# Edexcel AS/A2 Mathematics M2 - 17th June 2016 - Official ThreadWatch

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2 years ago
#481
(Original post by zandneger)
lambda(i-j) = (lambda)i+(-lambda)j
Since you know horizontal speed is constant, using speed = distance/time you get
distance(horizontal) = 3t
Hence, lambda = 3t and -lamba = -3t
So you want the value for t when displacement(vertical) = -3t
s=-3t u=4 v=/ a=-9.8 t=t
s=ut+1/2at^2
-3t=4t-4.9t^2
7t-4.9t^2=0
t(7-4.9t)=0
s=-3t when t=0 or when t=7/4.9
7/4.9 = 10/7
hence lambda = 3t = 30/7
I write lambda equals 42/g would I lose any marks for writing it in that form.
0
2 years ago
#482
(Original post by zandneger)
lambda(i-j) = (lambda)i+(-lambda)j
Since you know horizontal speed is constant, using speed = distance/time you get
distance(horizontal) = 3t
Hence, lambda = 3t and -lamba = -3t
So you want the value for t when displacement(vertical) = -3t
s=-3t u=4 v=/ a=-9.8 t=t
s=ut+1/2at^2
-3t=4t-4.9t^2
7t-4.9t^2=0
t(7-4.9t)=0
s=-3t when t=0 or when t=7/4.9
7/4.9 = 10/7
hence lambda = 3t = 30/7
Thank you it kinda does make sense
Could you please go through part b and c too?
Part b was asking for speed when particle passes through A and part c was direction of the motion of particle when particle passes through A
I hope I can get few method marks from here although I had wrong value for lambda
0
2 years ago
#483
How many marks would I lose if I put λ into suvat instead of -λ, then carried it through and assuming it's all correct and if I had made a mistake in calculating the speed of A after collision and carried it through the whole question?
0
2 years ago
#484
(Original post by CoolDocA)
I write lambda equals 42/g would I lose any marks for writing it in that form.
I did the same before giving it in decimal form. To be honest it should be marked correct, it didn't specify to give a decimal answer and it's accurate.
0
2 years ago
#485
(Original post by cjlh)
Because it's in terms of vectors you start at the origin (0i+0j). There's no ground so when the particle goes up and back down it'll pass through the i vector (x axis). This means that'll it'll go into the -j region, and as A was at lambda(i-j), the position vector of A is (lambda i, -lambda j). I don't really have time to go through the question again right now but that's how the position vectors work. You then have to work through using displacement and time in terms of lambda (it goes to (lambda i, -lambda j) so the i displacement will be lambda and j will be -lambda).
So it actually go below the x-axis! Right now I understand thank you
0
2 years ago
#486
(Original post by angelitajin)
Thank you, but still really don't understand what's going on tbh I don't even understand how come position vector can be minus(-)

Could you please go through more details? Or if you are going to solve in on paper and post the picture I will wait for it

Thank you very much in advance

Dont have any paper to hand but I can try explaining it better.

Ok, so we know that the position vector of P at time t seconds, is lambda(i-j)
This means the displacement in the horizontal direction(i) is equal to the negative of the displacement in the vertical direction(j)
So, from this, we can see the particle has been moved Lambda metres to the right, and moved lambda metres vertically downwards.

For the horizontal distance, Lambda metres to the right, lets first look at the initial velocity of the particle. It starts with a velocity of (3i+4j) metres per second. Which means initially every second, it travels 3 metres to the right, and 4 metres upwards. Now, since we know horizontal speed does not change during a projection, we can use the equation distance = speed*time to calculate lambda in terms of t.

We take the 3i from the initial velocity and multiply it by the time, t seconds, to get that the distance travelled horizontally is ALWAYS 3ti metres, and varies as t does.

So, because we now know the horizontal displacement is always 3ti metres, this means lambda must be 3t, as 3ti/i = 3t

Now, going back to the start, we know that the particle has been displaced the same amount downwards as it has to the right. This means it has a vertical displacement of -3tj metres (lambda lots of -j).

Next you must set up an equation using SUVAT to create an expression for the vertical displacement at a time t.
Using s = ut + 1/2at^2
With s = -3t u= 4 a = -9.8 t = t

-3t = 4t - 4.9t^2

Carrying over the -3t gives
7t - 4.9t^2 = 0
Factorising gives:
t(7-4.9t)=0

This is expected, as when t=0 the displacement is also zero.

The other value for t is taken from inside the bracket as 7/4.9 , which when typed into a calculator gives 10/7.

Now, from earlier in the question, you remember that lambda is equal to 3t right? Therefore:

10/7 * 3 = 30/7 = lambda

Hope this was a good enough explanation
2
2 years ago
#487
Guys, I accidentally substituted 3 and 4 into the limits of integration in 1b instead of 2 and 3. Calculations are correct, the formulas are also right. How many marks do you think I would loose?
0
2 years ago
#488
(Original post by Nathanbh321)
How many marks would I lose if I put λ into suvat instead of -λ, then carried it through and assuming it's all correct and if I had made a mistake in calculating the speed of A after collision and carried it through the whole question?
Exactly what I did; even drew a diagram with (i+j) on it! If it's explicit that you have misread the question, in the mark scheme it says deduct two marks as long as the subsequent work is correct, and treat all following A marks as follow throughs. If it seems to the examiner that you have misunderstood the concept, then you will probably lose the A marks.
0
2 years ago
#489
(Original post by zandneger)
...
It's better to keep the equation in terms of g - otherwise you need to round as a decimal.

lambda = 42/g
0
2 years ago
#490
(Original post by angelitajin)
Thank you it kinda does make sense
Could you please go through part b and c too?
Part b was asking for speed when particle passes through A and part c was direction of the motion of particle when particle passes through A
I hope I can get few method marks from here although I had wrong value for lambda
b)

Solve using suvat to work out the vertical velocity was 10ms downwards. The horizontal velocity is still 3ms to the right.

The resultant velocity is just found using pythagoras, so root of 10^2 plus 3^2, root 109 , 10.4ms-1 to 3 sig figs

c) You needed to give an angle to either the vertical or horizontal vectors i or j
I said the particle is travelling at an angle theta to the vector i, where theta = arctan (10/3)
0
2 years ago
#491
(Original post by zandneger)
Dont have any paper to hand but I can try explaining it better.

Ok, so we know that the position vector of P at time t seconds, is lambda(i-j)
This means the displacement in the horizontal direction(i) is equal to the negative of the displacement in the vertical direction(j)
So, from this, we can see the particle has been moved Lambda metres to the right, and moved lambda metres vertically downwards.

For the horizontal distance, Lambda metres to the right, lets first look at the initial velocity of the particle. It starts with a velocity of (3i+4j) metres per second. Which means initially every second, it travels 3 metres to the right, and 4 metres upwards. Now, since we know horizontal speed does not change during a projection, we can use the equation distance = speed*time to calculate lambda in terms of t.

We take the 3i from the initial velocity and multiply it by the time, t seconds, to get that the distance travelled horizontally is ALWAYS 3ti metres, and varies as t does.

So, because we now know the horizontal displacement is always 3ti metres, this means lambda must be 3t, as 3ti/i = 3t

Now, going back to the start, we know that the particle has been displaced the same amount downwards as it has to the right. This means it has a vertical displacement of -3tj metres (lambda lots of -j).

Next you must set up an equation using SUVAT to create an expression for the vertical displacement at a time t.
Using s = ut + 1/2at^2
With s = -3t u= 4 a = -9.8 t = t

-3t = 4t - 4.9t^2

Carrying over the -3t gives
7t - 4.9t^2 = 0
Factorising gives:
t(7-4.9t)=0

This is expected, as when t=0 the displacement is also zero.

The other value for t is taken from inside the bracket as 7/4.9 , which when typed into a calculator gives 10/7.

Now, from earlier in the question, you remember that lambda is equal to 3t right? Therefore:

10/7 * 3 = 30/7 = lambda

Hope this was a good enough explanation
that is just brilliant now I fully understand the question thank you so much A person like you deserve good grades Such a clear explanation!
0
2 years ago
#492
(Original post by SHJBHB)
Has to be 3sf in mechanics doesn't it, or can you get away with 4sf or greater for all questions?
I was taught to be as exact as possible unless the question asked differently. So fractions/surds>4sf>3sf>2sf etc

I dont see how an answer can be simply marked wrong, even if correct, based on the fact that you haven't rounded it when the question hasn't asked you to. Nowhere does it suggest you should be rounding to the same as the numbers you put into the formulas to calculate your answers.
0
2 years ago
#493
(Original post by notrodash)
Seemed alright, looking at this thread does inspire confidence. The wording in 1(b) threw me off though and I integrated between t=0 and t=3. Probably a loss of 2-3 marks if they give marks for the method and integration. In this question I found the determinant just to be sure that it didn't cross the t-axis, so maybe that's a method mark too?
There's no need to find the determinant since you know that the minimum speed of the particle was 11ms^-1 (given in the question) and since it is a positive quadratic graph it never goes through the x axis
0
2 years ago
#494
Did quite a lot better in m3 than this haha which is ironic.
1
2 years ago
#495
(Original post by cjlh)
It was out of 6 I think so I'd say 2 or 3 marks lost.
Someone posted the marks a few pages back and they said it was 5 marks?
0
2 years ago
#496
(Original post by 1asdfghjkl1)
Someone posted the marks a few pages back and they said it was 5 marks?
yeah seems I was mistaken, sorry
0
2 years ago
#497
How many marks will i get for 1b) for integrating between 0 and 3 and getting 15m as my final answer?
0
2 years ago
#498
(Original post by imclaren)
Guys, I accidentally substituted 3 and 4 into the limits of integration in 1b instead of 2 and 3. Calculations are correct, the formulas are also right. How many marks do you think I would loose?
Probably two. One for wrong limits and another for the accuracy mark.

(Original post by zandneger)
I was taught to be as exact as possible unless the question asked differently. So fractions/surds>4sf>3sf>2sf etc

I dont see how an answer can be simply marked wrong, even if correct, based on the fact that you haven't rounded it when the question hasn't asked you to. Nowhere does it suggest you should be rounding to the same as the numbers you put into the formulas to calculate your answers.
If you look in a lot of the mark schemes it often only credits answer to 2 or 3 sgf.
0
2 years ago
#499
Anyone know what they got for the value of d on the rod question? And anyone know what they got for the distance traveled on the work-energy question?
0
2 years ago
#500
Hey, so I made a couple of stupid mistakes: used limits of 3 and 4 instead of 2 and 3 in Q1, and got lambda as 7/10 (0.7) instead of 10/7 or whatever... I got all the other answers right - do you think I can still get 90+UMS? How many marks will I lose do you think?
0
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