You are Here: Home >< Maths

# Finding the range of values for the Coefficient of Restitution watch

1. I have been faced with many scenarios where I have to calculate the range of values for 'e' stating the answer in the form of an inequality.

In the example below particle A reverses direction after the FIRST impact. (the markscheme is in the second attachment).

In part (b) I am ABLE to prove the e<9/16.

My problem is with proving e>1/4

What I did is :

the velocity of A after impact 1 < 0

Then I solve to get e<1/4

However, the MS says that it should be :

the velocity of A after impact 1 > 0

Why should the velocity be GREATER than zero ? Since it has reversed direction it is travelling in the negative direction hence it should be LESS than zero.

Can someone help me out ?
Attached Images

2. velocity has to be greater than 0 because after B strikes the wall, both of the particles move in the same direction, thus both velocities must be considered positive. Alternatively, if u get confused, you can take velocities of both A(after srtiking B) and B(after striking the wall) to be negative as the direction of both particles get reversed...... Hope that helps.
3. We did this question the other day.

IIRC, the 9/16 is when one particle is moving faster than the other, and the 1/4 comes from the second particle moving with velocity greater than 0:

So v = u ((4e-1)/5))

=> 4e -1 > 0, e > 1/4

EDIT: It is greater than 0, not less than 0 because you took its direction as left. Negative would mean you drew it as left, but it goes the other way. You drew it as left, and it does go left, so the velocity is positive.
4. (Original post by aakashsaif)
velocity has to be greater than 0 because after B strikes the wall, both of the particles move in the same direction, thus both velocities must be considered positive. Alternatively, if u get confused, you can take velocities of both A(after srtiking B) and B(after striking the wall) to be negative as the direction of both particles get reversed...... Hope that helps.
The very fact we are dealing with velocities should mean that the calculated values are negative/positive, no ?
5. (Original post by H3rrW4rum)
EDIT: It is greater than 0, not less than 0 because you took its direction as left. Negative would mean you drew it as left, but it goes the other way. You drew it as left, and it does go left, so the velocity is positive.
Ah, yeah that makes sense.

Thanks

### Related university courses

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: April 11, 2011
The home of Results and Clearing

### 3,152

people online now

### 1,567,000

students helped last year
Today on TSR

Hang on, have Edexcel's come out already?

### University open days

1. SAE Institute
Animation, Audio, Film, Games, Music, Business, Web Further education
Thu, 16 Aug '18
2. Bournemouth University
Fri, 17 Aug '18
3. University of Bolton
Fri, 17 Aug '18
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