Finding the equation for final velocity in a perfectly elastic collision

[hmorenom111]

Homework Statement

Ball 1, with a mass of 140 g and traveling at 15.0 m/s , collides head on with ball 2, which has a mass of 350 g and is initially at rest. What are the final velocities of each ball if the collision is perfectly elastic?

Relevant Equations

Conservation of momentum and conservation of kinetic energy equations

So after not being able to solve this problem I did some researching online. I was looking around and came across this video, where they give the following equations for solving for the final velocities of both balls:

v_1f=((m₁-m₂)/(m₁+m₂))*v_1i

v_2f=(2m₁/(m₁+m₂))*v_1i

I plugged in my numbers and got the correct answers v_1f=-6.43 m/s and v_2f=8.57 m/s.

In the video they don't show any of the algebra towards finding these equations and I can't seem to figure them out.

Any help is appreciated. Thanks!

 

[RPinPA]

Well, as you say it's conservation of momentum and conservation of kinetic energy. So write down those two equations. The initial velocity of object 2 is 0.
##m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}##
##m_1 v_{1i}^2 = m_1 v_{1f}^2 + m_2 v_{2f}^2##

I multiplied both sides of the KE equation by 2 because there's no point in keeping that (1/2) around. The two unknowns are ##v_{1f}## and ##v_{2f}## so let's see if we can do a substitution or elimination of one of those. First I'm going to simplify a little more, reduce some of the clutter. Let's divide everything by ##m_1## and define a parameter ##\mu = m_2/m_1.## So now we have
##v_{1i} = v_{1f} + \mu v_{2f}##
##v_{1i}^2 = v_{1f}^2 + \mu v_{2f}^2##

OK, ##v_{1f} = v_{1i} - \mu v_{2f}## so
##v_{1i}^2 = (v_{1i} - \mu v_{2f})^2 + \mu v_{2f}^2
= v_{1i}^2 - 2\mu v_{1i} v_{2f} + \mu^2 v_{2f}^2 + \mu v_{2f}^2##
##0 = - 2\mu v_{1i} v_{2f} + (\mu^2 + \mu) v_{2f}^2 = v_{2f} \left [ -2\mu v_{1i} + (\mu^2 + \mu) v_{2f} \right ]##
which leads to ##v_{2f} = 0## or ##v_{2f} = 2v_{1i}/(\mu + 1) = 2 v_{1i} m_1 / (m_1 + m_2)##

That's your second equation. You can then find ##v_{1f}## by calculating ##v_{1f} = v_{1i} - (m_2/m_1) v_{2f}##

I'm missing something obvious for why ##v_{2f} = 0## is not an allowed solution, i.e. ball 1 retains all of its original momentum and KE. Perhaps simply that it would require ball 1 to pass through ball 2 in order to keep going forward.

Here is another derivation.

 

[hmorenom111]

Well, as you say it's conservation of momentum and conservation of kinetic energy. So write down those two equations. The initial velocity of object 2 is 0.
##m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}##
##m_1 v_{1i}^2 = m_1 v_{1f}^2 + m_2 v_{2f}^2##

I multiplied both sides of the KE equation by 2 because there's no point in keeping that (1/2) around. The two unknowns are ##v_{1f}## and ##v_{2f}## so let's see if we can do a substitution or elimination of one of those. First I'm going to simplify a little more, reduce some of the clutter. Let's divide everything by ##m_1## and define a parameter ##\mu = m_2/m_1.## So now we have
##v_{1i} = v_{1f} + \mu v_{2f}##
##v_{1i}^2 = v_{1f}^2 + \mu v_{2f}^2##

OK, ##v_{1f} = v_{1i} - \mu v_{2f}## so
##v_{1i}^2 = (v_{1i} - \mu v_{2f})^2 + \mu v_{2f}^2
= v_{1i}^2 - 2\mu v_{1i} v_{2f} + \mu^2 v_{2f}^2 + \mu v_{2f}^2##
##0 = - 2\mu v_{1i} v_{2f} + (\mu^2 + \mu) v_{2f}^2 = v_{2f} \left [ -2\mu v_{1i} + (\mu^2 + \mu) v_{2f} \right ]##
which leads to ##v_{2f} = 0## or ##v_{2f} = 2v_{1i}/(\mu + 1) = 2 v_{1i} m_1 / (m_1 + m_2)##

That's your second equation. You can then find ##v_{1f}## by calculating ##v_{1f} = v_{1i} - (m_2/m_1) v_{2f}##

I'm missing something obvious for why ##v_{2f} = 0## is not an allowed solution, i.e. ball 1 retains all of its original momentum and KE. Perhaps simply that it would require ball 1 to pass through ball 2 in order to keep going forward.

Here is another derivation.

Thank you for the quick and detailed response. I had to write it down to understand it but I got it now.

Yeah, that ##v_{2f} = 0## also made me wonder, I will ask other people about it.

Thanks!

 

[jbriggs444]

There is another approach that depends less on algebra and more on physical intuition.

Shift to a frame of reference where the center of mass of the system is motionless. This will require computing the velocity of the center of mass, $$v_{cm}=\frac{m_{1i}v_{1i} + m_{2i}v_{2i}}{m_{1i}+m_{2i}}$$

[I've not taken advantage of the fact that ##v_{2i} = 0##]

For an elastic collision in one dimension, symmetry demands that the rebound will be a mirror image of the approach. Let us use capital V's for the velocities relative to the center of mass.

$$V_{1i} = v_{1i} - v_{cm}$$
$$V_{1f} = -V_{1i}$$
$$v_{1f} = V_{1f} + v_{cm}$$

Similarly for ##v_2##.

[Edit, repaired a sign error]

 

[PeroK]

I'm missing something obvious for why ##v_{2f} = 0## is not an allowed solution, i.e. ball 1 retains all of its original momentum and KE. Perhaps simply that it would require ball 1 to pass through ball 2 in order to keep going forward.

This is an allowed solution. It's just the boring one that the balls do not collide. Let's generalise the problem to two dimensions, where the balls collide and scatter at an angle to the original direction of motion. For a glancing blow, the object ball may have an arbitrarily small final velocity. The solution where ##v_{2f} = 0## is the limiting case when the balls just miss each other.

 

[PeroK]

Homework Statement: Ball 1, with a mass of 140 g and traveling at 15.0 m/s , collides head on with ball 2, which has a mass of 350 g and is initially at rest. What are the final velocities of each ball if the collision is perfectly elastic?
Homework Equations: Conservation of momentum and conservation of kinetic energy equations

So after not being able to solve this problem I did some researching online. I was looking around and came across this video, where they give the following equations for solving for the final velocities of both balls:

v_1f=((m₁-m₂)/(m₁+m₂))*v_1i

v_2f=(2m₁/(m₁+m₂))*v_1i

I plugged in my numbers and got the correct answers v_1f=-6.43 m/s and v_2f=8.57 m/s.

In the video they don't show any of the algebra towards finding these equations and I can't seem to figure them out.

Any help is appreciated. Thanks!

Physics is often taught as though every problem of this type is different. Change the speed of the first ball to ##20m/s## and the mass of the second ball to ##250g## and you have a new problem. But, really, these are all the same physical problem just with different numbers. The "real" physics is that the ratio of the masses is the key factor. That's why ##\mu = m_2/m_1## is introduced. That is the key (dimensionless) physical quantity. "Dimensionless" means it's the same whether you measure mass in ##kg##, pounds or whatever.

The real physics involves statements like:

If ##m_1## is less than ##m_2## then you get a rebound (##v_{1f}## is negative).

If ##m_1 = m_2##, then ##m_1## stops and ##m_2## continues with the original velocity.

If ##m_1## is much greater than ##m_2## then the motion of ##m_1## is not much affected and ##m_2## gets propelled at up to twice the orginal velocity.

This is the debate that I assume goes on in educational circles between understanding and analysing formulas against simply plugging numbers into formulas. Sadly, in my opinion, the "plug-and-chug" educators seem to be winning the day.

 

[hmorenom111]

Physics is often taught as though every problem of this type is different. Change the speed of the first ball to ##20m/s## and the mass of the second ball to ##250g## and you have a new problem. But, really, these are all the same physical problem just with different numbers. The "real" physics is that the ratio of the masses is the key factor. That's why ##\mu = m_2/m_1## is introduced. That is the key (dimensionless) physical quantity. "Dimensionless" means it's the same whether you measure mass in ##kg##, pounds or whatever.

The real physics involves statements like:

If ##m_1## is less than ##m_2## then you get a rebound (##v_{1f}## is negative).

If ##m_1 = m_2##, then ##m_1## stops and ##m_2## continues with the original velocity.

If ##m_1## is much greater than ##m_2## then the motion of ##m_1## is not much affected and ##m_2## gets propelled at up to twice the orginal velocity.

This is the debate that I assume goes on in educational circles between understanding and analysing formulas against simply plugging numbers into formulas. Sadly, in my opinion, the "plug-and-chug" educators seem to be winning the day.

That makes a lot of sense, thank you!

 

[neilparker62]

Another alternate solution. This one rests on the fact that during collision , the colliding bodies experience equal and opposite forces and hence equal and opposite impulses. For a perfectly elastic collision, one can algebraically determine an expression for this equal and opposite impulse (change in momentum): $$ Δp = 2μΔv $$ where μ is the reduced mass of the colliding masses: $$ μ=\frac{m_1m_2}{m_1+m_2} $$ and Δv their relative velocity. Simply add the impulse to the (momentum of) the stationary mass and subtract it from the (momentum of) the moving mass to obtain final momentum for both. Divide by respective masses to obtain final velocity for both.

Note that in post #2, we solve a pair of simultaneous equations. When we use the equation Δp = 2μΔv, we have essentially 'pre-solved' the simultaneous equations. The resultant formula is a little simpler to remember and use than those obtained algebraically for ##v_{1f}## and ##v_{2f}##. It is also 'intuitive' in so far as we are directly applying Newton 3 to the collision problem by calculating the exact value of the equal and opposite impulse experienced by the colliding masses.

 

[hmorenom111]

There is another approach that depends less on algebra and more on physical intuition.

Shift to a frame of reference where the center of mass of the system is motionless. This will require computing the velocity of the center of mass, $$v_{cm}=\frac{m_{1i}v_{1i} + m_{2i}v_{2i}}{m_{1i}+m_{2i}}$$

[I've not taken advantage of the fact that ##v_{2i} = 0##]

For an elastic collision in one dimension, symmetry demands that the rebound will be a mirror image of the approach. Let us use capital V's for the velocities relative to the center of mass.

$$V_{1i} = v_{1i} - v_{cm}$$
$$V_{1f} = -V_{1i}$$
$$v_{1f} = V_{1f} + v_{cm}$$

Similarly for ##v_2##.

[Edit, repaired a sign error]

So why is it exactly that in the frame of reference of the center of mass the rebound will be a mirror image of the apporach?

 

[jbriggs444]

So why is it exactly that in the frame of reference of the center of mass the rebound will be a mirror image of the apporach?

In the absence of energy loss (dissipation, the second law of thermodynamics and such), the laws of mechanics are invariant with respect to time reversal. If a definite outcome is predicted at all, it must have the same arrival and departure velocities. Anything else would be asymmetric.

 

[PeroK]

So why is it exactly that in the frame of reference of the center of mass the rebound will be a mirror image of the apporach?

You should be able to show that for yourself. The total momentum is zero, so the objects must be moving in opposite directions. The speed of one mass determines the speed of the other. Hence the total KE depends on that one variable. If the KE is conserved then the speed of each mass must be the same after the collision as before.

Try doing the algebra to double check this.

 

[haruspex]

The speed of one mass determines the speed of the other. Hence the total KE depends on that one variable. If the KE is conserved then the speed of each mass must be the same after the collision as before.

There's a little more to it than that.
First, one has to note that these two constraints are algebraically independent. Since momentum conservation is linear and energy (irreducibly) quadratic, that follows.
Next, there is the possibility of multiple solutions. With one linear and one quadratic there are in principle two solutions, but no more. The original velocities (as though they missed) form one. Since negating both velocities is a second solution, those are the two.

 

[PeroK]

There's a little more to it than that.
First, one has to note that these two constraints are algebraically independent. Since momentum conservation is linear and energy (irreducibly) quadratic, that follows.
Next, there is the possibility of multiple solutions. With one linear and one quadratic there are in principle two solutions, but no more. The original velocities (as though they missed) form one. Since negating both velocities is a second solution, those are the two.

In other words:

Try doing the algebra to double check this.

 

[haruspex]

In other words:

Not really.
Doing the algebra would check that only those two solutions exist, but it would not validate the argument I responded to. My point was to refine that argument just to the point where it stood by itself, and further algebraic details were unnecessary.