Momentum & Elastic Collision: 2 Balls, Velocity After Collision?

[Mushroom79]

Homework Statement

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

Homework Equations

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

Momentum before = Momentum after

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.

Thanks in advance!

 

[Infinitum]

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.

Thanks in advance!

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

 

[Mushroom79]

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

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?

 

[Infinitum]

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?

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...

 

[Mushroom79]

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...

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'

 

[cepheid]

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'

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.

 

[Infinitum]

Um, don't you know quadratic equations?

http://en.wikipedia.org/wiki/Quadratic_equation

Quadratic in u means the variable is u, instead of x.

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 v_f2

 

[Mushroom79]

Um, don't you know quadratic equations?

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 v_f2

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

 

[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 interested in a non-perfect collision if it is possible, meaning two velocities instead of one2 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

 

[Infinitum]

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 interested in a non-perfect collision if it is possible


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

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.

 

[Mushroom79]

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 of the impact.

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.

 

[Infinitum]

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.

If it says inelastic, its most definitely perfectly inelastic.

 

[Mushroom79]

If it says inelastic, its most definitely perfectly inelastic.

Oh, good.
Should be easy enough to solve it

 

[ehild]

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

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

 

[Infinitum]

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

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

 

[Mushroom79]

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

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

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

 

[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