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OCR MEI M1 9th June 2015 4761/01

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Original post by AlexParmenter
Can someone help me with 8v on this paper I don't understand the mark scheme and I don't know what the axis should be and how you know what to plot? Thanks
http://www.mei.org.uk/files/papers/m109jn_86dj.pdf


Wow, I literally just finished this question 5 minutes ago. You're meant to plot the x and y coordinates, and just draw out the path of the boat; imagine what it would look like from above.
Original post by AlexParmenter
Can someone help me with 8v on this paper I don't understand the mark scheme and I don't know what the axis should be and how you know what to plot? Thanks
http://www.mei.org.uk/files/papers/m109jn_86dj.pdf


you put t=0, t=1 and t=2 into x and y that have been given and then plot these coordinates
in questions regarding what happens to something on a slope attached to a rope and if the rope breaks, how do you attempt those type of questions
Original post by AlexParmenter
Can someone help me with 8v on this paper I don't understand the mark scheme and I don't know what the axis should be and how you know what to plot? Thanks
http://www.mei.org.uk/files/papers/m109jn_86dj.pdf


havent looked at mark scheme yet but I assume i on x axis and j on y axis. I would go atleast on intervals of 0.5 starting from 0 and find the vaule of xi and bj (sub the t, e.g. t= 0 into r= (8t... ))
Original post by docdoc02
in questions regarding what happens to something on a slope attached to a rope and if the rope breaks, how do you attempt those type of questions


If its on a slope and a mass is hanging away from the slope and the system is accelerating down the slope (the mass upwards) if the string breaks you would use g= -9.8 as it still travels distance to rise (dunno of this helps soz)
Original post by lizard54142
Wow, I literally just finished this question 5 minutes ago. You're meant to plot the x and y coordinates, and just draw out the path of the boat; imagine what it would look like from above.


Original post by docdoc02
you put t=0, t=1 and t=2 into x and y that have been given and then plot these coordinates


Original post by RemainSilent
havent looked at mark scheme yet but I assume i on x axis and j on y axis. I would go atleast on intervals of 0.5 starting from 0 and find the vaule of xi and bj (sub the t, e.g. t= 0 into r= (8t... ))


thanks guys, i kept thinking it wanted time on the x axis but it makes sense now :smile:
Original post by AlexParmenter
thanks guys, i kept thinking it wanted time on the x axis but it makes sense now :smile:


Think parametric equations if you've come across those yet.
Original post by docdoc02
in questions regarding what happens to something on a slope attached to a rope and if the rope breaks, how do you attempt those type of questions


Just consider what the object does if it is by itself. There's no friction in M1, so if something is sliding up a slope because of something it's attached to then it slides back down when the rope breaks. If it's already sliding down it'll keep sliding down.
Original post by Duskstar
Just consider what the object does if it is by itself. There's no friction in M1, so if something is sliding up a slope because of something it's attached to then it slides back down when the rope breaks. If it's already sliding down it'll keep sliding down.


i thought that makes sense but a question in the mei book says it doesnt slide back down in one question :/
Original post by Duskstar
Just consider what the object does if it is by itself. There's no friction in M1, so if something is sliding up a slope because of something it's attached to then it slides back down when the rope breaks. If it's already sliding down it'll keep sliding down.


Yes there is... I think you mean coefficient of friction?
Original post by docdoc02
i thought that makes sense but a question in the mei book says it doesnt slide back down in one question :/


Context I guess. Where there still forces keeping the object up the slope? There could be a small concept of friction or something. I'd be interested to see it, because most of the time I think it's obvious :/
Original post by Duskstar
Context I guess. Where there still forces keeping the object up the slope? There could be a small concept of friction or something. I'd be interested to see it, because most of the time I think it's obvious :/


this question
20150608_221159.jpg509.5KB
Original post by Duskstar
Context I guess. Where there still forces keeping the object up the slope? There could be a small concept of friction or something. I'd be interested to see it, because most of the time I think it's obvious :/


This question says it doesn't slide back
20150608_222540.jpg527.8KB
Original post by docdoc02
this question


It should continue up the slope, stop and then slide back down? They might have just answered the first part then. In that case, it is a result of the rope puling the crate up then snapping like my first example. The crate won't immediately go back down the slope, so it will continue up, but it would eventually go back down with no friction.
Original post by docdoc02
This question says it doesn't slide back


In this one there is limiting friction. Because the component of the boat's weight down the slope is less than the limiting friction (3500N) when the rope snaps, the boat will not move back down. Imagine the friction as a force that pushes back with equal force (such that the body remains in equilibrium) until it's limit is reached, at which point it acts as any other force.
http://www.mei.org.uk/files/papers/M1_Paper_Jan13.pdf

Last question. 7(vi)

How?
Original post by Duskstar
In this one there is limiting friction. Because the component of the boat's weight down the slope is less than the limiting friction (3500N) when the rope snaps, the boat will not move back down. Imagine the friction as a force that pushes back with equal force (such that the body remains in equilibrium) until it's limit is reached, at which point it acts as any other force.


but isnt the friction working with the weight down the slope
or do you need to take positive direction as down the slope
Original post by iAmanze


Someone else asked this, and I've answered; scroll up a bit.
Original post by lizard54142
Someone else asked this, and I've answered; scroll up a bit.


Okay thanks.
(edited 8 years ago)
Original post by docdoc02
but isnt the friction working with the weight down the slope
or do you need to take positive direction as down the slope


In the question it says that it is moving at a constant speed. Therefore working parallel the slope: Tension - 500 ×9.8×sin 10 - friction =0.
As soon as the rope snaps, friction keeps working down the plane creating a negative acceleration. As soon as the velocity turns negative, friction then acts upward, which then creates equilibrium. Remember friction allows acts opposite to motion.At first when the rope snaps, tension is removed and acceleration is negative, however velocity is still positive for a very short period of time.

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