MEI Mechanics (M1) Help

On your diagram, draw on the forces for weight (vertically downwards) and normal contact force (perpendicular to the slope). You also have the frictional force parallel to the slope but you don't know its direction yet.

Now resolve parallel to the slope. Using trigonometry, you can calculate the components of the weight force and the 40N. You know that the net force on the block must be zero since the block is in equilibrium. See if you can work out the friction from here. If you get stuck, please post back.

As for tension, the tension of a rope is the same as the force it's exerting on the object. For example, if you had a ball hanging from a rope in equilibrium with weight 10N then the rope must be exerting an upward force of 10N on the ball, so the tension in the rope is 10N.

Another useful thing about tension is that if you had a rope joined from one object to another, then the rope's tension is the same as the force it's exerting on the first object, which is the same as the force it's exerting on the second object.

Reply 3

claret_n_blue

OP

17

My exam is on 27th Jan.

I am really bad at M1. I have done quite a few past papers and the best I have got is 49%. I have seen a similar question before but I just left it out and I couldn't ask online because my internet was broken.

ttoby

Now resolve parallel to the slope. Using trigonometry, you can calculate the components of the weight force and the 40N. You know that the net force on the block must be zero since the block is in equilibrium. See if you can work out the friction from here. If you get stuck, please post back.

Does that mean I do 40cos(35) to get the parallel component:

40cos(35) = 32.766 (is the unit Newtons?)

So the Friction = F - 32.766 = 0

F = 32.766N

Is that the right answer?

(edited 4 years ago)

Reply 4

claret_n_blue

OP

17

I also can't do the last three parts of Q6. I thought for part (ii) that I have to integrate, but I don't know where T will come in.

Reply 5

claret_n_blue

Does that mean I do 40cos(35) to get the parallel component:

40cos(35) = 32.766 (is the unit Newtons?)

Yes, that's correct.

So the Friction = F - 32.766 = 0

F = 32.766N

Is that the right answer?

No, you still need to take in to account the component of the weight force acting on the block, which is 100sin35.

Reply 6

claret_n_blue

OP

17

ttoby

Yes, that's correct.
No, you still need to take in to account the component of the weight force acting on the block, which is 100sin35.

So, becuase the system is in equilibrium F + 40cos(35) = 100sin(35)

Friction = 100sin(35) - 40cos(35)
F = 24.59N

Is that right?

Reply 7

claret_n_blue

So, becuase the system is in equilibrium F + 40cos(35) = 100sin(35)

Friction = 100sin(35) - 40cos(35)
F = 24.59N

Is that right?

Yes, that's correct :smile:

Reply 8

claret_n_blue

I also can't do the last three parts of Q6. I thought for part (ii) that I have to integrate, but I don't know where T will come in.

Would you be able to either post the question or link to the paper, and we should be able to help you with it.

Reply 9

sorry for intrude, but does anyone know where can I find past paper -->MEI M1<-- with WORKOUT SOLUTION... I got some M1 past papers but I can't solve some of the questions.... that's why I'm asking for WORKOUT SOLUTION... thx

Reply 10

tytanek1232

sorry for intrude, but does anyone know where can I find past paper -->MEI M1<-- with WORKOUT SOLUTION... I got some M1 past papers but I can't solve some of the questions.... that's why I'm asking for WORKOUT SOLUTION... thx

Have a look at http://www.mei.org.uk/index.php?section=papers&page=alevelpapers#M1 and click the link for the year you want. They put the question paper, mark scheme and examiner's report into one pdf.

Reply 11

ttoby

Have a look at http://www.mei.org.uk/index.php?section=papers&page=alevelpapers#M1 and click the link for the year you want. They put the question paper, mark scheme and examiner's report into one pdf.

I knew that website...but... What I am looking for is a WORKOUT SOLUTION for those past papers. Thanks :smile:

Reply 12

I knew that website...but... What I am looking for is a WORKOUT SOLUTION for those past papers. Thanks :smile:

Reply 13

tytanek1232

I knew that website...but... What I am looking for is a WORKOUT SOLUTION for those past papers. Thanks :smile:

Which particular questions are you finding difficult? It might be easier to just answer them here if the mark scheme is not giving you enough help.

Reply 14

Ok... if you look on the past paper MEI, M1, 23 May 2008. 4761/01... Section A, question 4 and Section B the whole questin 7 especially part (iii) and (iv).
Thanks a lot.

Reply 15

4. (i) We want the constant acceleration formulae for this. We are dealing with displacement, time, initial velocity and acceleration, so the relevant formula here is

s=ut+\frac{1}{2}at^2

.

For P, we have

s_P=0t+\frac{1}{2}(0.5)t^2=\frac{1}{4}t^2

For Q, we have

s_Q=10t+\frac{1}{2}(0)t^2=10t

Notice that these two distances are measured from different points.

(ii) The particles will catch up with each other when P has travelled 125m further than how far Q has travelled. Look at the diagram to try and convince yourself of this. So we want to solve

s_P=s_Q+125

\frac{1}{4}t^2=10t+125

t^2-40t-500=0

(t-50)(t+10)=0

so t=50 or t=-10. We are not interested in negative time so we take t=50 as our answer. To find out how far P has travelled in this time, substitute our answer into our equation for

s_P

.

Therefore,

s_P=\frac{1}{4}t^2=\frac{1}{4}50^2=625

So it takes 50s for P to catch up with Q, and P has travelled 625m in this time.

You wouldn't need to put all this explanation into your answer, this is just to help you understand it. I need to go now though so I'll have a look at question 7 later.

Reply 16