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# M1 - Edexcel, 21st May - ULTIMATE THREAD Tweet

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

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IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
Part c)

When the speed of the vehicles is the towbar breaks (OH NOEZ!). Assuming that the resistance to the motion of the car remains of constant magnitude 300N...

find the distance moved by the car from the moment the towbar breaks to the moment when the car comes to rest...

Alright, again, let's draw our diagram to help us visualise this...

Firstly... we need to know the deceleration of the car. This is going to have to be another F=ma job. Let's resolve again, because that's the direction the car is going.

So

Okay, so the car is accelerating at (I.e. it's decelerating at )

Notice I bolded the word "CONSTANT" in the question, this is a clue you can use SUVAT equations as SUVATs only work when the forces are constant.

Right, so now we know the (de)acceleration of the car, we can find out the distance moved before it comes to a halt..

So what do we know overall?
Looking back at the question, it was travelling at 6m/s, it rests at 0m/s and we've found the acceleration.

Initial velocity = u = 6
Final velocity = v = 0
Acceleration = a = -1/3
Displacement = s = ?

We can use

Part d) to follow if requested.

If not, I'll leave it at that.

Hope this solves your connected particle trouble, Junker.
Last edited by EvenStevens; 17-05-2008 at 18:22.
2. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD

Ask me in a PM to takeover the helper role when you get bored or tired
3. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
That's amazing EvenStevens. Yeah, you've helped so much! Thanks for taking your time to help. More +rep when I can.
Last edited by Junker; 17-05-2008 at 18:33.
4. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
I was just doing some practice from my M1 book on moments, and I got stuck on this question. (Heinemann exc 6B q9 p143)

"A uniform rod AB has length 6m and mass 4 kg. It is resting in equilibrium in a horizontal position on supports at points X and Y where AX = 2m and AY = 4.5m. A particle of mass M kg is placed at point C where AC = 5m. Given that the rod is on the point of tilting about Y, calculate the value of M."

I guess i'm stumped because no forces have been given except the downwards 4g.

Any help would be much appreciated as other than moments, i feel i am pretty solid on M1!
5. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
Okay. So rod is at point of tilting about Y therefore reaction is at X=0 so you can ignore that and you do have another downward force, Mg.
6. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
So, (4g x 3) + (5mg) = something

I really have no idea.
7. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
Okay you don't really need to know the reactions so it's best if you take moments at Y. Remember X=0 so ignore that. So:
Moments about Y is (0.5*Mg) = (1.5*4g)
Equal because clockwise moment = anticlockwise moment.
0.5Mg = 6g
"g" can cancel out because it is in each term.
0.5M=6
M = 12
I hope this helps.
8. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
A boy kicks a football vertically upwards from a height of 0.6m above the ground with a speed of 10.5m/s.

a) Find the greatest height above the ground reached by the ball

For this I got 6.225m

b) Calculate the length of time for which the ball is more than 2m above the ground

I did this part like so: to find the velocity of the particle at 2m (9.1m/s) and then found the time from this point to the top to be 0.93 which I multiplied by two to get an answer of 1.86s. Is there a quicker way than this?

Thanks
9. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
If it's on the point of tilting but actually not tilting, it's still in equilibrium but the normal on the X pivot is 0.

So if you take moments around pivot Y clockwise, you should get...

, cancel the 'g's,

Therefore m = 12.

Is that right?
10. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
(Original post by Gimothy)
I have a hard copy. I don't really want to type all the questions out, but if you tell me a topic ill find the relevant question for you.
Never mind , I just found it on TSR.
11. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
Any more questions??

You can find papers and respective MS's on www.freewebs.com/sohanshah
12. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
Thank you Junker and Even Stevens, the two end of the rod A and B and all of the distances were confusing me.

Thanks!
13. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
Ooh, I'm going to join the help squad.
14. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
Ooh, I'm going to join the help squad.
Me you and EvenStevens it is. We'll keep a rota .

Its my shift currently

Having said that, I will also do it until 3 am approx
15. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
I've forgotten moments! My question is on moments where there are 2 supports with reactions or vertical cables, how do you tell the clockwise moments and anticlockwise moments apart. I can't believe I forgot this but i'm just blank...
16. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
I'm giving sohanshah a chance to answer before I pounce
17. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
I'm giving sohanshah a chance to answer before I pounce
Please pounce, I really need to know. Is it just the way the arrows are?
18. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
You should always take moments from one of the supports. Any forces acting in opposite directions on opposite sides of the support from which you're taking the moment, act to tilt the rod in the same way. You treat the other support as you would treat any other force, and it acts in the 'obvious' direction - i.e. the direction that would hold the rod up. Normally it's this force you're expected to find initially.
Last edited by Adjective; 17-05-2008 at 21:06.
19. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD
You should always take moments from one of the supports. Any forces acting in opposite directions on opposite sides of the support from which you're taking the moment, act to tilt the rod in the same way. You treat the other support as you would treat any other force, and it acts in the 'obvious' direction - i.e. the direction that would hold the rod up. Normally it's this force you're expected to find initially.
Good job you did pounce. I've received saddening news and am trying to conquer it: freeexampapers is trying not to be shut down
20. Re: M1 - Edexcel, 21st May - ULTIMATE THREAD

Here, the question asks to find out what X and Y are.

As it is in Limiting equilibrium (question says), Sum of moments = 0.

3g + 5g - X - Y = 0

(this is resolving vertically)

So, 8g = X + Y

Then, around X, (3g * 1) - (5g * 3) + (2 * Y) = 0

So, 2Y = 12g
So, Y = 6g

Then, sub into 8g = X + Y

So, 8g = X + 6g

X = 2g.

Ans: X = 19.6133N Y = 58.8399N
Now, which bit are you confused about, because some of those minuses can be tricky.

I think i have explained that correctly, I hope this helps.
Last edited by 12ian34; 17-05-2008 at 21:20.
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