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# M2 - Jan 2014 IAL edexcel question 6a help! Watch

1. I dont understand what the mark scheme or the model answers are saying for this paper. Exam solutions doesnt do it either.

6a.
Paper:
https://ca99c64778b62ba7e7b339967029...%20Edexcel.pdf

MS:
https://ca99c64778b62ba7e7b339967029...%20Edexcel.pdf
2. (Original post by fpmaniac)
I dont understand what the mark scheme or the model answers are saying for this paper. Exam solutions doesnt do it either.

6a.
Paper:
https://ca99c64778b62ba7e7b339967029...%20Edexcel.pdf

MS:
https://ca99c64778b62ba7e7b339967029...%20Edexcel.pdf
What is the kinetic energy at the beginning? It has a kinetic of 1/2 m(3^2 + v^2) at the beginning by adding the KE of the horizontal and vertical component separately.

Not use suvat, horizontally the speed stats the same throughout the motion. So horizontal component is still 3 ms^(-1).

Vertically, use v = u +at, so the new speed is v - gt since the initial speed is v and the acceleration is -g.

Now the this means the K.E at this point is 1/2 m (3^2 + (v-gt)^2).

But you know the KE at this point must be half the initial K.E. That is the initial K.E is twice the new KE.

So 1/2m (3^2 + v^2) = 2 * (1/2 m (3^2 + (v-gt)^2)

so cancelling out the 1/2m we get:

3^2 + v^2 = 2(3^2 + (v-gt)^2)

now expand, get a quadratic in v and solve.
3. (Original post by Zacken)
What is the kinetic energy at the beginning? It has a kinetic of 1/2 m(3^2 + v^2) at the beginning by adding the KE of the horizontal and vertical component separately.

Not use suvat, horizontally the speed stats the same throughout the motion. So horizontal component is still 3 ms^(-1).

Vertically, use v = u +at, so the new speed is v - gt since the initial speed is v and the acceleration is -g.

Now the this means the K.E at this point is 1/2 m (3^2 + (v-gt)^2).

But you know the KE at this point must be half the initial K.E. That is the initial K.E is twice the new KE.

So 1/2m (3^2 + v^2) = 2 * (1/2 m (3^2 + (v-gt)^2)

so cancelling out the 1/2m we get:

3^2 + v^2 = 2(3^2 + (v-gt)^2)

now expand, get a quadratic in v and solve.
Oh ok. Thanks
4. (Original post by fpmaniac)
Oh ok. Thanks
Cool

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