You are Here: Home >< Maths

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

Announcements Posted on
Four hours left to win £100 of Amazon vouchers!! Don't miss out! Take our short survey to enter 24-10-2016
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

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: June 14, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### Who is getting a uni offer this half term?

Find out which unis are hot off the mark here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams