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tricky MCQs help

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The second one is about circular motion. The net force on the object at any given time has to be the centripetal force i.e. towards the center of the circle. So the correct answer is A.

The third one also refers to circular motion. This means the net force is equal to the centripetal force, so the net force (not a separate force) has to be towards the center of the circle. The only figure where the net force is towards the center of the circle is D (the others make it seem as if the centripetal force is an actual force not the net force). 

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6 minutes ago, thecsstudent said:

The second one is about circular motion. The net force on the object at any given time has to be the centripetal force i.e. towards the center of the circle. So the correct answer is A.

The third one also refers to circular motion. This means the net force is equal to the centripetal force, so the net force (not a separate force) has to be towards the center of the circle. The only figure where the net force is towards the center of the circle is D (the others make it seem as if the centripetal force is an actual force not the net force). 

For the third one, i answered C as it seemed like the horizontal force there is pointing towards the center of the circle. Is the answer D because the airplane is tilting, hence the center of the circle is tilted diagonally downwards to the left?

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2 minutes ago, IBdoingnothing said:

For the third one, i answered C as it seemed like the horizontal force there is pointing towards the center of the circle. Is the answer D because the airplane is tilting, hence the center of the circle is tilted diagonally downwards to the left?

I think there shouldn't be a force pointing towards the center but if you add the forces the net force should be towards the center (which is horizontally to the left). In D the diagonal force has a y-axis component and an x-axis component. The y-axis component cancels the downward force, so what remains is the a force to the left in the x axis. 

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1 hour ago, thecsstudent said:

I think there shouldn't be a force pointing towards the center but if you add the forces the net force should be towards the center (which is horizontally to the left). In D the diagonal force has a y-axis component and an x-axis component. The y-axis component cancels the downward force, so what remains is the a force to the left in the x axis. 

hahaha i get it, its because in the question it said "vertical plane" 

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  1. Electron question: The answer is B. First you have to realize that there are 2 forces acting on the particles, the electric force (Fe=qE) and the magnetic force (Fm=qvB), where E is the electric field, B is the magnetic field, and q is the charge. Note that both the electric and the magnetic force are pointing in vertical direction. And we know that the electron is not deflected; this means that the electric force acting on the electron is equal and opposite  to the magnetic force acting on the electron, in other words: eE=evB -->  E=vB (where 'e' is the charge of the electron). So now, if you replace the charge of the alpha particle into q, you'll get the same answer for the alpha (in other words, the net-force on the alpha particle is also 0).
  2. Ball-string question: The answer is A, because the question asks for the net force and when something is moving in a circle, the net-force is always the centripetal force.
  3. Airplane question: The answer is D. @thecsstudent That was an excellent answer that you provided. Great that you pointed out that the centripetal force is not an "actual" force, but it is rather the result of some other forces.
  4. Ball-drop question: The answer is C. This is because both of the balls experience the same acceleration due to gravity. So they must move the same distance during the same amount of time. Thus the distance between them must stay the same.
5 hours ago, IBdoingnothing said:

For the third one, i answered C as it seemed like the horizontal force there is pointing towards the center of the circle. Is the answer D because the airplane is tilting, hence the center of the circle is tilted diagonally downwards to the left?

No, the center of the circle is not tilted diagonally downwards to the left. If you look closely at the diagram D, you'll notice that the vector pointing vertically downwards is shorter than the other vector. So if you add those two vectors together, you will get the net force (i.e. the resultant vector) to be pointing horizontal towards the center of the circle.

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44 minutes ago, Vioh said:
  1. Ball-drop question: The answer is C. This is because both of the balls experience the same acceleration due to gravity. So they must move the same distance during the same amount of time. Thus the distance between them must stay the same.

In my revision session I have gotten this question wrong multiple times. In fact it appeared in a physics contest, which I also got wrong.
The answer is distance increases with time, A.

See explanation below. Another way to think of this is x = x0 + ut + 0.5at2 for the first mass dropped, but x = x = x0 + u(t-ΔT)  + 0.5a(t-ΔT)2 for the second ball dropped where ΔT is the time difference between releasing first ball and second. You can substitute values in for ΔT, or do some algebraic manipulation and the first graph will increase faster than the second one.

Notably, when both balls reach terminal velocity, the EDIT: [displacement difference] is constant. This is in the v-t graph when the two curves meet. 
7xiHi2O.png

Edited by kw0573
previously "time difference"
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33 minutes ago, kw0573 said:

See explanation below. Another way to think of this is x = x0 + ut + 0.5at2 for the first mass dropped, but x = x = x0 + u(t-ΔT)  + 0.5a(t-ΔT)2 for the second ball dropped where ΔT is the time difference between releasing first ball and second. You can substitute values in for ΔT, or do some algebraic manipulation and the first graph will increase faster than the second one.

Notably, when both balls reach terminal velocity, the EDIT: [displacement difference] is constant. This is in the v-t graph when the two curves meet. 
7xiHi2O.png

 

 Thats a very interesting approach you used to answer the question. However you can also show that the distant between them increase by taking the difference between x1 and x2 where x1=x0+vt +at^2 and x2 = x0 + v(t-t0) + a(t-t0)^2 where t0 is the the time after which the second ball was dropped.  

 

 

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1 hour ago, Vioh said:
  1.  
1 hour ago, Vioh said:
  1. Electron question: The answer is B. First you have to realize that there are 2 forces acting on the particles, the electric force (Fe=qE) and the magnetic force (Fm=qvB), where E is the electric field, B is the magnetic field, and q is the charge. Note that both the electric and the magnetic force are pointing in vertical direction. And we know that the electron is not deflected; this means that the electric force acting on the electron is equal and opposite  to the magnetic force acting on the electron, in other words: eE=evB -->  E=vB (where 'e' is the charge of the electron). So now, if you replace the charge of the alpha particle into q, you'll get the same answer for the alpha (in other words, the net-force on the alpha particle is also 0)

wow thanks so much. For the electron question, I did exactly the same method and I wasn't sure whether it didn't get deflected at all or it got deflected slightly upwards. My logic was that an alpha particle would get deflected upwards as it had a +2 charge. This is because this same electric field managed to cancel out the force on the magnetic field for an electron particle, so won't it slightly deflect an alpha particle upwards due to a heavier charge? 

Edited by IBdoingnothing

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@kw0573 Thanks for pointing that out. The question doesn't look as easy as it seems, and I was too quick to choose an answer.

45 minutes ago, IBdoingnothing said:

wow thanks so much. For the electron question, I did exactly the same method and I wasn't sure whether it didn't get deflected at all or it got deflected slightly upwards. My logic was that an alpha particle would get deflected upwards as it had a +2 charge. This is because this same electric field managed to cancel out the force on the magnetic field for an electron particle, so won't it slightly deflect an alpha particle upwards due to a heavier charge? 

@IBdoingnothing As I have posted earlier, the electric force and the magnetic force are always opposite each other. In fact, for the electron, the magnetic force is pointing upwards (from the right-hand rule), while the electric force is pointing downwards (that's the reason why the electron isn't deflected). For the alpha particle, it is the opposite: the magnetic force will be pointing downwards, while the electric force will be pointing upwards.

I have also established earlier that E=vB. Now if we multiply both sides by the charge of the alpha particle (Qa), we will have QaE=QavB. Now the left-hand side will represent the electric force acting on the alpha particle, and the right-hand side will represent the magnetic force acting on the alpha particle. And they are equal! So the alpha particle will not get deflected.

So in short, even though the alpha has a heavier charge, it won't get deflected. That's because with the heavier charge, you will not only increase the magnetic force, but you will also increase the electric force as well. And these 2 forces will cancel out each other (as they have opposite directions), giving no deflection for the alpha particle.

Feel free to ask if it's still unclear.

And PS to @kw0573 A big thumb up for the graph that you drew. That really got the point across :)

Edited by Vioh
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11 minutes ago, Vioh said:

@kw0573 Thanks for pointing that out. The question doesn't look as easy as it seems, and I was too quick to choose an answer.

@IBdoingnothing As I have posted earlier, the electric force and the magnetic force are always opposite each other. In fact, for the electron, the magnetic force is pointing upwards (from the right-hand rule), while the electric force is pointing downwards (that's the reason why the electron isn't deflected). For the alpha particle, it is the opposite: the magnetic force will be pointing downwards, while the electric force will be pointing upwards.

I have also established earlier that E=vB. Now if we multiply both sides by the charge of the alpha particle (Qa), we will have QaE=QavB. Now the left-hand side will represent the electric force acting on the alpha particle, and the right-hand side will represent the magnetic force acting on the alpha particle. And they are equal! So the alpha particle will not get deflected.

So in short, even though the alpha has a heavier charge, it won't get deflected. That's because with the heavier charge, you will not only increase the magnetic force, but you will also increase the electric force as well. And these 2 forces will cancel out each other (as they have opposite directions), giving no deflection for the alpha particle.

Feel free to ask if it's still unclear.

Yes, I get it now. The equations you stated really helped. Thanks!

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