You are Here: Home >< Maths

# edexcel i-al june 2015 m1 watch

1. I'm stuck on the first part. I managed to combine the equation and get k=3u/3u + v. how can i prove k < 1 from there?
Attached Images
2. WME01_01_que_20150603 (dragged) 1.pdf (52.1 KB, 104 views)
3. (Original post by jamesw5555)
I'm stuck on the first part. I managed to combine the equation and get k=3u/3u + v. how can i prove k < 1 from there?
V must be greater than 0 traveling in the positive direction so your equation can never be greater than 1.
4. (Original post by jamesw5555)
I'm stuck on the first part. I managed to combine the equation and get k=3u/3u + v. how can i prove k < 1 from there?
Here is how I would do it:

[EDIT: Made a mistake, see corrected version below]
5. (Original post by Middriver)
You lost a K from the Z on the second step?
Aargh, you're right. Let's try again:

-Initial momentum of A = 2mu
-Initial momentum of B = -3kmu

-Final momentum of A = -mu
-Final momentum of B = zkmu

where z is a positive constant (it has to be positive as particle B is travelling in the opposite direction to A).

So, using conservation of momentum:

Initial momentum = final momentum
=> 2mu - 3kmu = zkmu - mu
=> 2 - 3k = kz - 1
=> 3 = k(z+3)
=>3/(z+3) = k

And since z > 0, then k < 1.
6. (Original post by Ollie231213)
Aargh, you're right. Let's try again:

-Initial momentum of A = 2mu
-Initial momentum of B = -3kmu

-Final momentum of A = -mu
-Final momentum of B = zkmu

where z is a positive constant (it has to be positive as particle B is travelling in the opposite direction to A).

So, using conservation of momentum:

Initial momentum = final momentum
=> 2mu - 3kmu = zkmu - mu
=> 2 - 3k = kz - 1
=> 3 = k(z+3)
=>3/(z+3) = k

And since z > 0, then k < 1.
Ah okay now i understand it, i got the same thing as your last step but it was 3u/3u+z but you cancelled the mu to make it easier to work with. Thx for the help!

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:
Updated: May 31, 2016
Today on TSR

### The most controversial member on TSR?

Who do you think it is...

### Is confrontation required?

Discussions on TSR

• Latest
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE