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

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.

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).

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

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?

...

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

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

And there's your answer.

Hope this was a good enough explanation

Has to be 3sf in mechanics doesn't it, or can you get away with 4sf or greater for all questions?

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?

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?

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?

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.

Did quite a lot better in m3 than this haha which is ironic.