# M1 Jan 2009

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#1
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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4 years ago
#2
(Original post by 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?
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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4 years ago
#3
(Original post by 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?
The tension in the string will be zero, which is where I find this question bad, as minute differences in the friction would change that.

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.
0
4 years ago
#4
(Original post by 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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#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 RogerOxon)
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#6
It was basically a gold paper!
(Original post by 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.
0
4 years ago
#7
(Original post by 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!
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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#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 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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4 years ago
#9
(Original post by 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?
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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#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 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.
0
4 years ago
#11
(Original post by 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?
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.
1
#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!!!!
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#13
[QUOTE=Ismarika;71852822]
(Original post by 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!!!!
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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4 years ago
#14
[QUOTE=Ismarika;71876560]
(Original post by 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
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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#15
[QUOTE=Towcestermaths;71880820]
(Original post by 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.
What do you mean by using the symmetry? I think I get the fact that the speeds are same thingy!! Also what about d)?
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4 years ago
#16
[QUOTE=Ismarika;71884434]
(Original post by 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)?
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.
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