Hsc 2013 q (1 Viewer)

[QZP]

http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-physics.pdf

Can someone explain 30a please? I don't understand their sample answers :/

 

[QZP]

Yeah I don't understand the newton's 3rd law part. What is the pair? My explanation involved centripetal force + centrifugal force but centrifugal isn't in the syllabus so yeah

 

[Tugga]

I think it's the floor pushing up against the astronaut (providing centripetal force). The reaction pair is then the astronaut pushing back against the floor. This is simulating gravity (him being pushed into the ground). Hope that helps :S

 

[itsalberttrinh]

Would you talk about how the rotation provides a normal reaction force on the astronaut. Thus a sense of gravity, as opposed to no reaction force the astronaut will appear to be "free falling". Also the centripetal force there is a force acting towards the centre of motion and according to newton's third law for every action there is an equal an opposite reaction. The equal and opposite reaction is the sense of 'gravity' the astronaut feels as he/she is pushed outwards. Think about those theme park rides like the 'Rotor' at Luna Park how you're like pushed outwards and pushed up against the wall.

 

[QZP]

Thanks guys i got it

Here's another one Q23b http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-physics.pdf

Sample solutions from BOSTES "In the frame of the train the acceleration of the mass is zero, hence there is zero net force. Therefore, apart from the force applied by the string there must be opposing forces. When the string breaks the force exerted by the string is removed and the mass must accelerate in the direction of those opposing forces, which must be along the line of the string."

1) I thought that it would undergo parabolic motion due to gravity? Why is this wrong?
2) General question: When the train accelerates (say to the left), what is the force that pushes the mass backwards? I've always understood it as "inertia" but they explain it as an "opposing force". What is this opposing force and where is it from? T_T

 

[QZP]

bump...

 

[Tugga]

Thanks guys i got it

Here's another one Q23b http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-physics.pdf

Sample solutions from BOSTES "In the frame of the train the acceleration of the mass is zero, hence there is zero net force. Therefore, apart from the force applied by the string there must be opposing forces. When the string breaks the force exerted by the string is removed and the mass must accelerate in the direction of those opposing forces, which must be along the line of the string."

1) I thought that it would undergo parabolic motion due to gravity? Why is this wrong?
2) General question: When the train accelerates (say to the left), what is the force that pushes the mass backwards? I've always understood it as "inertia" but they explain it as an "opposing force". What is this opposing force and where is it from? T_T

To the best of my understanding...

1)I'll explain after 2)
2)It is inertia. If there was say a mass on a frictionless surface within the train, it would, in the frame of reference of the train, move backwards. The opposing force that they're talking about is the tension in the string keeping it from doing so.
i.e. in the frame of reference of the train, there is an effective force pushing the mass backwards and the string is providing the opposing force to allow the net force to be 0 and the position to remain constant.

They don't use the opposing force to explain the displacement of the mass. They use it to explain what happens afterwards [accelerate in the opposite direction of the force (opposing force of Tension in the string) that was removed].

Now regarding 1) you've forgotten to include the relative acceleration of the mass to the train in the horizontal direction, which makes the motion linear, since the original velocity is zero (this took me while to wrap my head around tbh, had to use 3u projectile motion to see that a straight line is plotted by constant acceleration in x and y axis). Parabolic motion only occurs when there's a beginning velocity not in the motion of resolved acceleration (if that makes sense :S). [If you think about it when the mass is stationary, the forces have been resolved in all directions, meaning once we cut the string and remove tension, we can resolve in that line of motion and see that there is a force resolved such that it is ONLY in that line and thus acceleration ONLY in that direction.] <--- not sure if this last sentence just makes things more confusing :\

Hope that helps!

 

[anomalousdecay]

What happens is that we have the following relative to the same reference frame as each other:




This occurs as we have two forces acting here, both of equal magnitude but opposite direction. One force is the Tension in the string acting on the ball and the other force is the train pulling onto the string. As a result the net force is zero.

So the ball accelerates at the same rate as the train with regards to any reference (so in the reference of the train, both have a = 0 as given by BOSTES answer or wherever you got that from) due to a force imposed by the Tension of the string with the corresponding Newton pair of the force of the train pulling the string.

So the net force due to this Newton pair is zero, as one would expect as per Newton 3. Since the net acceleration is zero, then all you will get is the ball falling in a straight line manner relative to the train as expected due to no relative acceleration between the ball and the train.

Does this help?

 

[itsalberttrinh]

Thanks guys i got it

Here's another one Q23b http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-physics.pdf

Sample solutions from BOSTES "In the frame of the train the acceleration of the mass is zero, hence there is zero net force. Therefore, apart from the force applied by the string there must be opposing forces. When the string breaks the force exerted by the string is removed and the mass must accelerate in the direction of those opposing forces, which must be along the line of the string."

1) I thought that it would undergo parabolic motion due to gravity? Why is this wrong?
2) General question: When the train accelerates (say to the left), what is the force that pushes the mass backwards? I've always understood it as "inertia" but they explain it as an "opposing force". What is this opposing force and where is it from? T_T

Addressing the parabolic motion, yes that would be correct to an external observer outside that frame of reference, however the questions states that the student is inside the train and will simple see ball fall straight in the line of motion. As for the opposing force, I think they just mean the force that is opposing the direction of the train, so the ball is experience a force in the opposite direction due to the train moving left... I think well thats my take on it.

 

[anomalousdecay]

QZP I just realised the simplest way to explain this to you (yep brainwaves of ideas peak at 2 am).

So in this set up while the train is moving at a constant velocity the relative motion of the ball to the train is zero. Hence the acceleration of the ball is zero as thle train is travelling at constant speed. Hence the sum of forces on the ball is zero. Now we have the tension force and reaction force acting at a certain angle to the vertical. So overall there is no net acceleration on the ball and hence you obtain the required graph.

 

[Tugga]

QZP I just realised the simplest way to explain this to you (yep brainwaves of ideas peak at 2 am).

So in this set up while the train is moving at a constant velocity the relative motion of the ball to the train is zero. Hence the acceleration of the ball is zero as thle train is travelling at constant speed. Hence the sum of forces on the ball is zero. Now we have the tension force and reaction force acting at a certain angle to the vertical. So overall there is no net acceleration on the ball and hence you obtain the required graph.

The train is accelerating though :S. When the string is cut the ball drops in a diagonal straight line, (in the frame of reference of train)

 

[anomalousdecay]

The train is accelerating though :S. When the string is cut the ball drops in a diagonal straight line, (in the frame of reference of train)

Oh my bad I was really unclear there.

Now when the string is cut the ball still experiences the opposing force to the tension, however there is no more tension in the string. So what happens is that the ball will end up accelerating to the right of the train. In an ideal situation this ball would hence fall in a straight line rather than a slightly parabolic motion as expected if you just simply let go of the ball from your hand. As a result, we get that approximately in this ideal situation:



The approximation comes about from the assumption is that there are no external forces doing work so mechanical energy is conserved (dw I learned this in uni not HSC so its already assumed in HSC in thsi situation). This assumption is very important however. Otherwise it will not actually go in a straight line but will be a bit skewed in a parabolic motion.

So what we get is that the train and ball relative to each other are not accelerating at all. In the end, this becomes a simple case of imagining that the train was moving at a constant velocity to ground and the ball was just dropped.

And yeah sorry if this is confusing and I'm over analysing the result for you guys but this is how I see it. Any way the depth I went to here is not required at all. The depth required is that outlined by the marking guidelines.

 

[BlugyBlug]

God, this question. I took diameter as radius in part b) last year ...