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# M1 string question

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1. Hi, I would like some help with part C of this question please. Basically, I am confused about the distance A has travelled. I have worked out the new acceleration which is
-2/3g ms-2. However since B has hit the ground, does this mean A has moved the same distance that B did. Distance between B and the ground is h. So when B hits the ground, does it mean A has moved m as well? Then I am also a bit confused about the 1/3h part, I don't really understand how to work this out as I am really unsure about the distance part.
Many thanks.
2. (Original post by coconut64)
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3. (Original post by Zacken)
I have just edited the post, please check that out! Thanks.
4. (Original post by Zacken)
I am not sure if the distance between A and P is also h m as the distance between B and the ground is h m. I can't really remember how that's done. Thanks.

Okay. I think I got the answer now. So bascially the distance between A and P is H + 1/3H right?
5. (Original post by coconut64)
I am not sure if the distance between A and P is also h m as the distance between B and the ground is h m. I can't really remember how that's done. Thanks.
A has moved a distance of h when B hits the ground. It then has another 1/3 h to go. You need to know the speed of A (and B) when B hits the ground, as that will be the initial speed for the next part of the motion, with the new acceleration.
6. (Original post by coconut64)
Hi, I would like some help with part C of this question please. Basically, I am confused about the distance A has travelled. I have worked out the new acceleration which is
-2/3g ms-2. However since B has hit the ground, does this mean A has moved the same distance that B did. Distance between B and the ground is h. So when B hits the ground, does it mean A has moved m as well? Then I am also a bit confused about the 1/3h part, I don't really understand how to work this out as I am really unsure about the distance part.
Many thanks.
I'll break it into stages for you.

(i) Both A and B accelerate at the same acceleration. B travels downwards h metres, so A travels rightwards h metres.

(ii) What is the speed of A after having travelled h metres with the acceleration of the string? (by acceleration of the string, I mean the acceleration that both A and B are falling at, d'you know what I mean?) - use SUVAT where the initial speed is from rest, the acceleration is the acceleration of both objects, the distance is h to find the 'final velocity' of A where the final velocity refers to the velocity of A after having travelled h.

(iii) B now hits the ground and the tension in the string disappears.

(iv) You now have that A is h/3 metres away from P, and the 'initial speed' is now the 'final speed' you got from (ii).

(v) Use SUVAT again with initial speed = final speed from (ii), acceleration = new acceleration, distance = h/3 to find the speed as A reaches P.

Basically, what we're doing here is splitting the motion into two, the first part of the motion is the part when there is a tension in the string an B accelerates downwards and A accelerates rightwards. Then B hits the ground and there is no more acceleration from the string, so the only acceleration acting on A is now the friction force providing a deceleration. We need to split the motion into 2 because we can only use SUVAT for constant acceleration. So we need to consider the motion with the first acceleration and then consider the motion for the second acceleration. In the first bit, we know A travels h and in the second bit, we want A to travel h/3.

Which bit of that confuses you?
7. (Original post by coconut64)
Okay. I think I got the answer now. So bascially the distance between A and P is H + 1/3H right?
Well, technically yes, but why on earth would anyone care about the initial distance between A and P? You need to split the motion into two bits. One where it travels h and then where it travels h/3.
8. (Original post by Zacken)
Well, technically yes, but why on earth would anyone care about the initial distance between A and P? You need to split the motion into two bits. One where it travels h and then where it travels h/3.

Well that person would be me then... I just didn't know if distance of A and P would be the same as B and the ground because if that's the case, why would A still be 1/3 h away from pulley?This was what confused me at the start. Okay so for this kind of question, the distance between the particle on the table and the pulley is never the same as the distance between the other particle and the ground? Thanks.
9. (Original post by coconut64)
I am not sure if the distance between A and P is also h m as the distance between B and the ground is h m. I can't really remember how that's done. Thanks.

Okay. I think I got the answer now. So bascially the distance between A and P is H + 1/3H right?
I don't think you're quite getting this.

For the first h that the particle travels it will have a certain force (tension - friction) acting on it. Immediately after, there is no tension, since the 2m particle has hit the floor and is not accelerating downwards anymore creating tension in the string.

So for the last 1/3 h the m particle is travelling under only the force of friction (this frictional force is i.e the same as before). So the acceleration will be different for the last 1/3 h distance it travels.

Therefore you need to consider the motion for the first h and the motion for the last 1/3 h travelled separately (different accelerations for each part which you need to work out).
10. (Original post by Zacken)
Well, technically yes, but why on earth would anyone care about the initial distance between A and P? You need to split the motion into two bits. One where it travels h and then where it travels h/3.
Also, please could you help me with this question? I really need help!
http://www.thestudentroom.co.uk/show....php?t=4038293
11. (Original post by coconut64)
Well that person would be me then... I just didn't know if distance of A and P would be the same as B and the ground because if that's the case, why would A still be 1/3 h away from pulley?This was what confused me at the start. Okay so for this kind of question, the distance between the particle on the table and the pulley is never the same as the distance between the other particle and the ground? Thanks.
Ah, okay. Yeah - the particle A would be further away from the pulley than the particle B is from the ground.
12. (Original post by IrrationalRoot)
I don't think you're quite getting this.

For the first h that the particle travels it will have a certain force (tension - friction) acting on it. Immediately after, there is no tension, since the 2m particle has hit the floor and is not accelerating downwards anymore creating tension in the string.

So for the last 1/3 h the m particle is travelling under only the force of friction (this frictional force is i.e the same as before). So the acceleration will be different for the last 1/3 h distance it travels.

Therefore you need to consider the motion for the first h and the motion for the last 1/3 h travelled separately (different accelerations for each part which you need to work out).

Hi, thanks for helping. I get it what I did wrong now. Thanks.
13. (Original post by coconut64)
Also, please could you help me with this question? I really need help!
http://www.thestudentroom.co.uk/show....php?t=4038293
By the way, I'm not surprised that you didn't get an answer to that question. You posted it in the wrong forum.

Remember, maths questions go in the maths forum, just like this thread is in the maths forum. Try and put them here in the future. :-)

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