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# M1 OCR (Not MEI) Exam - 9/06/2015 watch

1. (Original post by Super199)
Care to help me with 1ii from June 2005. I'm really bad at mechanics
I'm actually doing that paper right now, and I have to say that I was stuck for quite a while, so don't worry!!

So, the first thing that you need to do is divide your forces up. Since the ring is in equilibrium, we know that the horizontal components of the forces have to equal each other.

The horizontal component of the 1.6N force is 1.6cos45. That's the simple part.

Now for the tension components. So, we can see that there seem to be 2 tensions, one acting at 60 degrees up, and one directly horizontal. Since they are both acting on the ring, we have to find the horizontal components of both of them!

This is T, for the horizontal one, and Tcos60, for the upwards one. (We know that it's acting 60 degrees up because we know 2 angles in that triangle, 30 and 90.)

Now, we can just set them equal to eachother.

T + Tcos60 = 1.6cos45
T = 0.754

Hope that helps! Let me know if there's anything else I'm doing this paper right now, so I'm happy to work through anything that you find slightly weird!
2. (Original post by kawehi)
I'm actually doing that paper right now, and I have to say that I was stuck for quite a while, so don't worry!!

So, the first thing that you need to do is divide your forces up. Since the ring is in equilibrium, we know that the horizontal components of the forces have to equal each other.

The horizontal component of the 1.6N force is 1.6cos45. That's the simple part.

Now for the tension components. So, we can see that there seem to be 2 tensions, one acting at 60 degrees up, and one directly horizontal. Since they are both acting on the ring, we have to find the horizontal components of both of them!

This is T, for the horizontal one, and Tcos60, for the upwards one. (We know that it's acting 60 degrees up because we know 2 angles in that triangle, 30 and 90.)

Now, we can just set them equal to eachother.

T + Tcos60 = 1.6cos45
T = 0.754

Hope that helps! Let me know if there's anything else I'm doing this paper right now, so I'm happy to work through anything that you find slightly weird!
ah brilliant. I am a bit confused with question 2i.
How come one of the tensions are positive and one of them is negative. Aren't both the particles moving downwards?
So tension would oppose the movement?
The equations I had were 0.2g-T-0.4=0.2a
and 0.3g-T-0.25=0.3a

But I have messed up the first equation apparently. Can you explain this question. Thanks for you help
3. (Original post by Super199)
ah brilliant. I am a bit confused with question 2i.
How come one of the tensions are positive and one of them is negative. Aren't both the particles moving downwards?
So tension would oppose the movement?
The equations I had were 0.2g-T-0.4=0.2a
and 0.3g-T-0.25=0.3a

But I have messed up the first equation apparently. Can you explain this question. Thanks for you help
Ok, so first consider the diagram

(hopefully you can see it) I've labelled the 2 tensions, and the direction of motion of the particles.

When doing these weird connected particle questions, it's super important to think about the direction of motion!!!

So for the first particle, A. We can see that the downwards forces (in the DOM) are the particles weight (02g) the tension (T). The upwards force is the drag (0.4).

All of these together equal the net force. In other words, its:

forces in DOM - forces against DOM = net force

So:

0.2g + T - 0.4 = F

And F=ma, so:

0.2g + T -0.4 = 0.2a

You can do the exact same thing for the other particle:

0.3g - T - 0.25 = 0.3a

Because now, T is acting against the DOM.
4. (Original post by kawehi)
Ok, so first consider the diagram

(hopefully you can see it) I've labelled the 2 tensions, and the direction of motion of the particles.

When doing these weird connected particle questions, it's super important to think about the direction of motion!!!

So for the first particle, A. We can see that the downwards forces (in the DOM) are the particles weight (02g) the tension (T). The upwards force is the drag (0.4).

All of these together equal the net force. In other words, its:

forces in DOM - forces against DOM = net force

So:

0.2g + T - 0.4 = F

And F=ma, so:

0.2g + T -0.4 = 0.2a

You can do the exact same thing for the other particle:

0.3g - T - 0.25 = 0.3a

Because now, T is acting against the DOM.
Care to explain how you know where the tensions are? That's the only bit that is confusing me
5. (Original post by Super199)
Care to explain how you know where the tensions are? That's the only bit that is confusing me
Tension is just basically the pulling force in taut strings. The tension will always be 'pulling' on whatever object the string is attached to, so you know that's how you know where T is on the diagram.
6. (Original post by kawehi)
Tension is just basically the pulling force in taut strings. The tension will always be 'pulling' on whatever object the string is attached to, so you know that's how you know where T is on the diagram.
ah brilliant thanks .

I am also stuck on 3ii.

7. Hi I don't really get what we are trying to workout in the first part? Tbh I don't understand any of the parts.
8. (Original post by Super199)
ah brilliant thanks .

I am also stuck on 3ii.

Have you drawn out the diagram for this question? I usually find that everything is much more clear when you've got one!

This question is basically just SUVAT. I presume that you got the first part right, finding that u=3? If so, then we know that the velocity of P is 3, and Q is 0.5.

We need to find the displacement of each of these particles until they come to rest, then add them together to find the total displacement.

Let's do P first:
s= ?
u=3
v=0
a=-5
t=

We can use
v2 = u2 + 2as
which will give us an answer of
s = 0.9

You can do the exact same thing for the second particle.
9. (Original post by sykik)

Hi I don't really get what we are trying to workout in the first part? Tbh I don't understand any of the parts.
Bearing 270 is 90 + 90 +90, so it's the negative x-direction. The question is basically asking you to find the resultant force in the negative x direction.

To do this, just sum up all of the horizontal components of the forces.

We've got 9 as the first one, but the other one is slightly harder.

We know that the 5N force will have a horizontal component. To find it, first find the angle that the force acts from the horizontal.

150 - 90 = 60 degrees

Therefore, the horizontal component is 5cos60, using some basic trig. We have to subtract it from the 9N force, because it's acting in the opposite direction.

so 9 - 5cos60 = 6.5N

Do you want me to do the other ones too? Hope this helps
10. (Original post by kawehi)
Bearing 270 is 90 + 90 +90, so it's the negative x-direction. The question is basically asking you to find the resultant force in the negative x direction.

To do this, just sum up all of the horizontal components of the forces.

We've got 9 as the first one, but the other one is slightly harder.

We know that the 5N force will have a horizontal component. To find it, first find the angle that the force acts from the horizontal.

150 - 90 = 60 degrees

Therefore, the horizontal component is 5cos60, using some basic trig. We have to subtract it from the 9N force, because it's acting in the opposite direction.

so 9 - 5cos60 = 6.5N

Do you want me to do the other ones too? Hope this helps
Thanks I don't really get the bit in bold! I am really bad with bearings
11. (Original post by kawehi)

Have you drawn out the diagram for this question? I usually find that everything is much more clear when you've got one!

This question is basically just SUVAT. I presume that you got the first part right, finding that u=3? If so, then we know that the velocity of P is 3, and Q is 0.5.

We need to find the displacement of each of these particles until they come to rest, then add them together to find the total displacement.

Let's do P first:
s= ?
u=3
v=0
a=-5
t=

We can use
v2 = u2 + 2as
which will give us an answer of
s = 0.9

You can do the exact same thing for the second particle.
How come the velocity isn't negative as the direction of motion has changed?
12. (Original post by sykik)
Thanks I don't really get the bit in bold!
Ok so you know how the total number of degrees around a graph is 360? Like how each of the quadrants are right angles, etc? This diagram might help:

The question is basically asking you to find out what the force is, for a bearing of 270 degrees. As you can see, 270 degrees is equivalent to the negative x direction, if you pretend that the diagram is a graph.
13. (Original post by Super199)
How come the velocity isn't negative as the direction of motion has changed?
Since we are just trying to find the total distance travelled by each of the particles, we can consider them individually, and give them each their own separate DOM. You can make it negative if you want, but it won't change the answer. All you need to do is just find out how far each particle travels before coming to rest, and then add those distances together.
14. (Original post by kawehi)
Ok so you know how the total number of degrees around a graph is 360? Like how each of the quadrants are right angles, etc? This diagram might help:

The question is basically asking you to find out what the force is, for a bearing of 270 degrees. As you can see, 270 degrees is equivalent to the negative x direction, if you pretend that the diagram is a graph.

If that's 270 then why is Resultant in negative x direction? I don't get why is it resultant
15. (Original post by kawehi)
Since we are just trying to find the total distance travelled by each of the particles, we can consider them individually, and give them each their own separate DOM. You can make it negative if you want, but it won't change the answer. All you need to do is just find out how far each particle travels before coming to rest, and then add those distances together.
Brilliant. If you don't get 100 ums Idek haha.

With question 4i.

I think it simultaneous equations. But I can't seem to solve haha
The one's I have got are 2=0.8u+1/2a(0.8^2)
and 6=1.2u+1/2a(0.2^2)
Are these right?
16. (Original post by Super199)
Brilliant. If you don't get 100 ums Idek haha.

With question 4i.

I think it simultaneous equations. But I can't seem to solve haha
The one's I have got are 2=0.8u+1/2a(0.8^2)
and 6=1.2u+1/2a(0.2^2)
Are these right?
It's on page 2. Assuming this is June 2005?
17. (Original post by sykik)

If that's 270 then why is Resultant in negative x direction? I don't get why is it resultant
Hmm, I'm not entirely sure what you mean. 270 is the same as the negative x-direction, and resultant is just the sum of the forces. I think that the question is just worded weirdly to throw people off, but don't worry about it too much!

Overall, the question is just asking you to find the horizontal resultant. You don't really need to make it any more complicated than that!
18. (Original post by sykik)

If that's 270 then why is Resultant in negative x direction? I don't get why is it resultant
Because the 9 N is in the negative x direction and the 5sin30 is in the positive. As we are resolving at 270 degrees,the direction we choose is the negative x direction we do 9 - 5sin30! I hope that's clear?
19. Thank you! Yh i figured it out after thinking about it
20. (Original post by Super199)
Brilliant. If you don't get 100 ums Idek haha.

With question 4i.

I think it simultaneous equations. But I can't seem to solve haha
The one's I have got are 2=0.8u+1/2a(0.8^2)
and 6=1.2u+1/2a(0.2^2)
Are these right?
Haha thanks

Yeah, I got the first one, but for the second one, I decided to use the data from point P to the 'end' so that the SUVAT looked like this:

s= 2 + 6 = 8
u= u
v=
a= a
t= .8 + 1.2 = 2

8 = 2u + 2a
a = 4 - u

Maybe if you try it with that, it would be easier to solve?

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