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Mechanics Question

Mudas

Edexcel - M1 - June 2013 (R) - Q4
Erm. Well. How on earth do you do that question?

Reply 1

Zacken

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?

Reply 2

Mudas

https://ce0f1d00673c9b9787bbeaa1aa089c01a93530c8.googledrive.com/host/0B1ZiqBksUHNYczJQc2YxN0hpWjA/Edexcel-Set-2/Ch.2%20Kinematics.pdf

Third question from the bottom!

Reply 3

B_9710

Original post by HarrisonGCSE

https://ce0f1d00673c9b9787bbeaa1aa089c01a93530c8.googledrive.com/host/0B1ZiqBksUHNYczJQc2YxN0hpWjA/Edexcel-Set-2/Ch.2%20Kinematics.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.

Reply 4

RMIM

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.

(edited 8 years ago)

Reply 5

Zacken

Original post by HarrisonGCSE

https://ce0f1d00673c9b9787bbeaa1aa089c01a93530c8.googledrive.com/host/0B1ZiqBksUHNYczJQc2YxN0hpWjA/Edexcel-Set-2/Ch.2%20Kinematics.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

h - 50 = 2T - \frac{1}{2}gT^2

Ball B:

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

That gives us

h = 20T - \frac{1}{2}gT^2

You know how two equations in

h

and

T

... simultaneous equations.

Reply 6

RMIM

Original post by Zacken

Okay,

don't hand him the answer Zac :wink:

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.

Reply 7

Zacken

Original post by RMIM

The hard bit is to produce that equation.

There isn't a hard bit.

Reply 8

RMIM

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.