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The Edexcel M1 (16/05/12 - AM) Revision Thread

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Distance

i'm guessing gradient represents acceleration?
Btw, could you please reply to my post: http://www.thestudentroom.co.uk/showpost.php?p=37532318&postcount=309
thanks.

Reply 361

arnab

Original post by emma3

Is the answer for d) 4.25 ? If it is, I'll explain how I did it, if not - I'll have to re-think, haha

yep its 4.25.....possible for you to break it down for me? step by step?

Reply 362

Pinkhead

Original post by arnab

Any help with this question?

At time t = 0, a football player kicks a ball from the point A with position vector (2i + j) m
on a horizontal football field. The motion of the ball is modelled as that of a particle
moving horizontally with constant velocity (5i + 8j) m s–1. Find
(a) the speed of the ball,

(b) the position vector of the ball after t seconds.

The point B on the field has position vector (10i + 7j) m.

(c) Find the time when the ball is due north of B.

At time t = 0, another player starts running due north from B and moves with constant
speed v m s–1. Given that he intercepts the ball,

d) find the value of v.

I worked out part a, b, c but have no idea as to how to do part d :'(

well, you've worked out the time when the ball is due north of B, so put that back into your position vector to find out where the ball actually is at that time.
Now just subtract your new position vector from the original position vector of B, and divide by the time it takes for that change to occur.
V=d/t
so v= change in position vector/t
Ask if you need more help.

Reply 363

Jukeboxing

Original post by nm786

i'm guessing gradient represents acceleration?
Btw, could you please reply to my post: http://www.thestudentroom.co.uk/showpost.php?p=37532318&postcount=309
thanks.

Yes, gradient is acceleration. Ive replied to your post, you probably missed it.

Reply 364

arnab

Original post by Pinkhead

sorry but y do you do the subtraction? can you explain that again? can you show it to me, step by step, as in the calculation?

Reply 365

ninegrandstudent

How would you determine if a 'particle' moves or not, when a force (ie a vertical force) is removed?

This has been asked in a few papers:
Jan 06 Q5
May 2002 Q4

I'm really confused, the markschemes not helpful. I know I need to find the new normal reaction...but from there I'm stuck :')

Reply 366

Pinkhead

Original post by arnab

sorry but y do you do the subtraction? can you explain that again? can you show it to me, step by step, as in the calculation?

your answer for time should be 1.6 seconds.
Put that back into your position vector for the ball after t seconds:

[br]r=(2i+j) + (5i+8j)t[br]r=(2+5(1.6))i + (1+8(1.6))j[br]r=10i + 13.8j[br]

That is the position vector of the ball at t=1.6. Call it rC
Now the distance between this position vector and the original point B is:

rC - rB

(10i+7j) - (10i+13.8j)

which gives us the distance of 6.8j
Now the other player moves with constant speed vms-1. If the distance 6.8j is covered in 1.6 seconds, then

v=6.8j/1.6

v=4.35ms^-1

Reply 367

nm786

Original post by Jukeboxing

Yes, if the force PN or XN is applied horizontally then you should use cos(alpha) and sin(alpha).

So what you've done is correct on the diagram.

thanks,

Reply 368

Pinkhead

Original post by ninegrandstudent

You basically work out the new friction force. If the friction force is greater than the force being applied in order to move the object, it will obviously not move.

Reply 369

ninegrandstudent

Original post by Pinkhead

You basically work out the new friction force. If the friction force is greater than the force being applied in order to move the object, it will obviously not move.

Which force is applied to move the object though? Say if a vertical force was keeping it in place, and that is removed? Probably being really blonde and missing something obvious!

Reply 370

Pinkhead

Original post by ninegrandstudent

Which force is applied to move the object though? Say if a vertical force was keeping it in place, and that is removed? Probably being really blonde and missing something obvious!

In question 5 from the jan 06 paper, the force pulling the object down the plane once the horizontal force is removed is the component of the weight, i.e. 10sin30.

Reply 371

arnab

Original post by Pinkhead

your answer for time should be 1.6 seconds.
Put that back into your position vector for the ball after t seconds:

[br]r=(2i+j) + (5i+8j)t[br]r=(2+5(1.6))i + (1+8(1.6))j[br]r=10i + 13.8j[br]

That is the position vector of the ball at t=1.6. Call it rC
Now the distance between this position vector and the original point B is:

rC - rB

(10i+7j) - (10i+13.8j)

which gives us the distance of 6.8j
Now the other player moves with constant speed vms-1. If the distance 6.8j is covered in 1.6 seconds, then

v=6.8j/1.6

v=4.35ms^-1

ahhh thank you so much. I sort of get it now, just need to get my head around it lol

Reply 372

arnab

In this question, the unit vectors i and j are due east and due north respectively.]
A particle P is moving with constant velocity (–5i + 8j) m s–1. Find
(a) the speed of P,

(b) the direction of motion of P, giving your answer as a bearing.

At time t = 0, P is at the point A with position vector (7i – 10j) m relative to a fixed origin O. When t = 3 s, the velocity of P changes and it moves with velocity (ui + vj) m s–1, where u and v are constants. After a further 4 s, it passes through O and continues to move with velocity (ui + vj) m s–1.

(c) Find the values of u and v.

(d) Find the total time taken for P to move from A to a position which is due south of A.

Any idea on how to do part C and D

Reply 373

nm786

Original post by arnab

what paper is this question from?

Reply 374

oli_G

Original post by arnab

So you know the position vector of P is (7i - 10j)m and it moves with a velocity of (-5i + 8j)m/s for the first 2 seconds. So the displacement of P when t=3 is

(7-10) i + (-10 + 16)j = (-3i + 6j) m

It then moves with velocity ui + vj for 4 seconds passing through the origin, so

s = -3i + 6j

4(ui + vj) = (3i - 6j)

u = 3/4

and

v = -3/2

(edited 11 years ago)

Reply 375

ninegrandstudent

Original post by Pinkhead

In question 5 from the jan 06 paper, the force pulling the object down the plane once the horizontal force is removed is the component of the weight, i.e. 10sin30.

I think I get it - will try again with a different question and see!

Reply 376

Jambone2

Soooo....who else is utterly screwed for this?

Reply 377

Pinkhead

Original post by arnab

c) when t =3, the position vector of P is

(7 – 15)i + (–10 + 24)j

= –8i + 14j

so if the origin is 0,0 the position vector must be 0.
position vector= original position + velocity x time
so

0=-8i+14j + (ui+vJ)*4

0=(4u-8)i + (14+4v)j

equate i and j to 0 to find u and v

4u-8=0 14+4v=0[br] u=2 v=-3.5

d) position vector after t seconds is

(–8i + 14j) + t(2i – 3.5j)

group the components:

(2t-8)i + (14-3.5t)j

if P is due south of A, you can equate the i components of the position vector with the i component of A, which is 7i:

2t-8=7

t=7.5

Now add this onto the time before the velocity changes (3 seconds) and you'll get 10.5seconds.

(edited 11 years ago)

Reply 378

nm786

Original post by oli_G

So you know the position vector of P is (7i - 10j)m and it moves with a velocity of (-5i + 8j)m/s for the first 2 seconds. So the displacement of P when t=3 is
(7-10) i + (-10 + 16)j = (-3i + 6j) m
It then moves with velocity ui + vj for 4 seconds passing through the origin, so