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

anyone doing solomon papers?

Reply 304

Jukeboxing

8

Original post by strawberrykoi

what grades are you guys aiming for in this one? :smile:

100 ums

Reply 305

Jukeboxing

8

Original post by nm786

can you please tell me how to do question 7 on this paper: http://www.kingsleyschoolbideford.co.uk/Images-(1)/SeniorSchool/Maths-PDFS/M1_20100524.aspx
and could you please tell me if my diagram is correct or is there anything else i should add.
Thanks, man. :smile:

Your diagram is perfect, all you need to do is resolve forces to form two equations which can be used to solve both part (a) and (b).

resolve perpendicluar to the plane

So R - 0.4gcos(alpha) - Psin(alpha) = 0

Parallel to the plane(Up)

Pcos(alpha) - (1/3)R - 0.4gsin(alpha) = 0

Now ill leave the solving to you :smile:

(edited 11 years ago)

Reply 306

Lackadaisical

Can anyone help me to part D to this question:
A coastguard station O monitors the movements of ships in a
channel. At noon, the station radar records two ships moving
with constant speed. Ship A is at the point with position vector
(−6i + 10j) km relative to O and has velocity (3i + 3j) km/h.
Ship B is at the point with position vector (4i + 5j) km and has
velocity (−3i + 6j) km/h, where i and j are unit vectors directed
due east and due north respectively.
(a) Given that the two ships maintain these velocities, show that
they collide.
The coast guard radios ship A and orders it to reduce its speed
to move with velocity (2i + 3j) km/h.
(b) Find an expression for the vector AB at time t hours after noon.
(c) Find the the distance between A and B at 1400 hours.
(d) Find the time at which B will be due north of A.

:s-smilie:

Reply 307

Original post by Lackadaisical

Can anyone help me to part D to this question:
A coastguard station O monitors the movements of ships in a
channel. At noon, the station radar records two ships moving
with constant speed. Ship A is at the point with position vector
(−6i + 10j) km relative to O and has velocity (3i + 3j) km/h.
Ship B is at the point with position vector (4i + 5j) km and has
velocity (−3i + 6j) km/h, where i and j are unit vectors directed
due east and due north respectively.
(a) Given that the two ships maintain these velocities, show that
they collide.
The coast guard radios ship A and orders it to reduce its speed
to move with velocity (2i + 3j) km/h.
(b) Find an expression for the vector AB at time t hours after noon.
(c) Find the the distance between A and B at 1400 hours.
(d) Find the time at which B will be due north of A.

:s-smilie:

Answer:
a) yes at time =5/3
b) AB= (10-5t)i + (3t-5)j
c) 1km
d) 10-5t=0
5t=10
t=2
so time =1400hours

pls correct me if im wrong

Reply 308

nm786

Original post by 99llewellyn99

Answer:
a) yes at time =5/3
b) AB= (10-5t)i + (3t-5)j
c) 1km
d) 10-5t=0
5t=10
t=2
so time =1400hours

pls correct me if im wrong

he wants you to explain to him how to do the question.

Reply 309

yea but i frst wnt 2 kno if i am right coz im doing mechanics aftr a long time

Reply 310

nm786

Original post by 99llewellyn99

yea but i frst wnt 2 kno if i am right coz im doing mechanics aftr a long time

ok

Reply 311

knowledgecorruptz

9

Original post by 99llewellyn99

Answer:
a) yes at time =5/3
b) AB= (10-5t)i + (3t-5)j
c) 1km
d) 10-5t=0
5t=10
t=2
so time =1400hours

pls correct me if im wrong

Got the same answer :smile:

Reply 312

so i guess i am right
well Lackadaisical so from part 'b' i component of vector AB is (10-5t) so for B to be north of A or vice versa the i component should equal 0
OR
the i component of A should equal the i component of vector B

Reply 313

knowledgecorruptz

9

Original post by Lackadaisical

Can anyone help me to part D to this question:
A coastguard station O monitors the movements of ships in a
channel. At noon, the station radar records two ships moving
with constant speed. Ship A is at the point with position vector
(−6i + 10j) km relative to O and has velocity (3i + 3j) km/h.
Ship B is at the point with position vector (4i + 5j) km and has
velocity (−3i + 6j) km/h, where i and j are unit vectors directed
due east and due north respectively.
(a) Given that the two ships maintain these velocities, show that
they collide.
The coast guard radios ship A and orders it to reduce its speed
to move with velocity (2i + 3j) km/h.
(b) Find an expression for the vector AB at time t hours after noon.
(c) Find the the distance between A and B at 1400 hours.
(d) Find the time at which B will be due north of A.

:s-smilie:

a) The position of ship A and ship B will be there original position (Ro) + (velocity x time) = Ro + vt

For A: -6i +10j + 3ti + 3tj
For B: 4i +5j -3tj +6tj

When they collide, their position vectors will be the same because they're at the same point (hence the collision). Therefore:

-6i +10j +3ti +3tj = 4i +5j -3tj +6tj
Simplify to: (-10 +6t)i + (5-3t)j = 0
If the distance is 0, the i (horizontal) distance and the j (vertical) distance must both be zero, so: -10 +6t = 0 ... t = 5/3 ...and you get the same answer if you work out "t" using the fact that j distance equals 0.

You should be able to finish the rest of them with that in mind. Remember for the last question, if B is due north of A, their i (horizontal) positions will be the same, ONLY there vertical (j) positions will differ. Hope that was useful :smile:

Reply 314

salvequamtu

Original post by Lackadaisical

Can anyone help me to part D to this question:
A coastguard station O monitors the movements of ships in a
channel. At noon, the station radar records two ships moving
with constant speed. Ship A is at the point with position vector
(−6i + 10j) km relative to O and has velocity (3i + 3j) km/h.
Ship B is at the point with position vector (4i + 5j) km and has
velocity (−3i + 6j) km/h, where i and j are unit vectors directed
due east and due north respectively.
(a) Given that the two ships maintain these velocities, show that
they collide.
The coast guard radios ship A and orders it to reduce its speed
to move with velocity (2i + 3j) km/h.
(b) Find an expression for the vector AB at time t hours after noon.
(c) Find the the distance between A and B at 1400 hours.
(d) Find the time at which B will be due north of A.

:s-smilie:

for part a)...
get the a & b in the form 'r0 +vt' [r0 being the position/starting vector]

So a= (-6i+10j) + t(3i+3j) <<---- expand and put in form i & j
= i(-6+3t) + j(10+3t)

b= (4i+5j) + t(-3i+6j)
= i(4-3t) + j(5+6t)
Now to show that they collide... you must make i=i and j=j & you should get the same time...if they collide:

-6+3t = 4-3t 10+3t=5+6t
6t = 10 3t = 5
t= 5/3 t = 5/3

now for part b) ...
to find the expression for the distance AB you quite simply subtract the 2 expression we made earlier in part a) [ so b-a] :

i(4-3t) + j(5+6t) - i(-6+3t) + j(10+3t) = i(10-6t) + j(-5+3t)

now part c) ... you quite simple substitute in t=2 (hours)
[as]

-2i + j
you can now form a triangle with this and use Pythagoras to find the distance
( you should end up with square root of 5 :smile:

)

now for d) in order for the time at which B to be due north of A, B will have to have the same i component of A, so...

I think the time will be the same as in part a) ...5/3 hours?? :frown:

not sure about that one sorry

Reply 315

nm786