Mechanics Questions: Answers to Exam Questions

[sparsh]

Hi
Could someone please help me with some questions that I got wrong in an exam recently...

1. Avessel is partly filled with a liquid whose expansion coefficient is equal to the volume expansion coefficient of the vessel on heating the vessel the intermal available volume of space 1) increases 2) decreases 3) remains same

2. A man x is sitting at the rear end of a long compartment of a train running at constant horizontal velocity, tosses a coin to a person y who is near the front end of the compartment. The trajectory of the ball, as seen by y and a person z on the ground, will have : 1.) Equal vertical and horizontal ranges 2. ) equal vertical ranges but different horizontal ranges 3.) Different vertical and horizontal ranges

3. What is the velocity of efflux of kerosene from a tank in which the pressure is 4x10² N/m² above the atmospheric pressure. Density of kerosene is 800kg/m³. The orifice is at the bottom....

4. The top of a lake frozen. Where the air in contact with the lake surface is at -13°C. The temperature of water at the bottom of the lake will be : (Answer given is -4degrees but i think it should be 4 degrees)

 

[lando45]

Hi, I don't quite understand what you've written out for Question 1?

And for Question 2, have you tried drawing it on a graph? That will make it easier to get your head around.

I believe that Question 3 just relates to pressure laws.

And what makes you think the answer to Question 4 is 4 degrees?

 

[sparsh]

Note : Just in case you dint figure out. All of these are multiple choice questions and I have given the options along with each question... (except for the third one)

The answer to question 4 is in degrees because the options are a) 13 b) -13 c) 4 d) -4 (all in degrees) Moreover we have to tell the temperature at the bottom so conveniently the answer can be given in degrees ...

 

[Hootenanny]

Sparsh, you should know that we can't help you if you do not show any working.

(1) Is elementry, just consider what would happen when the vessel is heated.

(2) I agree with lando on this one a sketch would be helpful.

(4) Perhaps if you could justify your answer..?

~H

 

[sparsh]

In the 4th one... density of water is max at 4degrees so it becomes heavy and sinks down ... All the layers of water except the frozen top will remian at 4 degrees (fishes in lakes in cold countries survive the cold winter)

The 1st one : - the relative expansion of liq = a(liq) - a(solid) where a is the coefficient of cubical expansion ... So according to me the volume available on heating should remian the same ...

The second one the sketch should be the same for both the frames as the frame of the train is not accelerated ... Am I correct in this regard ?

And please gimme some idea in the 3rd one as well . I tried Bernollis Equation but can't get anywhere.

 

[Hootenanny]

In the 4th one... density of water is max at 4degrees so it becomes heavy and sinks down ... All the layers of water except the frozen top will remian at 4 degrees (fishes in lakes in cold countries survive the cold winter)

I would agree with you here, perhaps the answer given is a typo?

The 1st one : - the relative expansion of liq = a(liq) - a(solid) where a is the coefficient of cubical expansion ... So according to me the volume available on heating should remian the same ...

This is correct also

The second one the sketch should be the same for both the frames as the frame of the train is not accelerated ... Am I correct in this regard ?

Consider relative velocities. This may be easier; if the man on the train just threw the ball straight up and down, compare the vertical and horizontal displacements of the man on the train and the stationary observer.

And please gimme some idea in the 3rd one as well . I tried Bernollis Equation but can't get anywhere.

I'm still thinking about this one. Perhaps some else could chip in?

~H

 

[arildno]

As for 3, use Bernoulli by considering a streamline from the yop of the tank to the bottom. When the tank is large, you may regard the velocity of the kerosene at the top to be zero.
Ignore the gravitational potential difference between the top and the bottom.

 

[sparsh]

@ arildno

Did exactly you said in the 1st place. Wasnt getting an answer matching with the options given . Must have gone wrong somewhere. .. Thanks ..

Consider relative velocities. This may be easier; if the man on the train just threw the ball straight up and down, compare the vertical and horizontal displacements of the man on the train and the stationary observer.

]

I dint get what to do here. Please explain a bit more...

Regards
Sparsh

http://www.alchemy-education.com

 

[Hootenanny]

OKay. Man X is sitting on a train. Man Y is standing as the side of the rail track. The train is traveling and a constant velocity of 5 m/s. The man on the train throws a ball straight up and catches it again, the ball is in the air for one second. As far as man X is concered that ball has a zero horizontal displacement. However, as man Y views it, the ball has traveled a horizontal distance at the same velocity as the train, hence, from Y's point of view the ball has a horizontal displacement of 5 meters.

Do you follow?

~H

 

[sparsh]

Hmmm.

I think you dint read the question properly. The man X is actually tossing the coin towards a man Y who is at the front end of the train...

 

[Hootenanny]

Hmmm.

I think you dint read the question properly. The man X is actually tossing the coin towards a man Y who is at the front end of the train...

I know. As I said in my OP, consider the relative velocities. Let the velocity of the train be v (with respect to the earth) and the velocity of the coin relative to the train be u. From person Y's perspective (at rest relative to the train) the velocity of the coin is simply u. However, from person Z's perspective (moving relative to the train), the velocity of the coin is u + v. The different perspectives assumes different stationary points.

~H

 

[loom91]

For 2, I think a simple Gallilean transformation of the reference frame gives the answer quantitatively as well as qualitatively and is the best way to go here. If you don't know how to do that, then intuitive logic is your only bet.

Firstly, in the vertical direction all concerned observers are moving at the same constant velocity, so that result should be easy to figure out. In the horizontal direction, consider that the length traveled by the coin as seen by Y is simply the distance between X and Y in the train's reference frame, but when seen from the ground the point where X released the coin is much further from the point where Y catches it than the length of th compartment. Is this making sense?

Incidentally, if you repeat this experiment with a beam of light instead of a coin, you get some really interesting results...

 

[Hootenanny]

As loom correctly pointed out (and he has given you the answer ) a Gallilean transformation is the best way to go, if you have been given this tool. As you didn't intially spot the relative velocities I assumed you hadn't been given this tool, apologies if I'm wrong. The problem can still be figured out from first principles. However, if you would like to apply Galilean transformations, there is a good tutorial located here; http://www.src.wits.ac.za/pages/teaching/Connell/phys284/2005/lecture-01/lecture_01/node5.html and I would be happy to walk you through it here if you have any trouble.

~H

 

[sparsh]

Thanks people...

 

[loom91]

Incidentally, this question has confused me somewhat. Does this mean in Gallilean relativity lengths are frame-dependent? If Y and Z were to independntly measure the distance between the point where the coin is released and the point where it is caught, they would get different answers? How can this be? Length is supposed to be invariant in classical mechanics!