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# M2 edexcel questions watch

1. Hi everyone!! I have afew questions on a solomon M2 paper (Paper D) i can't do. The paper is found on the following address:

http://www.rdoh.com/exams/maths/M2/S...Papers/M2D.pdf

If it asks for a user name and password, the user name is exams and the password is ukl

The questions i am stuck on are 3b, 5b and 6a. For 3b I got a different answer to what they got in the markscheme (i did a different method, and i can't really see where i am going wrong), whereas for 5b and 6a, i can't do these parts of the question at all , the thing is i don't understand what they have done in the markscheme for 5b and 6a.

The markscheme is found on

http://www.rdoh.com/exams/maths/M2/S...s/M2Dmarks.pdf

Thanks everyone!!!
2. 5b)
The particle is moving forward until t = 1/3, at which point its velocity is zero. Then it starts moving in the opposite direction. So to calc the total dist travelled you have to add the two distances travelled separately. i.e from t=0 to t=1/3 plus from t=1/3 to t=2.
If you did it with trying to just put in t=2 in the eqn for distance, you get the distance from the start point after 2 secs which is NOT the distance travelled.
3. (Original post by Fermat)
5b)
So to calc the total dist travelled you have to add the two distances travelled separately. i.e from t=0 to t=1/3 plus from t=1/3 to t=2.
When you say the dist between t=1/3 and t=2, do you take the difference between t=2 and t=1/3?
4. With 3b they've said at the max speed up the hill, the thrust from the engine equals the resistive force and component of the weight parallel to the slope. For the thrust of the engine they've done P = Fv, F = P/v If you say how you did it someone can try and work out where you went wrong.

In 5a you've worked out where v = 0 which are the times when it changes direction. In 5b you need to work out the distance travelled in the first 2 seconds. By integrating the equation for velocity you get an equation for displacement. To calculate the distance travelled you need to find the displacement at t = 1/3 (when it changes direction) and at t = 2. The particle initially moves 13/27 m in the positive direction, then moves back to displacement -6 m. In this second part it moves a distance of (6 + 13/27)m, total distance travelled is (6 + 2*13/27)m

6a) for the minimum value of a (can't type alpha!) the ball will travel just over 12m horizontally and fall 0.6m.
Horizontal component of velocity is 14cosa, no accn so s = ut, 12 = 14t*cosa, t = 6/7*seca
Vertical component is 14sina up, taking up as positive, s = ut + 1/2*at^2
-0.6 = 14t*sina - 4.9t^2
-0.6 = 12*tana - 3.6*(seca)^2
Use the identity 1 + (tana)^2 = (seca)^2 and let tana = t
-0.6 = 12t - 3.6(1 + t^2)
3.6t^2 - 12t + 3 = 0
Use quadratic formula to get t = 0.27 or 3.06
For smallest a, tana = 0.27, a = 15 degrees to nearest degree
5. 6a)
Resolve the velocity into horizontal and vertical components, viz 14.cosø and 14.sinø respectively.
The distance to M is 12m and you know the horizontal speed, so find the time t for the ball to travel that horizontal dist.
The ball must travel, in the same amount of time, up in the air and then back down again until it is 0.6m (vertically) below where it started. So, get an eqn of motion for vertical movement.
You have an init velocity, and a time travelled (from the horizontal movement), so solve for ø!
6. For 6a, the ball does not hit the ground (0.6m below the original point), 12 m horizontally from the original point. So how can you use the times for both horizontal and vertical components?
7. (Original post by Silly Sally)
When you say the dist between t=1/3 and t=2, do you take the difference between t=2 and t=1/3?
Think of it like displacement from origin.

(don't get confused between distance and displacement )

At t= 1/3, displacement is 13/27.
at t=2, displacement is -6m.
i.e the particle has moved backwards to a position 6m behind the origin.
After t= 1/3, the particle moves back to the origin, travelling another distance of 13/27.
And of course then it moves yet another 6m to give a dislacement of -6m
So total movement is,
13/27 forward from origin
13/27 back to origin
6m further backwards from origin

total = 13/27 + 13/27 + 6
total = 6(26/27)
==========
8. This is what i did for 3b)

If car is travelling at the maximum speed,

driving force = friction + resistance due to weight
= 1800 + 1200.g.sin(sin-1 1/14)
= 1800 + 840
= 2640

So to find maximum speed:

Power = Driving force x max. speed
90000 = 2640 x max. speed

so max. speed = 34.090909...
= 34.1 ms-1

Can anyoe see what i am doing worng?
9. (Original post by Silly Sally)
For 6a, the ball does not hit the ground (0.6m below the original point), 12 m horizontally from the original point. So how can you use the times for both horizontal and vertical components?
M is the goal line.
You have to find out ø so that the ball just passes over the goal line.
So start by assuming the ball hits the goal line (dist = 12m) and work out ø for this.
Does that make sense?
10. 3b) Resistive force is 36v - you've used 1800
11. (Original post by Fermat)
M is the goal line.
You have to find out ø so that the ball just passes over the goal line.
So start by assuming the ball hits the goal line (dist = 12m) and work out ø for this.
Does that make sense?

So basically just assume that it touches the ground - so i basically ignore the diagram then. So the diagram is more confusing than helping!!!!
12. (Original post by Silly Sally)
So basically just assume that it touches the ground - so i basically ignore the diagram then. So the diagram is more confusing than helping!!!!
When you look at the diagram, imagine that the parabola is extended, such that it curves down and touches the point M.
The curve of the parabola will describe the physical movement of the ball.
13. (Original post by Bezza)
3b) Resistive force is 36v - you've used 1800
Oh i see!!!! I used the resistive force from part a - when i had to use the max speed of 50!!! I am a silly billy!!!
14. Oh Fermat and Bezza!!! You are STARS!!!

I am eternally grateful for unconfuddling me!!!

Bezza - I can't add to your rep because i gave you rep yesterday - so you'll have to wait for 10 days I'm sorry!!!

Thanks once again
Sally
15. (Original post by Silly Sally)

Sally
16. (Original post by Silly Sally)
Oh Fermat and Bezza!!! You are STARS!!!

I am eternally grateful for unconfuddling me!!!

Bezza - I can't add to your rep because i gave you rep yesterday - so you'll have to wait for 10 days I'm sorry!!!

Thanks once again
Sally
That's ok, don't worry about the rep though.

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