# Velocity and Center of mass question?

• nukeman
The kinetic energy of the two balls after the collision is greater than their kinetic energy before the collision.f

#### nukeman

Velocity and Center of mass question?

## Homework Statement

Having some issues with this one.

"Ball A has a mass of .200kg and a velocity of 1.50 m/s. Ball B has a mass of .300kg and a celocity of -0.400 m/s. The balls meet in a head on collision. After the collision, the velocity of ball A is -0.780 m/s.

A) Assuming that the two balls form an isolated system, determine whether or not this was an elastic collision.

and

B) (Mainly this one) - Find the velocity of the centre of mass of the system, before and after the collision. ?

## The Attempt at a Solution

How do you find the velocity of the centre of mass of a system??

Any help would be great!

I might get banned if I help you too much before you have given an attempt. I'm guessing you already know the definition of the centre of mass, so try to think logically how you would define the velocity of the centre of mass, in terms of the variables.

Ok by solving for ball B, I got 2.00 m/s - Is that correct?

As its stated ball A is -.780 m/s

So, was this an elastic collision then? How can I justify if it was or not ?

## Homework Statement

Having some issues with this one.

"Ball A has a mass of .200kg and a velocity of 1.50 m/s. Ball B has a mass of .300kg and a celocity of -0.400 m/s. The balls meet in a head on collision. After the collision, the velocity of ball A is -0.780 m/s.

A) Assuming that the two balls form an isolated system, determine whether or not this was an elastic collision.

and

B) (Mainly this one) - Find the velocity of the centre of mass of the system, before and after the collision. ?

## The Attempt at a Solution

How do you find the velocity of the centre of mass of a system??

Any help would be great!

One way to find the velocity of the c of m would be to establish an initial position, then see where it is 1 second later.

eg. let the two masses be 6m apart at time t = 0 - say position 0m, and position +6m
Where will their c of m be?
Where will they be in 1 seconds time?
Where will the c of m be in 1 seconds time?
How far did the c of m move in 1 second?
What is the velocity of the c of m?

Ok by solving for ball B, I got 2.00 m/s - Is that correct?

As its stated ball A is -.780 m/s

So, was this an elastic collision then? How can I justify if it was or not ?

No, I got a different answer for the speed of ball B after collision. You're using the equation for the conservation of momentum, right? Maybe you made a calculation error?

Try again for the speed of the second ball. Now think in terms of energy, for a collision to be considered elastic the ____________ equals the ______________. Hope that helps.

Oh strange...

My calculation for Ball B is: 1.28 m/s - is that correct?

Total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. That is the deifinition I know for Elastic Collision.

With that definition, how does the data from the collision make this elastic?

Oh strange...

My calculation for Ball B is: 1.28 m/s - is that correct?

Total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. That is the deifinition I know for Elastic Collision.

With that definition, how does the data from the collision make this elastic?

1.28 m/s is different to my answer. You're using conservation of momentum to get the speed of ball B, right?

Bruce, here is the formula I am using.

V2f = (2m1/(m1 + m2))V1i + ((m2 - m1)/(m1 + m2))V2i

Bruce, here is the formula I am using.

V2f = (2m1/(m1 + m2))V1i + ((m2 - m1)/(m1 + m2))V2i

Anyone?

are you starting out with...
$\sum p_ i= \sum p_f \Leftrightarrow m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{f}v_{2f} ?$

are you starting out with...
$\sum p_ i= \sum p_f \Leftrightarrow m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{f}v_{2f} ?$

umm, no :( is that the formula I should be using?

But this IS an elastic collision correct?

yes, all the crazy summation notation is saying is that assuming there is no net external force on the system, the sum of the product of the mass and velocity of each particle in the system is a constant.

and yes, I believe if you work out all the values you see that kinetic energy is conserved

What is the velocity of ball b? - I can't seem to get it right. I know the velocity of ball a is -.780

just rearrange the formula above for $v_{bf}$

so I get 1.12 m/s

Is that correct?

and how would I prove its elastic?

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so I get 1.12 m/s

Is that correct?

and how would I prove its elastic?

Yes that is correct for the velocity of B

It is elastic if the kinetic energy after = kinetic energy before.

It is inelastic if the kinetice energy after is less than the kinetic energy before.

It is impossible if the kinetic energy after is greater than the kinetic energy before.

Calculate the kinetic energy before and after and see which one of those three applies.

I agree with PeterO. To make it super-clear: In an elastic collision, the sum of the individual kinetic energies is the same before and after.

So for a collection of masses, you must calculate each of their kinetic energies individually, then sum them up.

It is impossible if the kinetic energy after is greater than the kinetic energy before.

If the kinetic energy is greater after the collision than before, that means some external force is acting on the system of particles, but in that case you could not use conservation of momentum.

From my calculations, it is elastic.

Is this correct?

Ball Ai = .23 J
Ball Af = .06 J

Ball Bi = .02 J
Ball Bf = .19 J

So, before and after =

Correct?

From my calculations, it is elastic.

Is this correct?

Ball Ai = .23 J
Ball Af = .06 J

Ball Bi = .02 J
Ball Bf = .19 J

So, before and after =

Correct?

correct