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EDEXCEL mechanics 3/ june 06 pm

.A.

lets get this right before I leave.

A to B it takes 1.5 sec. So A to 0 it takes 0.75 sec. The question said that the ball is directed towards A initally. So it takes 0.75 s to go to A from 0. Then another 0.75 to come back to 0. So far 1.5 s.

The question asked for t=2s. So we are left with 0.5s now.

Its at centre position.

X = asinwt. a=amplitude (0.12) w=2pi/3 t=0.5. This will give you distance away from O in 0.5s. OB=0.12

P = particle position.

PB = OB - PO

= 0.016

.A.

Man
actually
if it STARTS at A then ur right...and if its starts at B then am right...
we both got the right answers just from different places...
see total distance = 0.24
0.24 - 0.016 = 0.224 which is what I get...
it never stated that the particle starts at A :s-smilie:

so for sum reason I assumed it started at B

Reply 141

valentine_strongrod

.A.

lets get this right before I leave.

A to B it takes 1.5 sec. So A to 0 it takes 0.75 sec. The question said that the ball is directed towards A initally. So it takes 0.75 s to go to A from 0. Then another 0.75 to come back to 0. So far 1.5 s.

The question asked for t=2s. So we are left with 0.5s now.

Its at centre position.

X = asinwt. a=amplitude (0.12) w=2pi/3 t=0.5. This will give you distance away from O in 0.5s. OB=0.12

P = particle position.

PB = OB - PO

= 0.016

.A.

ur right...i didnt read that part where it says at t=0 its goin towards A..
**** thats more marks gone

Reply 142

saklut

IChem

Well, that's what I got, so and therefore, it's probably wrong :p:

I am 99% sure it's correct :wink:

Reply 143

Sepulamb

About the question about T=3/2 dont u guys think it should be multiplied by 4 for because T=(4m)gcos(theta)
Because the 2 particles have a joint mass of 4m instead of m?

Reply 144

Swayum

18

For the question with vertical circular motion, what was the tension in the string when the string made a 60 degree angle?

Reply 145

saklut

Swayum

For the question with vertical circular motion, what was the tension in the string when the string made a 60 degree angle?

3MG/2

Reply 146

Swayum

18

Damn, I got like T = mg(1 + root3/2). Obviously, I used either sin instead of cos or cos instead of sin.

Reply 147

saklut

Sepulamb

About the question about T=3/2 dont u guys think it should be multiplied by 4 for because T=(4m)gcos(theta)
Because the 2 particles have a joint mass of 4m instead of m?

No, because it combines afterwards when it is vertically below the center

Reply 148

saklut

Swayum

Damn, I got like T = mg(1 + root3/2). Obviously, I used either sin instead of cos or cos instead of sin.

I see that you used sin60 instead of cos

Reply 149

ukstudent2011

Swayum

Damn, I got like T = mg(1 + root3/2). Obviously, I used either sin instead of cos or cos instead of sin.

I did what everyone else did, but I'm not sure that it was right. How did you decide what the tension actually must be equal to?

Reply 150

Swayum

18

LJoll

I did what everyone else did, but I'm not sure that it was right. How did you decide what the tension actually must be equal to?

Only had about 5 mins to do the entire question (what was it like 15 marks?!) and I just couldn't figure out where the angle goes so I guessed it. The 50-50-90 rule screwed me over again (whenever you have a 50-50 chance of getting something right, there's a 90% chance you'll get it wrong).

Basicly just resolved towards the centre of the circle, so that's T - component_of_mg = ma. And the component of mg is always sin(theta) or cos(theta). I got it as sin(theta).

Reply 151

saklut

LJoll

I did what everyone else did, but I'm not sure that it was right. How did you decide what the tension actually must be equal to?

I used this for part a)
T-mgcos60= mv^2/a

mgh=1/2mv^2
mgacos60 = 1/2 mv^2
v^2 = ga

since T-mgcos60= mv^2/a

T-mgcos60= mg
T= mg(1+cos60) = 3mg/2

Reply 152

Sepulamb

saklut

I used this for part a)
T-mgcos60= mv^2/a

mgh=1/2mv^2
mgacos60 = 1/2 mv^2
v^2 = ga

since T-mgcos60= mv^2/a

T-mgcos60= mg
T= mg(1+cos60) = 3mg/2

I did the exact same thing and got 6mg... just because instead of mg i put 4mg (the combined mass of the two particles)

Do u guys think it will be wrong?

Reply 153

republican1990

Not sure about the paper... last year you needed like 65/75 to even get an A so im hoping this year the boundaries are lower.

Q3 where you had to show (omega)^2 is greater than or equal to h/g (may have been less than or equal to)... I got to show omega was equal to that but why would it also be less / greater than? I'm guessing you had to explain that bit to get the full 8 marks...

Reply 154

saklut

republican1990

Not sure about the paper... last year you needed like 65/75 to even get an A so im hoping this year the boundaries are lower.

Q3 where you had to show (omega)^2 is greater than or equal to h/g (may have been less than or equal to)... I got to show omega was equal to that but why would it also be less / greater than? I'm guessing you had to explain that bit to get the full 8 marks...

It is in contact with the table, hence there might be a normal reaction force from the table. The maximum value of tension is when there is no reaction force. I started by saying Tcosθ<=mg and then you can proceed to get the final inequality.

Reply 155

republican1990

saklut

It is in contact with the table, hence there might be a normal reaction force from the table. The maximum value of tension is when there is no reaction force. I started by saying Tcosθ<=mg and then you can proceed to get the final inequality.

Ah yes I should have realised that cause I started with a normal reaction R and realised I wasnt getting anywhere so i redid it assuming R = 0. Although I did find somewhere to randomly put a <= sign and continue using it from there. Not sure if the examiner will fall for it though!

Reply 156

alex_hk90

19

republican1990

Q3 where you had to show (omega)^2 is greater than or equal to h/g (may have been less than or equal to)... I got to show omega was equal to that but why would it also be less / greater than? I'm guessing you had to explain that bit to get the full 8 marks...

Q3.(a):

Resolving vertically:

T\cos\theta = mg - R

N2L horizontally:

T\sin\theta = mr\omega^2

Therefore:

\tan\theta = \displaystyle\frac{mr\omega^2}{mg - R}

By trigonometry:

r = h\tan\theta

Substituting in:

\tan\theta = \displaystyle\frac{mh\tan\theta\omega^2}{mg - R}

Giving:

mg - R = mh\omega^2

Therefore:

R = m(g - h\omega^2)

The particle is on the table so:

R \geq 0

Giving:

g - h\omega^2 \geq 0

Therefore:

h\omega^2 \leq g

And finally:

\displaystyle\omega^2 \leq \frac{g}{h}

Reply 157

republican1990

alex_hk90

Q3.(a):

Resolving vertically:

T\cos\theta = mg - R

N2L horizontally:

T\sin\theta = mr\omega^2

Therefore:

\tan\theta = \displaystyle\frac{mr\omega^2}{mg - R}

By trigonometry:

r = h\tan\theta

Substituting in:

\tan\theta = \displaystyle\frac{mh\tan\theta\omega^2}{mg - R}

Giving:

mg - R = mh\omega^2

Therefore:

R = m(g - h\omega^2)

The particle is on the table so:

R \geq 0

Giving:

g - h\omega^2 \geq 0

Therefore:

h\omega^2 \leq g

And finally:

\displaystyle\omega^2 \leq \frac{g}{h}

Yeah I did something kind of similar but as I said I just randomly put a => in somewhere! Thanks!

Reply 158

jcarlisle

the last question was awful!
and i got that it didn't topple as well with the angle of 12.9
i really hope the boundaries are low.......

Reply 159

brkr

how low do u guys think the boundary will go... 50 i hope?? :s

EDEXCEL mechanics 3/ june 06 pm

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