If fat converts to energy at a rate of 38MJkg^-1, how much mass are you likely to lose?

• This answer seems the worst so I'll refrain from saying
  

[Drake Glau]

You walk up a 1000m high mountain, which takes you 5 hours. Walking on the flat consumes about 400W. You are hoping to lose some of your 70kg mass with all this exercise but on the way you eat 4 small chocolate bars, which according to the wrapper will each supply 1130KJ

A) Calculate the energy you expend during the 5 hour walk. While you're climbing your body only works at a 15% efficiency.

• I got 48 MJ as the total
  

B) Calculate the energy that must be obtained from your body after the energy of the chocolate has been used.

• I got 4520KJ (but i'm not sure whether to account for the efficiency)
  

C) If fat converts to energy at a rate of 38MJkg^-1, how much mass are you likely to lose?

• This answer seems the worst so I'll refrain from saying
  

400W, 15%=60W

5*60*60=18000

A watt is J/s so in 5 hours you would use 60J*18000s=1,080,000J or 1.08MJ

So for the walk you need 1.08MJ, the bars will provide 4*1130KJ which is 4520KJ (4.52MJ) but you will only use 15% of that so the bars provided an actualy energy output of 0.678MJ which means your body will provide 1.08-0.678=0.402MJ

(If you didn't account for efficient in this part you would have gotten a negative number )

0.402/38=0.0106Kg.

[Award Winning Boss]

You walk up a 1000m high mountain, which takes you 5 hours. Walking on the flat consumes about 400W. You are hoping to lose some of your 70kg mass with all this exercise but on the way you eat 4 small chocolate bars, which according to the wrapper will each supply 1130KJ

A) Calculate the energy you expend during the 5 hour walk. While you're climbing your body only works at a 15% efficiency.

• I got 48 MJ as the total
  

B) Calculate the energy that must be obtained from your body after the energy of the chocolate has been used.

• I got 4520KJ (but i'm not sure whether to account for the efficiency)
  

C) If fat converts to energy at a rate of 38MJkg^-1, how much mass are you likely to lose?

• This answer seems the worst so I'll refrain from saying
  

400W, 15%=60W

5*60*60=18000

A watt is J/s so in 5 hours you would use 60J*18000s=1,080,000J or 1.08MJ

So for the walk you need 1.08MJ, the bars will provide 4*1130KJ which is 4520KJ (4.52MJ) but you will only use 15% of that so the bars provided an actualy energy output of 0.678MJ which means your body will provide 1.08-0.678=0.402MJ

(If you didn't account for efficient in this part you would have gotten a negative number )

0.402/38=0.0106Kg.

I did 400*18000 to get 7.2 MJ as the energy you use then divided it by 0.15 to get the total of 48 MJ. Your way makes more sense but I'm confused as to why I went wrong. I saw the question differently than you I think

thanks for the help

[Drake Glau]

Efficiency would apply to the rate of energy being done. Just like when you make cars or something. If you make 30 cars per day that would be 150 cars in a 5 day work week. If you are working at 50% efficiency then you don't make that 150 cars then get rid of 50% of them, you jut work at a rate 50% less than normal, so that would be 15 cars per day in this case.

^Hope that helps see efficiency questions a little simpler.

[IB Smacher Glau]

(d) As a result of noise in electric circuits, digital pulses can often lose their shape and hence distort the information that they carry. The pulses can be re-shaped using a circuit called a Schmitt trigger. The diagram shows a Schmitt trigger that incorporates an operational amplifier.

(i) State two essential properties of an operational amplifier.

(2)

(ii) In the situation shown the output voltage V0 of the amplifier is at its minimum value of – 6.0 V. The voltage at the non-inverting input to the amplifier is equal to 1.0 V and at the inverting input it is VX. The output voltage will switch to its maximum value + 6.0 V if the voltage VX just exceeds +1.0 V. Determine the minimum voltage Vin that will result in an output voltage of + 6.0 V.

(4)

(Total 16 marks)

Answer :

(ii) pd across 10kΩ =

V

in – 1.0 = 104

I

;

pd across 50 kΩ = 7.0 = 5.0 × 104

I

;

I

= 1.4 × 10–4 A;

V

in = 1.0

+

1.4 × l0–4 × 104 = 2.4V; 4

or

pd across 50 kΩ =

Vin + 6.0;

= 7.0V;

50

V

in + 300 = 420;

V

in = 2.4 V;

Award full marks if third marking point is omitted and

answer is correct.

Award

[4]

for any other valid method that gives correct answer.

[16]

Examiner Comment :

The question on operational amplifiers was nearly always answered poorly. Very few candidates really demonstrated a familiarity with this section of the syllabus usuallyshowing no skills to solve problems involving circuits incorporating operational amplifiers.

[fan]

a snowball rolls off a barn roof that slopes downwards at an angle of 40 degrees. the edge of the roof is 14m above the ground and the snowball has a speed of 7ms-1 as it rolls off the roof

(a) how far from the edge of the barn does the snowball strike the ground if it doesnt strike anything else.

cant seem to get the answer which is 6.93m :\

Edited April 17, 2012 by fan

[Award Winning Boss]

a snowball rolls off a barn roof that slopes downwards at an angle of 40 degrees. the edge of the roof is 14m above the ground and the snowball has a speed of 7ms-1 as it rolls off the roof

(a) how far from the edge of the barn does the snowball strike the ground if it doesnt strike anything else.

cant seem to get the answer which is 6.93m :\

This question made my brain hurt... great for procrastination though and I had to draw the diagram a million times

You need to figure out the time it'll take for it to hit the ground first:

s = ut + 0.5at^2

14 = sin(40)*7t + 0.5*9.8*t^2

14 = 4.5t + 4.9t^2

(then I cried for a little bit)

4.9t^2 + 4.5t - 14 = 0 (I put this into some calculator thingy)

t = 1.29 or t = -2.209

(then the normal s = d/t = d = s*t)

d = (cos(40)*7) * 1.29 = 6.92

I guessed a bunch of different cos tan etc because just '7' isn't the right component.

that was difficult.

[Drake Glau]

Well, I tried and didn't get the answer but I don't have a calculator on me so maybe this will help anyway...

First thing I did was set up a right triangle with the hypotenuse being 7m/s and the upper angle being 40 degrees as the problem says.

Used some trig to find the initial downward velocity (cos(40)=u/7). This gave me u, a, and s for a suvat equation. In this case I used s=ut+at²/2 and solved for t which i got to be 1.22965s

Next I used some more trig to find the horizontal velocity. sin(40)=u/7, sin because now its the other side of the triangle.

Since the horizontal velocity does not have any force being applied to (only force here is gravity which is affecting the vertical velocity) this speed would stay constant for the entire duration of the flight.

This yielded a distance of 5.53m though.

He beat me, but I'm confused how you used sin(40) in your suvat equation, I used cos(40) because the component you are looking for is adjacent to the angle isn't it? O.o

Edited April 17, 2012 by Drake Glau

[fan]

Thanks makes sense now.

@Dranke Glau from the way you are looking at it you would use cos(50). if you use cos you have to use the angle between the roof and the wall (90degrees - 40 degrees). Since sin(40) is the same as cos(50) you end up with the same answer as award winning boss.

[Award Winning Boss]

Thanks makes sense now.

@Dranke Glau from the way you are looking at it you would use cos(50). if you use cos you have to use the angle between the roof and the wall (90degrees - 40 degrees). Since sin(40) is the same as cos(50) you end up with the same answer as award winning boss.

Where did you get that question from?

[Drake Glau]

Thanks makes sense now.

@Dranke Glau from the way you are looking at it you would use cos(50). if you use cos you have to use the angle between the roof and the wall (90degrees - 40 degrees). Since sin(40) is the same as cos(50) you end up with the same answer as award winning boss.

I see. I thought about that a little after I had edited my post but had to leave to class. Thanks for clarifying though