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# AQA Mechanics 2 Unofficial Mark Scheme Watch

1. Does anybody happen to remember their other answers for question five? I struggled with the last question a lot so I'm banking on the marks from question five to keep my grade up! Thanks a lot
2. Has anyone got a copy of the paper or the last question I can have a look at?
3. (Original post by A Slice of Pi)
Has anyone got a copy of the paper or the last question I can have a look at?
There was a relatively difficult question before the last one that I remember.

A ladder of weight W lies against a rough vertical wall and a rough horizontal floor at an angle theta to the horizontal. The co-efficient of friction at the floor is 2mu and the co-efficient of friction at the wall is mu.

Draw a diagram displaying the all the forces acting on the ladder. [2 marks]

Find tan(theta) in terms of mu. [7 marks]
4. (Original post by A Slice of Pi)
Has anyone got a copy of the paper or the last question I can have a look at?
The one that you'll be helping us the most on is this one though:

A ball attached to a string in a vertical plane is projected with velocity u at the bottom of the circle. The radius of the circle is l, express u in terms of l and g needed for complete revolutions to take place. [4 marks]

It's very similar to Q8a on June 2011 but involves a string attached to a ball instead of a bead on a wire. I put my answer as u > 2root(gl) as many others have which is identical to the answer in the aforementioned past paper, however others are saying that the answer is u > root(5gl) (and it looks like that might be the correct answer). This question is by far the most controversial question out of them all.
5. (Original post by TheLifelessRobot)
The one that you'll be helping us the most on is this one though:

A ball attached to a string in a vertical plane is projected with velocity u at the bottom of the circle. The radius of the circle is l, express u in terms of l and g needed for complete revolutions to take place. [4 marks]

It's very similar to Q8a on June 2011 but involves a string attached to a ball instead of a bead on a wire. I put my answer as u > 2root(gl) as many others have which is identical to the answer in the aforementioned past paper, however others are saying that the answer is u > root(5gl) (and it looks like that might be the correct answer). This question is by far the most controversial question out of them all.
Ah, I see what the confusion is. Note that the case in June 2011 was a bead threaded on a wire. There is always going to be a reaction force on the bead in this scenario, which is why you need to consider kinetic energies instead of forces. In this question, you require the tension in the string at the top to be positive for full revolutions. I agree with the answer for the ladder question as well.
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