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# Can someone see what I've done wrong with this M1 question. watch

1. I was doing the May 2003 Edexcel M1 paper, but I seem to have got a really stupid answere for this question (no.1).

A uniform plank AB has a mass 40kg and length 4m. It is supported in a horizontal position by two smooth pivots, one at the end A, the other a the point C of the plank where AC = 3m, as shown in Fig. 1. A man of mass 80kg stands on the plankwhich remains in equilibrium. The magnitudes of the reactions at the two picots are each equal to R newtons. By modelling the plank as a rod and the man as a particle, find

a) the value of R.
b) the distance of the man from A.

This is my working out:

a)2R = 80g + 40g
R = 1176/2
= 588N

(80d) + (40g * 2) = 3R
80d + 784 = 1764
d = 12.25m ------the plank is only supposed to be 4m though.

I probably done something really stupid so if anyone can spot what it is I'll be thankfull.

I was also stuck on this question, I don't know what to do.

A particle P moves with constant acceleration (2i - 3j) ms^-2. At time t seconds, its velocity is V ms^-1. When t=0, V= -2i + 7j

a) Find the value of t when P is moving parallel to the vector i.
b) Find the speed of P when t = 3
c) Find the angle between the vector j and the direction of motion of P when t=3
2. call x length A to man:
About A: 80gx + 3R = 80g
About C: 40g = (3-x)70g + 3R

40g + 80gx = 240g - 80gx + 80g
160x = 280
x = 1.75m

3R = 80g - 80gx
3R = 80[g(1-x)]
= 80(-0.75g)
= -60gN

so 60gN, or 588N upwards.

I haven't checked these but they seem quite likely numbers.
I'll do the vectors one after lunch because they take a while.
3. (Original post by Bebop)
a) the value of R.
b) the distance of the man from A.

This is my working out:

a)2R = 80g + 40g
R = 1176/2
= 588N

(80gd) + (40g * 2) = 3R
80gd + 784 = 1764
d = 1.25m
you just forgot to put the 'g' in for the man
4. For the vector question...

a)

First you need to write an expression for velocity at time t.

You know u=(-2i+7j) and a=(2i-3j). Seeing as you don't need to know a value for t and you are looking for v you can use v=u+at. Rearrange so you have a expressions for i and j in terms of t.

When the particle is parallel to i the j component of it's vector will be 0.

Therefore work out at what value of t the j component is equal to 0.

b)

Use the expression you already found to find the velocity vector. Then use Pythagoras to find the speed.

c)

The j vector is vertically upwards. Draw this and the velocity vector at t=3. Then use tanx=opp/adj and add the answer to 90.

Hope that helps! If you need some more explicit answers just ask.
5. (Original post by nooneknows)
For the vector question...

a)

First you need to write an expression for velocity at time t.

You know u=(-2i+7j) and a=(2i-3j). Seeing as you don't need to know a value for t and you are looking for v you can use v=u+at. Rearrange so you have a expressions for i and j in terms of t.

When the particle is parallel to i the j component of it's vector will be 0.

Therefore work out at what value of t the j component is equal to 0.

b)

Use the expression you already found to find the velocity vector. Then use Pythagoras to find the speed.

c)

The j vector is vertically upwards. Draw this and the velocity vector at t=3. Then use tanx=opp/adj and add the answer to 90.

Hope that helps! If you need some more explicit answers just ask.

When the particle is parallel to i the j component of it's vector will be 0. Why is this?
6. (Original post by Illuvatar)
When the particle is parallel to i the j component of it's vector will be 0. Why is this?
i is the x-direction, j is the y-direction. When the j-component is 0, the vector is not moving in the y-direction and is therefore parallel to the x-direction, i.
7. (Original post by Illuvatar)
When the particle is parallel to i the j component of it's vector will be 0. Why is this?
because vector i is a unit vector along the x-axis, and the unit vector j is along the y-axis. Therefore, for a vector to be parallel to i, then it must not have a vertical ( j ) component
8. (Original post by ZJuwelH)
i is the x-direction, j is the y-direction. When the j-component is 0, the vector is not moving in the y-direction and is therefore parallel to the x-direction, i.
ah ofcourse thanks!

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