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Edexcel AS/A2 Mathematics M1 - 8th June 2016 - Official Thread

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Original post by L'Evil Wolf
A particle A starts at the point with position vector 12i+12j. The initial velocity of A is (-i+j) ms^-1, and it has constant acceleration (2i-4j)ms^-2. Another particle B has initial velocity i ms^-1 and constant acceleration 2j ms^-2. After 3 seconds the two particles collide.

Find
A. The position vector of the point where the two particles collide
B. The position vector of B at the starting point

=========

I know where in the case velocity is constant the position vector is (initial position vector +vt) but in this case there is an acceleration and I dont know how to take it into account

This is supposed to be done without any calculus by the way


What paper was this? Link pls?
Original post by thelegend99
What paper was this? Link pls?


Edexcel 6D Q12 or omething
Hi, Just wondering if anyone can help with Jan06 Q1!

It says to find the value of h, which is a distance above the ground, I've attached a picture showing this, However what I don't understand is, the mark scheme says to use the suvat shown on the right (shown in pic), but doesn't that suvat correspond to the area highlighted in red I've selected in the pic? How does it find h? Thanks
Unknown.png66.7KB
Original post by ryandaniels2015
Hi, Just wondering if anyone can help with Jan06 Q1!

It says to find the value of h, which is a distance above the ground, I've attached a picture showing this, However what I don't understand is, the mark scheme says to use the suvat shown on the right (shown in pic), but doesn't that suvat correspond to the area highlighted in red I've selected in the pic? How does it find h? Thanks


S is displacement, so the displacement is just h and not the whole of the red thing as the displacement when it goes to its peak and goes to the same kevel as it started cancel.



Also another way yoy can think of it is that when it goes to its peak and falls to the same level as it was thrown from, it would have the same speed
(edited 7 years ago)
Original post by metrize
S is displacement, so the displacement is just h and not the whole of the red thing as the displacement when it goes to its peak and goes to the same kevel as it started cancel.



Also another way yoy can think of it is that when it goes to its peak and falls to the same level as it was thrown from, it would have the same speed


Thanks for the reply, I've attached a pic, so do the 2 displacement cancel (shown in red) so you are left with the blue h shown in pic?
(edited 7 years ago)
Unknown2.png84.3KB
Original post by ryandaniels2015
Thanks for the reply, I've attached a pic, so do them 2 displacement cancel (shown in red) so you are left with the blue h shown in pic?


Yes, displacement is a vector quantity so let us model down as the positive direction, as it goes up, it would have a negative displacement going up and when it goes down to the same level it would have the positive displacement of the same value. When it falls it will have a further displacement of h, so when added all you have left is h.

Hope this makes sense, if it doesnt i can try to explain it mkre clearly
Original post by metrize
Yes, displacement is a vector quantity so let us model down as the positive direction, as it goes up, it would have a negative displacement going up and when it goes down to the same level it would have the positive displacement of the same value. When it falls it will have a further displacement of h, so when added all you have left is h.

Hope this makes sense, if it doesnt i can try to explain it mkre clearly


Makes a lot of sense, Thanks a lot!!
Can someone please tell me any hard questions in M1 past papers? I don't think I'll have time to go through all of them
Reply 188
Avatar for Marccs
Marccs
4
Hi
Can anyone help me with regards to solving questions about connected particles on a pulley or on a table and when one falls and the rope becomes slack or breaks. I get very confused with how to start to approach the problem to work out maximum height, total distance etc
Thank You
Reply 189
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candol
7
Original post by Marccs
Hi
Can anyone help me with regards to solving questions about connected particles on a pulley or on a table and when one falls and the rope becomes slack or breaks. I get very confused with how to start to approach the problem to work out maximum height, total distance etc
Thank You


General idea is this.
Stage 1. The distance the particle travels until it hits the floor is the same as the other particle will travel along the table
Stage 2. The speed that the particle hits the floor (v) will be the initial speed for the other particle's next stage of travel
This next stage also needs a new calculation for a, as F=ma no longer includes the force from the other particle
Use suvat to calculate new distance (v=0, u=old v ,a = new, s=what we are looking for
Stage 3 Finally add both distances travelled etc

hope this helps
Reply 190
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Marccs
4
Original post by candol
General idea is this.
Stage 1. The distance the particle travels until it hits the floor is the same as the other particle will travel along the table
Stage 2. The speed that the particle hits the floor (v) will be the initial speed for the other particle's next stage of travel
This next stage also needs a new calculation for a, as F=ma no longer includes the force from the other particle
Use suvat to calculate new distance (v=0, u=old v ,a = new, s=what we are looking for
Stage 3 Finally add both distances travelled etc

hope this helps


Ohh so as the first particle drops the second particle will move up at velocity of 0 and then final velocity of the other particle?
Reply 191
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candol
7
Original post by Marccs
Ohh so as the first particle drops the second particle will move up at velocity of 0 and then final velocity of the other particle?


The particle will move up with same velocity as the other is going down. The moment the second particle hits the floor, the velocity at that moment becomes the starting velocity (u) for the next stage of motion
Reply 192
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Marccs
4
Original post by candol
The particle will move up with same velocity as the other is going down. The moment the second particle hits the floor, the velocity at that moment becomes the starting velocity (u) for the next stage of motion


But sometimes the othher particles velocity is immediately the final velocity of the particle that fell and not the second stage ?
f88a22f91af8423ffef551e0f2aea0a3.pngJune 2015, not sure how to do part b.
Reply 194
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candol
7
Original post by Marccs
But sometimes the othher particles velocity is immediately the final velocity of the particle that fell and not the second stage ?


the key is that final velocity becomes the initial velocity for the next stage of motion
Original post by Thomith
f88a22f91af8423ffef551e0f2aea0a3.pngJune 2015, not sure how to do part b.


Watch this
Somone please help with this q

http://www.thestudentroom.co.uk/showthread.php?t=1255487
Reply 197
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candol
7
Original post by L'Evil Wolf


sorry cant see the question
(edited 7 years ago)
Reply 198
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Marccs
4
Original post by candol
the key is that final velocity becomes the initial velocity for the next stage of motion


Ok that makes sense thanks
Reply 199
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candol
7
Original post by L'Evil Wolf


try sketching the diagram first.
After that we can look at puting in the angles/sides
I would then suggest finding the angle at B (use the cosine rule as you have all 3 sides)
(edited 7 years ago)

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