Help with elastic collisions please

[iwsc]

Homework Statement

Take the general case of a body of mass m1 and velocity v1 elastically striking a stationary (v2=0) body of mass m2 head-on. Show that the final velocity v1' is given by v1'= ((m1-m2)/(m1+m2)) v1'.

Relevant Equations

m1v1 + m2v2 = m1v1' + m2v2' (conservation of momentum)
0.5m1v1^2 + 0.5m2v2^2 = 0.5m1v1'^2 + 0.5m2v2'^2 (conservation of energy)

After simplifying the equations, I got:

m1(v1-v1') = m2v2' (momentum) and

m1(v1-v1')(v1+v1') = m2v2'^2 (kinetic energy)

From there, I'm not sure what to do. I referred to a textbook and it said to divide the energy equation by the momentum equation (the simplified versions) and then do a little algebra. I don't understand how to divide an equation by another equation nor do I know why we do the division in the first place. Help with this or another approach to the problem would be most appreciated!

 

[haruspex]

divide the energy equation by the momentum equation

It means divide the LHS of one by the LHS of the other, and the RHSs likewise.
You will soon see why.

 

[George Jones]

First, form the fraction
$$\frac{m_1 \left( v_1 - v_1' \right) \left( v_1 + v_1' \right)}{m_1 \left( v_1 - v_1' \right) }$$
Now, without changing the value of the fraction, use the kinetic energy equation to replace the numerator by $m_2 {v'}_2^2$, an use the momentum equation to replace the denominator by $m_2 {v'}_2$. Since they are equal, set the two equal to each other.

@haruspex already posted while I was typing.

 

[iwsc]

It means divide the LHS of one by the LHS of the other, and the RHSs likewise.
You will soon see why.

So:

(m1(v1 - v1')(v1 + v1')) / m1(v1 - v1') = (m2v2'^2) / (m2v2')
v1 + v1' = v2' ??

 

[haruspex]

So:

(m1(v1 - v1')(v1 + v1')) / m1(v1 - v1') = (m2v2'^2) / (m2v2')
v1 + v1' = v2' ??

Yes. That equation can also be got directly if you know Newton’s "experimental law". The general form of that is $v_{1f}-v_{2f}=-e(v_{1i}-v_{2i})$, where i and f stand for initial and final, and e is the coefficient of elasticity.
In the present case, e is 1 and $v_{2i}=0$.

 

[iwsc]

Yes. That equation can also be got directly if you know Newton’s "experimental law". The general form of that is $v_{1f}-v_{2f}=-e(v_{1i}-v_{2i})$, where i and f stand for initial and final, and e is the coefficient of elasticity.
In the present case, e is 1 and $v_{2i}=0$.

Okay, that part I'm good with, but I still don't understand how to get those equations to v1'= ((m1-m2)/(m1+m2)) v1'.

 

[haruspex]

Okay, that part I'm good with, but I still don't understand how to get those equations to v1'= ((m1-m2)/(m1+m2)) v1'.

First, that is not the correct target. It should be $v_1'= \frac{m_1-m_2}{m_1+m_2} v_1$. Doesn’t make sense to have v1' both sides.
The benefit of dividing the energy equation by the momentum equation is that it gives you a linear equation. You can now combine this with your original momentum equation, also linear, in the usual way of solving linear simultaneous equations.
You have two unknowns, v1' and v2'. You are only interested in v1', so you need to manipulate them to eliminate v2'.

 

[iwsc]

First, that is not the correct target. It should be $v_1'= \frac{m_1-m_2}{m_1+m_2} v_1$. Doesn’t make sense to have v1' both sides.
The benefit of dividing the energy equation by the momentum equation is that it gives you a linear equation. You can now combine this with your original momentum equation, also linear, in the usual way of solving linear simultaneous equations.
You have two unknowns, v1' and v2'. You are only interested in v1', so you need to manipulate them to eliminate v2'.

Sorry, the v1' at the end was a typo.

So then with

v1 + v1' = v2' and
m1(v1 - v1') =m2v2' which becomes (m1(v1 - v1')) / m2 = v2'

I would do v2' = v2' to get

v1 + v1' = (m1(v1 - v1')) / m2
v1' = ((m1(v1 - v1')) / m2 ) - v1

Is this correct? And from here, how do I go on to the wanted equation?

 

[haruspex]

Sorry, the v1' at the end was a typo.

So then with

v1 + v1' = v2' and
m1(v1 - v1') =m2v2' which becomes (m1(v1 - v1')) / m2 = v2'

I would do v2' = v2' to get

v1 + v1' = (m1(v1 - v1')) / m2
v1' = ((m1(v1 - v1')) / m2 ) - v1

Is this correct? And from here, how do I go on to the wanted equation?

You need to get it into the form v1'=(an expression that does not refer to v1').
So collect the v1' terms together.

 

[iwsc]

You need to get it into the form v1'=(an expression that does not refer to v1').
So collect the v1' terms together.

Sorry, I'm lost. I don't know what I'm supposed to be doing.

 

[haruspex]

Sorry, I'm lost. I don't know what I'm supposed to be doing.

You have v1' on both sides of the equation at the end of post #8.
You need to move the terms around so that all references to v1' are on the same side, and all terms that do not have a factor v1' are on the other side.

 

[neilparker62]

Homework Statement:: Take the general case of a body of mass m1 and velocity v1 elastically striking a stationary (v2=0) body of mass m2 head-on. Show that the final velocity v1' is given by v1'= ((m1-m2)/(m1+m2)) v1'.
Relevant Equations:: m1v1 + m2v2 = m1v1' + m2v2' (conservation of momentum)
0.5m1v1^2 + 0.5m2v2^2 = 0.5m1v1'^2 + 0.5m2v2'^2 (conservation of energy)

After simplifying the equations, I got:

m1(v1-v1') = m2v2' (momentum) and

m1(v1-v1')(v1+v1') = m2v2'^2 (kinetic energy)

From there, I'm not sure what to do. I referred to a textbook and it said to divide the energy equation by the momentum equation (the simplified versions) and then do a little algebra. I don't understand how to divide an equation by another equation nor do I know why we do the division in the first place. Help with this or another approach to the problem would be most appreciated!

Alternate Method.

When we solve quadratic equations, there's a well known formula that can be used so that one does not have to continually go through the process of completing the square. Much less well known is that for elastic collisions (with minor modification - any collision in fact) , there is also a formula that can be used that saves you wading through conservation of momentum and energy (probably good for you to do anyway!)

The formula relates to the equal and opposite impulse that both masses experience upon impact. If you know what that impulse $\Delta P$ is, you can simply subtract from the original momentum of the moving mass and divide by same mass to obtain final velocity. For perfectly elastic collisions: $$\Delta P=2\mu\Delta v$$ where $\mu=\frac{m_1m_2}{m_1+m_2}$ is the 'reduced mass' of the colliding bodies and $\Delta v$ is their relative velocity ($=v_1$ in this case).

 

[iwsc]

You have v1' on both sides of the equation at the end of post #8.
You need to move the terms around so that all references to v1' are on the same side, and all terms that do not have a factor v1' are on the other side.

Alright, I got it! Thank you for the help!