# OCR MEI M3 May 2016 - Post Exam discussion Watch

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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

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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#5

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 ^^

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!

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

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

**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

(Original post by

How did you guys find K?

**pepperben**)How did you guys find K?

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#7

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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#8

(Original post by

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!

**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!

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#9

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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#10

(Original post by

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?

**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?

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#11

(Original post by

I got 16/9 for k using the conservation of energy, wtf

**VlAd x**)I got 16/9 for k using the conservation of energy, wtf

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#12

(Original post by

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.

**-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.

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#14

**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!

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#15

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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#16

(Original post by

Are u guys sure u did the value of k right ? Cos i had (1+k)^2 term needed to be solved...

**-Gifted-**)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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#17

(Original post by

I had that but had it equal to 25/9 so we could find k as 2/3

**DraigGoch**)I had that but had it equal to 25/9 so we could find k as 2/3

So k was equal to 16/9

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**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!

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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#20

(Original post by

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

**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

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

(Original post by

U don't need to use the momentum

**VlAd x**)U don't need to use the momentum

(Original post by

I believe K was 16/9 too, most people from my school got it too

**Davi6336**)I believe K was 16/9 too, most people from my school got it too

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