# M1 Jan 2009

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Ismarika

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Two particles P and Q, of mass 2 kg and 3 kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. A constant force F of magnitude 30 N is applied to Q in the direction PQ, as shown in Figure 4. The force is applied for 3 s and during this time Q travels a distance of 6 m. The coefficient of friction between each particle and the plane is µ .

Find

(a) the acceleration of Q - 4/3

(b) the value of µ - 10/21

(c) the tension in the string - 320 N

When the particles have moved for 3 s, the force F is removed.

(e) Find the time between the instant that the force is removed and the instant that Q comes to rest.

For part e) So I get that v would be 4ms and you use that as u and for

deceleration you resolve but I don't understand why the tension is removed and not modelled when considering the motion of Q and also can anyone tell me how to know which particles to consider each time when resolving?

Find

(a) the acceleration of Q - 4/3

(b) the value of µ - 10/21

(c) the tension in the string - 320 N

When the particles have moved for 3 s, the force F is removed.

(e) Find the time between the instant that the force is removed and the instant that Q comes to rest.

For part e) So I get that v would be 4ms and you use that as u and for

deceleration you resolve but I don't understand why the tension is removed and not modelled when considering the motion of Q and also can anyone tell me how to know which particles to consider each time when resolving?

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Towcestermaths

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#2

(Original post by

Two particles P and Q, of mass 2 kg and 3 kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. A constant force F of magnitude 30 N is applied to Q in the direction PQ, as shown in Figure 4. The force is applied for 3 s and during this time Q travels a distance of 6 m. The coefficient of friction between each particle and the plane is µ .

Find

(a) the acceleration of Q - 4/3

(b) the value of µ - 10/21

(c) the tension in the string - 320 N

When the particles have moved for 3 s, the force F is removed.

(e) Find the time between the instant that the force is removed and the instant that Q comes to rest.

For part e) So I get that v would be 4ms and you use that as u and for

deceleration you resolve but I don't understand why the tension is removed and not modelled when considering the motion of Q and also can anyone tell me how to know which particles to consider each time when resolving?

**Ismarika**)Two particles P and Q, of mass 2 kg and 3 kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. A constant force F of magnitude 30 N is applied to Q in the direction PQ, as shown in Figure 4. The force is applied for 3 s and during this time Q travels a distance of 6 m. The coefficient of friction between each particle and the plane is µ .

Find

(a) the acceleration of Q - 4/3

(b) the value of µ - 10/21

(c) the tension in the string - 320 N

When the particles have moved for 3 s, the force F is removed.

(e) Find the time between the instant that the force is removed and the instant that Q comes to rest.

For part e) So I get that v would be 4ms and you use that as u and for

deceleration you resolve but I don't understand why the tension is removed and not modelled when considering the motion of Q and also can anyone tell me how to know which particles to consider each time when resolving?

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RogerOxon

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#3

**Ismarika**)

Two particles P and Q, of mass 2 kg and 3 kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. A constant force F of magnitude 30 N is applied to Q in the direction PQ, as shown in Figure 4. The force is applied for 3 s and during this time Q travels a distance of 6 m. The coefficient of friction between each particle and the plane is µ .

Find

(a) the acceleration of Q - 4/3

(b) the value of µ - 10/21

(c) the tension in the string - 320 N

When the particles have moved for 3 s, the force F is removed.

(e) Find the time between the instant that the force is removed and the instant that Q comes to rest.

For part e) So I get that v would be 4ms and you use that as u and for

deceleration you resolve but I don't understand why the tension is removed and not modelled when considering the motion of Q and also can anyone tell me how to know which particles to consider each time when resolving?

For (e), we will, in this ideal world, have P and Q decelerate at the same rate. Their speeds will therefore match and their separation is constant. Therefore, we can treat them individually, or as a single particle.

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RogerOxon

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#4

(Original post by

(a) the acceleration of Q - 4/3

(b) the value of µ - 10/21

(c) the tension in the string - 320 N

When the particles have moved for 3 s, the force F is removed.

(e) Find the time between the instant that the force is removed and the instant that Q comes to rest.

**Ismarika**)(a) the acceleration of Q - 4/3

(b) the value of µ - 10/21

(c) the tension in the string - 320 N

When the particles have moved for 3 s, the force F is removed.

(e) Find the time between the instant that the force is removed and the instant that Q comes to rest.

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Ismarika

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#5

Yes sorry I got c) wrong it was 12N i got confused with another question. But sending you the working out now as well as the question!

(Original post by

Please post your working. (c) cannot be right.

**RogerOxon**)Please post your working. (c) cannot be right.

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Ismarika

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#6

It was basically a gold paper!

(Original post by

I need to see the diagram. Can you post it or say exactly where the question came from. I have looked at MEI, AQA and Edexcel M1 Jan 2009 but can't find it.

**Towcestermaths**)I need to see the diagram. Can you post it or say exactly where the question came from. I have looked at MEI, AQA and Edexcel M1 Jan 2009 but can't find it.

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Towcestermaths

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(Original post by

Yes sorry I got c) wrong it was 12N i got confused with another question. But sending you the working out now as well as the question!

**Ismarika**)Yes sorry I got c) wrong it was 12N i got confused with another question. But sending you the working out now as well as the question!

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#8

That's the thing i'm confused on; why? How can I use that info for other scenarios? I basically understand that if you're working out tension you have to consider a single particle, and if you're finding accelaration you use only one particle as using tension for a single particle would be too much hassle in terms of simultaneous equations but I dont think I understand why there is a change in tension or even the thrust when there is different changes of motion on the particle?

(Original post by

As Rogeroxon says, there will be no tension in the string as each particle will be decelerating at the same rate. and you can just consider Q on its own. Then the only force acting is the friction, so the acceleration can be easily worked out and then its just a simple suvat question.

**Towcestermaths**)As Rogeroxon says, there will be no tension in the string as each particle will be decelerating at the same rate. and you can just consider Q on its own. Then the only force acting is the friction, so the acceleration can be easily worked out and then its just a simple suvat question.

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Towcestermaths

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(Original post by

That's the thing i'm confused on; why? How can I use that info for other scenarios? I basically understand that if you're working out tension you have to consider a single particle, and if you're finding accelaration you use only one particle as using tension for a single particle would be too much hassle in terms of simultaneous equations but I dont think I understand why there is a change in tension or even the thrust when there is different changes of motion on the particle?

**Ismarika**)That's the thing i'm confused on; why? How can I use that info for other scenarios? I basically understand that if you're working out tension you have to consider a single particle, and if you're finding accelaration you use only one particle as using tension for a single particle would be too much hassle in terms of simultaneous equations but I dont think I understand why there is a change in tension or even the thrust when there is different changes of motion on the particle?

If you don't like this explanation. then consider the two particles and the string as one body. The total force acting on the combined body will be 2g*10/21 + 3g *10/21 = 5g *10/21. Then using F =ma, a=F/m = (5g *10/21)/ 5 =10/21 * g. You then just use this value of a in a suvat calculation.

If you want to you could then consider just one of the particles and assume that there is a tension of T in the string. By considering the total of the forces and the acceleration you will be able to show that T = 0.

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#10

So as a whole if two particles are accelerating or decelerating in this case, the tensions cancel out right? I don't do physics so these things kinda hard for me to reason with especially the fact that considering one particle will also cancel out the tension but the tensions are in opposite hence they cancel right?

(Original post by

Its sort of intuitive really as each particle will have the same acceleration. This is because the frictional force acting on each particle will be equal to the weight * 10/21 and the acceleration will be the force divided by the mass. So the acceleration from the friction on each particle will be equal to 10/21 *g. As the acceleration due to friction will be the same for each particle then there will be no tension in the string.

If you don't like this explanation. then consider the two particles and the string as one body. The total force acting on the combined body will be 2g*10/21 + 3g *10/21 = 5g *10/21. Then using F =ma, a=F/m = (5g *10/21)/ 5 =10/21 * g. You then just use this value of a in a suvat calculation.

If you want to you could then consider just one of the particles and assume that there is a tension of T in the string. By considering the total of the forces and the acceleration you will be able to show that T = 0.

**Towcestermaths**)Its sort of intuitive really as each particle will have the same acceleration. This is because the frictional force acting on each particle will be equal to the weight * 10/21 and the acceleration will be the force divided by the mass. So the acceleration from the friction on each particle will be equal to 10/21 *g. As the acceleration due to friction will be the same for each particle then there will be no tension in the string.

If you don't like this explanation. then consider the two particles and the string as one body. The total force acting on the combined body will be 2g*10/21 + 3g *10/21 = 5g *10/21. Then using F =ma, a=F/m = (5g *10/21)/ 5 =10/21 * g. You then just use this value of a in a suvat calculation.

If you want to you could then consider just one of the particles and assume that there is a tension of T in the string. By considering the total of the forces and the acceleration you will be able to show that T = 0.

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(Original post by

So as a whole if two particles are accelerating or decelerating in this case, the tensions cancel out right? I don't do physics so these things kinda hard for me to reason with especially the fact that considering one particle will also cancel out the tension but the tensions are in opposite hence they cancel right?

**Ismarika**)So as a whole if two particles are accelerating or decelerating in this case, the tensions cancel out right? I don't do physics so these things kinda hard for me to reason with especially the fact that considering one particle will also cancel out the tension but the tensions are in opposite hence they cancel right?

Imagine what would happen if the connecting string was cut. Basically the 2 particles would stay at the same distance from each other as they both start at the same velocity and decelerate at the same rate. Hence the string is not actually doing anything and therefore the tension is zero.

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#12

[QUOTE=Towcestermaths;71852176]If they are accelerating at the same rate, they are travelling at the same speed and the accelerating force is not coming from the other particle... then yes

Imagine what would happen if the connecting string was cut. Basically the 2 particles would stay at the same distance from each other as they both start at the same velocity and decelerate at the same rate. Hence the string is not actually doing anything and therefore the tension is zero.[/QUO

Yes i think i get it now thank you!!!!

Imagine what would happen if the connecting string was cut. Basically the 2 particles would stay at the same distance from each other as they both start at the same velocity and decelerate at the same rate. Hence the string is not actually doing anything and therefore the tension is zero.[/QUO

Yes i think i get it now thank you!!!!

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#13

[QUOTE=Ismarika;71852822]
I also needed help with part c) and d) of this question i dont undersand the mark scheme and I got the accelaration as 10/27 because speed of B was 100/3 which is correct

(Original post by

If they are accelerating at the same rate, they are travelling at the same speed and the accelerating force is not coming from the other particle... then yes

Imagine what would happen if the connecting string was cut. Basically the 2 particles would stay at the same distance from each other as they both start at the same velocity and decelerate at the same rate. Hence the string is not actually doing anything and therefore the tension is zero.[/QUO

Yes i think i get it now thank you!!!!

**Towcestermaths**)If they are accelerating at the same rate, they are travelling at the same speed and the accelerating force is not coming from the other particle... then yes

Imagine what would happen if the connecting string was cut. Basically the 2 particles would stay at the same distance from each other as they both start at the same velocity and decelerate at the same rate. Hence the string is not actually doing anything and therefore the tension is zero.[/QUO

Yes i think i get it now thank you!!!!

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#14

[QUOTE=Ismarika;71876560]
From your graph you can see that the two trains are doing the same speed whilst train A is at the constant velocity of 20 m per sec. So for train B use v = u + at, with u =0, a = 10/27 and v = 20 to calculate the time that train B is also doing 20 m per sec. Then use the symmetry that you can see in the graph to calculate the next time that the trains are at the same speed.

(Original post by

I also needed help with part c) and d) of this question i dont undersand the mark scheme and I got the accelaration as 10/27 because speed of B was 100/3 which is correct

**Ismarika**)I also needed help with part c) and d) of this question i dont undersand the mark scheme and I got the accelaration as 10/27 because speed of B was 100/3 which is correct

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#15

[QUOTE=Towcestermaths;71880820]
What do you mean by using the symmetry? I think I get the fact that the speeds are same thingy!! Also what about d)?

(Original post by

From your graph you can see that the two trains are doing the same speed whilst train A is at the constant velocity of 20 m per sec. So for train B use v = u + at, with u =0, a = 10/27 and v = 20 to calculate the time that train B is also doing 20 m per sec. Then use the symmetry that you can see in the graph to calculate the next time that the trains are at the same speed.

**Ismarika**)From your graph you can see that the two trains are doing the same speed whilst train A is at the constant velocity of 20 m per sec. So for train B use v = u + at, with u =0, a = 10/27 and v = 20 to calculate the time that train B is also doing 20 m per sec. Then use the symmetry that you can see in the graph to calculate the next time that the trains are at the same speed.

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#16

[QUOTE=Ismarika;71884434]
Look at the graph that you have drawn. The lines for both trains are symmetrical about the t = 90 seconds line. Thus, once you have calculated the first value of t for when the trains are doing the same speed, you can see that the second value of t will be the same amount of time after 90s as the first one was before.

For d) just calculate the area under the graph for each train at t = 96s and see which train has travelled the furthest and by how much.

(Original post by

What do you mean by using the symmetry? I think I get the fact that the speeds are same thingy!! Also what about d)?

**Towcestermaths**)What do you mean by using the symmetry? I think I get the fact that the speeds are same thingy!! Also what about d)?

For d) just calculate the area under the graph for each train at t = 96s and see which train has travelled the furthest and by how much.

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