x Turn on thread page Beta
 You are Here: Home >< Maths

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

1. (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.
2. (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?
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.
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 Since we have stated that must be positive then the equation becomes .

For same speed:

Displacement SUVATS

Equate SUVATS

Sub in

So
4. (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.
5. (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 .
6. (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.
7. (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?
8. (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.
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?
10. (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.
11. (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.
12. (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?
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?
14. (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
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.
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
17. 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!
18. (Original post by Marxist)
Yes it would work.
What would be the calculation then for R? I can't seem to get 19.6N when I resolve for R
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.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 30, 2018
Today on TSR

### Loughborough better than Cambridge

Loughborough at number one

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams