Author Topic: All Mechanics DOUBTS HERE!!!  (Read 96363 times)

Offline astarmathsandphysics

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Re: All Mechanics DOUBTS HERE!!!
« Reply #435 on: April 30, 2011, 03:04:11 pm »
Half of 10 take 2

Offline Weaam

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Re: All Mechanics DOUBTS HERE!!!
« Reply #436 on: May 03, 2011, 02:37:41 pm »
Can someone send me a link where I can find edexcel papers and mark schemes for C1 , C2 and M1 ?
Most importantly M1 Jan 09 mark schemee
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Offline astarmathsandphysics

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Offline wakemeup

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Re: All Mechanics DOUBTS HERE!!!
« Reply #438 on: May 07, 2011, 09:05:20 pm »
Can someone please clearly explain 8d of this question paper to me? The mark scheme isn't making much sense at all.


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Re: All Mechanics DOUBTS HERE!!!
« Reply #439 on: May 08, 2011, 08:33:57 am »
Can someone please clearly explain 8d of this question paper to me? The mark scheme isn't making much sense at all.

The displacement equation for 0<t<4 :

4t2 -0.5t3

From t>4 :

16t - t2 -16

In the first 4 seconds the displacement is : 32m

It is at instantaneous rest at 8 seconds. At this point it changes direction such that its displacement from the origin will begin to DECREASE.

At 8 seconds displacement is : 48

At 10 seconds : 44 m        NOTE : The displacement has decreased as previously stated.

Hence, it travels 48m FORWARDS and 4 m BACKWARDS.

Thus, 48 + 4 = 52m




Offline wakemeup

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Re: All Mechanics DOUBTS HERE!!!
« Reply #440 on: May 08, 2011, 12:16:50 pm »
Thanks, I understand it now.

Offline iluvme

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Re: All Mechanics DOUBTS HERE!!!
« Reply #441 on: May 11, 2011, 12:23:59 pm »
June 2010 Paper 41 Question 6 i)

I got a=6ms-2

How do you show the speed as 3.95 ms-1?
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elemis

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Re: All Mechanics DOUBTS HERE!!!
« Reply #442 on: May 11, 2011, 01:01:56 pm »
June 2010 Paper 41 Question 6 i)

I got a=6ms-2

How do you show the speed as 3.95 ms-1?

CIE ? Give me a minute.

elemis

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Re: All Mechanics DOUBTS HERE!!!
« Reply #443 on: May 11, 2011, 01:11:37 pm »
0.45*10 - T = 0.45a

T - 0.3*0.2*10 = 0.2a

Adding the two equation to eliminate T gives :

0.45*10 - 0.3*0.2*10 = 0.65a

Solving for a :

a = 6

Now. If you read the first part of the question you'll notice a very important piece of information which I think you've missed :

Quote
Particles A and B, of masses 0.2 kg and 0.45 kg respectively, are connected by a light inextensible
string of length 2.8m
. The string passes over a small smooth pulley at the edge of a rough horizontal
surface, which is 2m above the floor. Particle A is held in contact with the surface at a distance of
2.1m from the pulley and particle B hangs freely (see diagram). The coefficient of friction between
A and the surface is 0.3. Particle A is released and the system begins to move.

That means the distance from the pulley to B is 0.7 m. Hence, B is 1.3 m above the ground.

Use v^2=u^2 +2as  to find the final speed.

Offline iluvme

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Re: All Mechanics DOUBTS HERE!!!
« Reply #444 on: May 11, 2011, 01:16:44 pm »
Thanks Ari.

Could you help me with second part too?
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elemis

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Re: All Mechanics DOUBTS HERE!!!
« Reply #445 on: May 11, 2011, 01:17:18 pm »
Thanks Ari.

Could you help me with second part too?

Okay. Hang on.

elemis

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Re: All Mechanics DOUBTS HERE!!!
« Reply #446 on: May 11, 2011, 01:25:46 pm »
When B hits the ground there will be no tension in the string.

Hence, for A :

-0.3*0.2*10 = 0.2*a

a = -3

However, by this time A has moved 1.3 metres towards the pulley (since B fell 1.3 m) and A has a speed of 3.95.

So using this data you should be able to calculate the final speed.


Offline iluvme

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Re: All Mechanics DOUBTS HERE!!!
« Reply #447 on: May 11, 2011, 01:31:40 pm »
Why is it -0.3?
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elemis

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Re: All Mechanics DOUBTS HERE!!!
« Reply #448 on: May 11, 2011, 01:38:51 pm »
Why is it -0.3?

Since friction is acting in the direction OPPOSITE to motion.

Offline iluvme

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Re: All Mechanics DOUBTS HERE!!!
« Reply #449 on: May 11, 2011, 01:41:48 pm »
Oh right, so stupid of me.

Thanks Ari. :)
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