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# Mechanics Question

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1. Edexcel - M1 - June 2013 (R) - Q4
Erm. Well. How on earth do you do that question?
2. (Original post by HarrisonGCSE)
Edexcel - M1 - June 2013 (R) - Q4
Erm. Well. How on earth do you do that question?
Can you link the question paper?
3. https://ce0f1d00673c9b9787bbeaa1aa08...Kinematics.pdf

Third question from the bottom!
4. (Original post by HarrisonGCSE)
https://ce0f1d00673c9b9787bbeaa1aa08...Kinematics.pdf

Third question from the bottom!
When A and B are at the same height, the vertical displacement that object B will have moved through will be the
height, h, but the vertical displacement that A will have moved through will h-50
Draw a diagram if you can't quite see why this is.
5. simultaneously solve.

1. diagram
2. look for an appropriate equation
3. simultaneously solve.

You have 2 things the same. They are both at h at time T.

(I think, without actually trying the questions.)

Do you have the answers? Madness to do these kind of questions if you cant check the answer after.
6. (Original post by HarrisonGCSE)
https://ce0f1d00673c9b9787bbeaa1aa08...Kinematics.pdf

Third question from the bottom!
Okay, so you have two ball A:

S = h-50 (because it starts off 50 metres above the ground so it's displacement at height h is h-50 metres)
U = 2 ms^(-1) (initial speed)
V =
A = -9.81 = -g
T = T

That gives us

Ball B:

S = h
U = 20
V =
A = -g = -9.81
T=T

That gives us

You know how two equations in and ... simultaneous equations.
7. (Original post by Zacken)
Okay,
don't hand him the answer Zac

let him fight it out - stays in your brain then.

you have now given him a gcse question.

The hard bit is to produce that equation.
8. (Original post by RMIM)

The hard bit is to produce that equation.
There isn't a hard bit.
9. (Original post by Zacken)
There isn't a hard bit.
ok, the main bit.

I dare say u might get even half marks or more just by producing the equations without even solving them

but anyway - Harrison, u have seen how it's done now. This type of question comes up all the time and the method stays the same.

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