OCR MEI M3 May 2016 - Post Exam discussion Watch

imsoanonymous123
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Last question: My friend keeps saying its -14/3 or something. I say it's -8/3. Who is right or are we both wrong?

My solution

T1 - mg = m(25g)/3
T2 - m(k+1)g = 3m(k+1)g
k+1 = 1 + 2/3 = 5/3
T2 - T1 = mg(5 + 5/3 - 1 - 25/3) = -8mg/3
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Xxxxxxxxy
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I've got the same answer. I hope you are right!
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pepperben
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How did you guys find K?
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¡Muy bien!
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Numbers I remember:
.384
2/3
72°
2517 N
17.5
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StrangeBanana
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I'll chip in

1ai) alpha = -1, beta = 1, gamma = -1
ii) 0.05 grams
bi) 17.5m
ii) 2520N

2i) Show that k = 2
ii) Show that CM is (0.625, 0)
iii) angle was 72.6 degrees

3i) Show that the diff. equation holds.
ii) equilibrium position is at x = 4
iii) amplitude = 0.2m, period = 2pi/sqrt(10)
iv) velocity = 0.384ms^-1 downward

4i) cos(alpha) = 1/3
ii) Show speed is sqrt(ag/3)
iii) k = 2/3
iv) -(8/3)mg

Please add in/correct as necessary ^^

(Original post by imsoanonymous123)
Last question: My friend keeps saying its -14/3 or something. I say it's -8/3. Who is right or are we both wrong?

My solution

T1 - mg = m(25g)/3
T2 - m(k+1)g = 3m(k+1)g
k+1 = 1 + 2/3 = 5/3
T2 - T1 = mg(5 + 5/3 - 1 - 25/3) = -8mg/3
Yeah, you get -14/3 if you don't take into account that the mass has changed after the collision

(Original post by pepperben)
How did you guys find K?
Conservation of energy to get the speed of P just before collision, then conservation of momentum to get the speed of the combined mass just after the collision (in terms of k). Then conservation of energy again, because you know its speed when it gets to A. Quite an involved question!
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VlAd x
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I got 16/9 for k using the conservation of energy, wtf
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DraigGoch
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What are people thinking for grade boundaries? It was weird because it was really easy up to the last 2 parts of the last q. So I'm thinking maybe slightly lower than usual maybe 66 or 67 for A* and 71 for full?
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VlAd x
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(Original post by StrangeBanana)
I'll chip in

1ai) alpha = -1, beta = 1, gamma = -1
ii) 0.05 grams
bi) 17.5m
ii) 2520N

2i) Show that k = 2
ii) Show that CM is (0.625, 0)
iii) angle was 72.6 degrees

3i) Show that the diff. equation holds
ii) amplitude = 0.2m, period = 2pi/sqrt(10)
iii) velocity = 0.384ms^-1 downward

4i) cos(alpha) = 1/3
ii) Show speed is sqrt(ag/3)
iii) k = 2/3
iv) -(8/3)mg

Please add in/correct as necessary ^^



Yeah, you get -14/3 if you don't take into account that the mass has changed after the collision



Conservation of energy to get the speed of P just before collision, then conservation of momentum to get the speed of the combined mass just after the collision (in terms of k). Then conservation of energy again, because you know its speed when it gets to A. Quite an involved question!
Do u remember the equation for k? Or just the steps of that cuz I got 16/9 using the same method, idk if it's a calculation mistake or not
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-Gifted-
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I got k to be( 5/ root 3) - 1 using momemtum and energy. It seemed right. But your value of k seems more reasonable. I dunno.
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VlAd x
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(Original post by DraigGoch)
What are people thinking for grade boundaries? It was weird because it was really easy up to the last 2 parts of the last q. So I'm thinking maybe slightly lower than usual maybe 66 or 67 for A* and 71 for full?
Frankly, loads of ppl got those two questions right
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DraigGoch
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(Original post by VlAd x)
I got 16/9 for k using the conservation of energy, wtf
Some energy would've been lost in the collision. I used conservation of energy to fine the speeds of P and P and Q at the bottom and conservation of momentum to find the mass.
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VlAd x
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(Original post by -Gifted-)
I got k to be( 5/ root 3) - 1 using momemtum and energy. It seemed right. But your value of k seems more reasonable. I dunno.
U don't need to use the momentum
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Davi6336
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I believe K was 16/9 too, most people from my school got it too
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pepperben
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(Original post by StrangeBanana)
I'll chip in

1ai) alpha = -1, beta = 1, gamma = -1
ii) 0.05 grams
bi) 17.5m
ii) 2520N

2i) Show that k = 2
ii) Show that CM is (0.625, 0)
iii) angle was 72.6 degrees

3i) Show that the diff. equation holds
ii) amplitude = 0.2m, period = 2pi/sqrt(10)
iii) velocity = 0.384ms^-1 downward

4i) cos(alpha) = 1/3
ii) Show speed is sqrt(ag/3)
iii) k = 2/3
iv) -(8/3)mg

Please add in/correct as necessary ^^



Yeah, you get -14/3 if you don't take into account that the mass has changed after the collision



Conservation of energy to get the speed of P just before collision, then conservation of momentum to get the speed of the combined mass just after the collision (in terms of k). Then conservation of energy again, because you know its speed when it gets to A. Quite an involved question!
Okay thank you. I couldn't see how to find velocity of p just before the collision .... Did they give us the angle at which p and q collided?
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-Gifted-
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Are u guys sure u did the value of k right ? Cos i had (1+k)^2 term needed to be solved...
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DraigGoch
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(Original post by -Gifted-)
Are u guys sure u did the value of k right ? Cos i had (1+k)^2 term needed to be solved...
I had that but had it equal to 25/9 so we could find k as 2/3
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Davi6336
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(Original post by DraigGoch)
I had that but had it equal to 25/9 so we could find k as 2/3
I had the mass as m(k+1), then I found it to be equal to 26/9m.

So k was equal to 16/9
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VlAd x
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Did the question say this is a I elastic collision
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imsoanonymous123
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(Original post by StrangeBanana)
I'll chip in

1ai) alpha = -1, beta = 1, gamma = -1
ii) 0.05 grams
bi) 17.5m
ii) 2520N

2i) Show that k = 2
ii) Show that CM is (0.625, 0)
iii) angle was 72.6 degrees

3i) Show that the diff. equation holds
ii) amplitude = 0.2m, period = 2pi/sqrt(10)
iii) velocity = 0.384ms^-1 downward

4i) cos(alpha) = 1/3
ii) Show speed is sqrt(ag/3)
iii) k = 2/3
iv) -(8/3)mg

Please add in/correct as necessary ^^



Yeah, you get -14/3 if you don't take into account that the mass has changed after the collision



Conservation of energy to get the speed of P just before collision, then conservation of momentum to get the speed of the combined mass just after the collision (in terms of k). Then conservation of energy again, because you know its speed when it gets to A. Quite an involved question!
you know you're in for a good mark when all your answers are the same as strangebanana's

also I reckon this years grade boundaries might be reasonably high simply because I felt like this paper was unusually short for an M3 paper (4 marks for writing amplitude and period LOL and 7 marks for the part (i) show that in question 3), so there won't be enough people running out of time to keep the grade boundaries down. In terms of the difficulty of the questions hard to say whether it is any different from the average though.
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StrangeBanana
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(Original post by VlAd x)
Do u remember the equation for k? Or just the steps of that cuz I got 16/9 using the same method, idk if it's a calculation mistake or not
I remember the speeds of the combined particle right after the collision was:
\frac{5}{k+1}\sqrt{\frac{ag}{3}}

And using conservation of energy got you a (k+1)^2 = 25/9, which you solve to get 2/3.

(Original post by VlAd x)
U don't need to use the momentum
You do, in the collision.

(Original post by Davi6336)
I believe K was 16/9 too, most people from my school got it too
I'm sure it's 2/3. I think this question tripped up most people.
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