# Homework Help: Momentum (velocity)

1. May 27, 2012

### Mushroom79

1. The problem statement, all variables and given/known data

2 balls, both moving to the right - what are their velocity after collision?
Elastic collision

Ball 1
M 60 kg
u0= 18 m/s

Ball 2
m=220 kg
v0= 7 m/s

2. Relevant equations

Law on momentum conservation
m*v0+M*u0= m*v+M*u

Momentum before = Momentum after

3. The attempt at a solution

Law on momentum conservation gives:
220*7+60*18=220*v+60*u (1)

As it is a elastic collision, it gives :
1/2(220)(7)^2+1/2(60)(18)^2=1/2(220)(v)^2+1/2(60)(u)^2 (2)

2620=220v*60u (1)

15110=1/2(220)(v)^2+1/2(60)(u)^2 (2)

v=(2620-60u)/220 (1)

I need to simplify equation (2) and put in (1), but I can't get further with that
Any help would be appreciated
Would be great if someone could double-check as well, not used with this.

2. May 27, 2012

### Infinitum

Or.... you can simplify (1) and put it in (2) . It works out easier. Just substitute the v you got from equation (1) into (2), which will give you 'u' without much trouble.

Your approach does look correct to me

3. May 27, 2012

### Mushroom79

Oh, okay
So you mean:

v=12-0,27u (1)

15110=110v^2+30u^2 (2) (further simplifed)

15110=110(12-0,27u)^2+30u^2

Now to get 'u' alone.. hmm.. any hint?

Last edited: May 27, 2012
4. May 27, 2012

### Infinitum

Haven't checked the exact numbers, but it looks correct to me. Now comes the boring part. You expand the square term, get a quadratic in u, solve it to get your answer. It surely doesn't look pretty but what are calculators for...

5. May 27, 2012

### Mushroom79

Expand the square term, ok:
15110=1320-29.7u^2+30u^2

Have a slight problem understanding what is meant by get a quadratic in 'u'

6. May 27, 2012

### cepheid

Staff Emeritus
A quadratic expression in u is a second order polynomial in which "u" is the variable that is being raised to increasing powers in each successive term. In this case, since it is a second order polynomial (also known as a quadratic), terms exist with powers of u up to and including the second power of u.

Edit: some powers of u can be missing, that's fine. (In your case you are missing the term that has u raised to the first power). What defines it as second order / quadratic is that there is a u-squared term, and this is the highest power of u in the expression.

7. May 27, 2012

### Infinitum

Um, dont you know quadratic equations?

You need to find the roots of your quadratic and find the corresponding value for v from (1).

---------------------

To simplify this for you(but still, learn quadratics, they're fundamental for most problems in physics), you can keep the terms in variable form, and you'll get,

$v_{f1} = ((m_1 - m_2)·v_{i1} + 2 ·m_2·v_{i2})/{(m_1 + m_2)}$

Now you can find out vf2

8. May 27, 2012

### Mushroom79

I will definitely look up quadratics.
I was able to get the answers, which seems to be correct -thank you for all your help :)

9. May 28, 2012

### Mushroom79

Another question while I'm at it - if the collision was inelastic, how do you think then?

Does it have to be a "perfect" collision? - I'm intrested in a non-perfect collision if it is possible, meaning two velocities instead of one

2 balls, both moving to the right - what are their velocity after collision?
INelastic collision

Ball 1
M 60 kg
u0= 18 m/s

Ball 2
m=220 kg
v0= 7 m/s

Law on momentum conservation is still true,
m*v0+M*u0= m*v+M*u

Kinetic energy is not conserved

10. May 28, 2012

### Infinitum

Is the collision perfectly inelastic? If yes, then the momentum conservation statement would be like

$mv_0 + Mu_0 = (M+m)v_f$

If its a partially elastic collision, you would need extra information i.e the knowledge about the co-efficient of restitution of the impact.

11. May 28, 2012

### Mushroom79

The question simply says "inelastic".
Also, it says to find out velocities (in plural), so I assume it's not perfect.

So I'll have to take a closer look at the formulas now, thank you.

12. May 28, 2012

### Infinitum

If it says inelastic, its most definitely perfectly inelastic.

13. May 28, 2012

### Mushroom79

Oh, good.
Should be easy enough to solve it

14. May 28, 2012

### ehild

Problems of elastic collision can be solved in an easy way without quadratics.

There are two equations, one for momentum, the other for kinetic energy.

m1v0+m2u0=m1v+m2u

m1v02+m2u02=m1v2+m2u2

Arrange both equation so as the terms belonging to the same object are on the same side.

m1(v0-v)=m2(u-u0) *

m1(v02-v2) = m2(u2-u02)-->

m1(v0-v)(v0+v)=m2(u-u0)(u+u0) **

Divide eq.(**) by eq.(*). You can do it as the velocities do change during the collision, so you do not divide by zero. You get a third equation:

v0+v=u+u0 ***

The system of linear equations (*) and (***) is easy to solve.

ehild

15. May 28, 2012

### Infinitum

This is a how I got my direct equation for $v_{f1}$ in one of my previous posts. I seem to have forgotten to post how I got it

16. May 28, 2012

### Mushroom79

m1(v0-(u+u0)-v0) = m2(((v0+v)/u0)-u0)

Or am I completely wrong? - I seem to lack some mathematical skills

17. May 28, 2012

### ehild

The unknowns are v (velocity of ball 1 after collision) and u (velocity of ball 2 after collision).

m1(v0-v)=m2(u-u0) *
v0+v=u+u0 ***

Isolate u from (***):

u=v0+v-u0.

Substitute for v in (*)

m1(v0-v)=m2(v0+v-u0-u0)

Collect the terms with v on one side of the equation:

v(m1+m2)=(m1-m2)v0+2m2u0

Isolate v:

$$v=\frac{(m_1-m_2)v_0+2m_1u_0}{m_1+m_2}$$

ehild