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What are the most difficult M1 questions you have come across?

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Reply 20
Original post by Dominator1
the acceleration is fine, since one is accelerating and the other is decelerating. I was wondering about the distance because if it's the case of displacement being a vector then the two balls wouldn't have had the same velocity (because they're travelling opposite directions), they would have the same speed.

The acceleration is not fine if you're setting the displacements equal to each other.

The displacements are only equal to each other at the point of contact if both displacement equations use the same positive directions. So g should be 9.8 in both or -9.8 in both.

For the velocities, whichever direction you take as positive, the velocities will be equal but have opposite signs since the balls are travelling in opposite directions but have the same speed. E.g. one could have v = 3 and the other could have v = - 3.
Original post by notnek
The acceleration is not fine if you're setting the displacements equal to each other.

The displacements are only equal to each other at the point of contact if both displacement equations use the same positive directions. So g should be 9.8 in both or -9.8 in both.

For the velocities, whichever direction you take as positive, the velocities will be equal but have opposite signs since the balls are travelling in opposite directions but have the same speed. E.g. one could have v = 3 and the other could have v = - 3.


I apollogise but I still don't get it. If and object is accelerating under gravity it aill have a positiv acceleration of g. negative if decelerating. how's using the same positive directions gonna make it any different?
Original post by notnek
The acceleration is not fine if you're setting the displacements equal to each other.

The displacements are only equal to each other at the point of contact if both displacement equations use the same positive directions. So g should be 9.8 in both or -9.8 in both.

For the velocities, whichever direction you take as positive, the velocities will be equal but have opposite signs since the balls are travelling in opposite directions but have the same speed. E.g. one could have v = 3 and the other could have v = - 3.


The point is that the ball travelling upwards will never have a negative velocity before the collision occurs since they collide with the same speed. If both balls travel in the same direction at the same speed then they will not collide.
Since it is an argument of speed the equation is really ugt=gt|u-gt|=|-gt| Since we have stated that ugtu-gt must be positive then the equation becomes ugt=gtu-gt=gt.

For same speed:
ugt=gtu-gt=gt
u=2gtu=2gt
Displacement SUVATS
s=4012gt2s=40 - \frac{1}{2}gt^{2}
s=ut12gt2s=ut-\frac{1}{2}gt^{2}
Equate SUVATS
ut12gt2=4012gt2ut-\frac{1}{2}gt^{2}=40-\frac{1}{2}gt^{2}
ut=40ut=40
Sub in u=2gtu=2gt
2gt2=402gt^{2}=40t2=402gt^{2}=\frac{40}{2g}
So
t=107=1.43t=\frac{10}{7}=1.43
(edited 7 years ago)
Original post by Dominator1
I apollogise but I still don't get it. If and object is accelerating under gravity it aill have a positiv acceleration of g. negative if decelerating. how's using the same positive directions gonna make it any different?


@notnek is correct. If I take upwards to be the positive direction and the weight of the object is the only force that acts upon it. Then the weight acts downwards (in the negative direction) so the acceleration will always be negative no matter what, and thus a falling object will have a negative velocity and an object moving upwards will have a positive velocity.

If you take downwards to be positive then the weight acts in the positive direction and thus the acceleration will always be positive. And a falling object will have a positive velocity and an object travelling upwards will have a negative velocity.
Reply 24
Original post by Cryptokyo
The point is that the ball travelling upwards will never have a negative velocity before the collision occurs since they collide with the same speed.

It will have negative velociity if you take downwards as positive.

Considering the magnitudes and setting them equal to each other like you have done is a good approach. But you could also say that the two ball's velocities have opposite signs. As long as you use the same direction as positive for both balls, you will end up with u=2gtu=2gt .
(edited 7 years ago)
Original post by Cryptokyo
@notnek is correct. If I take upwards to be the positive direction and the weight of the object is the only force that acts upon it. Then the weight acts downwards (in the negative direction) so the acceleration will always be negative no matter what, and thus a falling object will have a negative velocity and an object moving upwards will have a positive velocity.

If you take downwards to be positive then the weight acts in the positive direction and thus the acceleration will always be positive. And a falling object will have a positive velocity and an object travelling upwards will have a negative velocity.


Thank you, that was helpful.
Original post by Cryptokyo
Q3 on this paper. It is great fun!

It tests almost everything you need to know for pulleys and scale pans


Hey, I had a question regarding part c

Would it work if I was to resolve for R to find the reaction force upwards? Is there any other way other than resolving the forces on Q?
img025.jpg
(edited 7 years ago)
Original post by Cryptokyo
Q3 on this paper. It is great fun!

It tests almost everything you need to know for pulleys and scale pans


Don't think that question was too bad.
Original post by Someboady
Hey, I had a question regarding part c

Would it work if I was to resolve for R to find the reaction force upwards? Is there any other way other than resolving the forces on Q?


Not sure what you mean?
Original post by Marxist
Don't think that question was too bad.


Part b and c would be impossible for a lot of people who might not understand what to do. But I think once you do know what to do, it isn't that bad.
Original post by KloppOClock
Part b and c would be impossible for a lot of people who might not understand what to do. But I think once you do know what to do, it isn't that bad.


To be honest - this is just applying what you already now. It shouldn't be difficult imo. But yeah I think I know what you mean.
Original post by Marxist
Not sure what you mean?


We're trying to find the reaction force on Q by R. Since there is an equal and opposite reaction force, would I be able to resolve the forces about R to find its reaction force?
Original post by Someboady
We're trying to find the reaction force on Q by R. Since there is an equal and opposite reaction force, would I be able to resolve the forces about R to find its reaction force?


Oh if you mean part b I can show you how I do it?
Original post by Marxist
Oh if you mean part b I can show you how I do it?

Yes, part b. I know how to do it, by resolving for Q...but I was thinking if it would work if I resolved for R
Original post by Someboady
Yes, part b. I know how to do it, by resolving for Q...but I was thinking if it would work if I resolved for R


Yes it would work. :smile:
Original post by Someboady
Yes, part b. I know how to do it, by resolving for Q...but I was thinking if it would work if I resolved for R


Capture.JPG
I know this is meant to be an easy question, but I keep getting a different sign. Has anyone done q1 on the international 2013 june paper. I'm sure it's 14=2v-10, but on the mark scheme it says 14=2v+10. Can someone please explain why it's positive, I thought Ft=mv-mu and mu is in the positive direction!
Original post by Marxist
Yes it would work. :smile:


What would be the calculation then for R? I can't seem to get 19.6N when I resolve for R
Original post by Someboady
What would be the calculation then for R? I can't seem to get 19.6N when I resolve for R


Just posted working. See previous post.

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