# M1 connected particles exam question help neededWatch

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#1
I can't seem to get the correct answer for 5c (Ans = 2.4m). I've been at it for hours so can someone please show me the method step by step thank you.

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6 years ago
#2
(Original post by Romisa_lovesA7X)
I can't seem to get the correct answer for 5c (Ans = 2.4m). I've been at it for hours so can someone please show me the method step by step thank you.

Posted from TSR Mobile
I've just worked through the problem and the answer drops out quote nicely!

Can you confirm that you've got the answers to parts a) and b) correct? (I got 2.24ms^-2 for the acceleration and 37.84N for the tension, using g = 9.81 for gravity)

For part c) remember that the particle on the surface has 2 phases to its motion: for the 1st second it is moving as part of a connected system attached to the other particle, and afterwards it moves as an independent particle being slowed down by friction. Therefore you need to calculate 2 separate distances and add them together.

See if you can set up equations for these 2 situations and post back
1
#3
(Original post by davros)
I've just worked through the problem and the answer drops out quote nicely!

Can you confirm that you've got the answers to parts a) and b) correct? (I got 2.24ms^-2 for the acceleration and 37.84N for the tension, using g = 9.81 for gravity)

For part c) remember that the particle on the surface has 2 phases to its motion: for the 1st second it is moving as part of a connected system attached to the other particle, and afterwards it moves as an independent particle being slowed down by friction. Therefore you need to calculate 2 separate distances and add them together.

See if you can set up equations for these 2 situations and post back
Yes those are the answers I'd calculated for parts a & b
Oh I see...I'm not very good with constructing equations and so far have only been able to work out the distance travalled during first second (1.12m?)...
EDIT: Got it!
For first second: s? u 0ms^-1 a 2.2ms^-2
s = 0xt + 1/2x2.2x1^2 = 1.12m

Remaining distance: s? u2.24 ms^-1 v0ms^-1 a= 17.64/9 = 1.96ms^-2
using v^2=u^2+2as
s= (-2.24)^2/2x1.96= 1.28m
1.12+1.28= 2.4 m
1
6 years ago
#4
(Original post by Romisa_lovesA7X)
Yes those are the answers I'd calculated for parts a & b
Oh I see...I'm not very good with constructing equations and so far have only been able to work out the distance travalled during first second (1.12m?)...
EDIT: Got it!
For first second: s? u 0ms^-1 a 2.2ms^-2
s = 0xt + 1/2x2.2x1^2 = 1.12m

Remaining distance: s? u2.24 ms^-1 v0ms^-1 a= 17.64/9 = 1.96ms^-2
using v^2=u^2+2as
s= (-2.24)^2/2x1.96= 1.28m
1.12+1.28= 2.4 m
Well done - I thought you could probably get it with a little prompting
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