How Do Elastic Collisions Affect the Velocities and Angles of Moving Masses?

[blixel]

Homework Statement

Mass 1 has a mass of 5kg and is traveling with an initial velocity of 20 m/s at a 45 degree angle. (Starting in Quadrant II and heading toward the origin.)

Mass 2 has a mass of 6kg and is traveling with an initial velocity of 15 m/s at a 45 degree angle. (Starting in Quadrant III and heading toward the origin.)

The final velocity of mass 2 is known and it is 25 m/s after the collision.

The collision is completely elastic.

I need to find the velocity of mass 1 after collision, and the final angles of both masses.

Homework Equations

I don't know all of the relevant equations. But I think I need to use:
∑mv_x(before)=∑mv_x(after)
∑mv_y(before)=∑mv_y(after)

The Attempt at a Solution

First I broke the initial velocity of mass 1 down into its x and y components. Since the initial angle is 45°, the x and y velocities are the same.
(20m/s)*cos(45°)≈14.14214

Then I broke the initial velocity of mass 2 down into its x and y components.
(15m/s)*cos(45°)≈10.60660

Next, I thought that since I knew the final velocity of mass 2, I should be able to calculate its angle using:
∑mv_x(before)=∑mv_x(after)
[(15m/s)*cos(45°)]/(25m/s)=(3√2)/2

Then cos^-1([3√2)/2])≈64.896°

I'm not sure if these steps are correct up to this point ... but I went on to try to calculate the final velocity of mass 1 using:
m1v1^2+m2v2^2=m1v'1^2+m2v'2^2

With that equation I calculated that the final velocity of mass 1 is ≈-3.1299m/s

I'm extremely confused on this problem and feel like everything I'm doing is wrong.

 

[TomHart]

Are you sure all of the values in the problem statement are correct? Maybe I messed up on the calculations, but it looks like the final kinetic energy of m2 is greater than the initial kinetic energy of m1 + m2.

 

[blixel]

Are you sure all of the values in the problem statement are correct? Maybe I messed up on the calculations, but it looks like the final kinetic energy of m2 is greater than the initial kinetic energy of m1 + m2.

Well, I copied the problem down as it was written on the board. I even took a picture of the chalkboard to make sure I didn't make any transcription errors. This problem was not out of the book ... it was made up as a take-home practice problem.

 

[TomHart]

Yeah, the more I think about it, the more I am convinced there is a problem with the problem. I just don't think it's possible for m2's speed to increase from 15 m/s to 25 m/s due to a collision of a lesser mass object traveling at 20 m/s.

 

[blixel]

Yeah, the more I think about it, the more I am convinced there is a problem with the problem. I just don't think it's possible for m2's speed to increase from 15 m/s to 25 m/s due to a collision of a lesser mass object traveling at 20 m/s.

Well it wouldn't surprise me ... but I still need to work the problem as much as I can in the mean time. Are you able to determine if the work I did is valid?

 

[TomHart]

∑mvx(before)=∑mvx(after)
[(15m/s)*cos(45°)]/(25m/s)=(3√2)/2

This did not make sense to me. I don't really understand what you are doing. I agree that the sum of momentums in the x direction before = sum after, but the equation that you wrote to implement that didn't look right at all.

Here is how I think it should look.
∑mvx(before)=∑mvx(after)
m₁v_1xinitial + m₂v_2xinitial = m₁v_1xfinal + m₂v_2xfinal

Then you should write a similar equation for the y direction.
Then there should be a before/after equation for kinetic energy.

But I don't think it's a good idea to try to work this problem. There is not going to be a valid solution.

 

[TomHart]

[(15m/s)*cos(45°)]/(25m/s)=(3√2)/2

Looking at this equation a little more, it appears that you are trying to equate m2's initial momentum to m2's final momentum. That is not valid. It is the sum of the momentums before that are equal to the sum of the momentums after. So you have to add all of the x-direction momentums. In this case, that is m1 and m2.

 

[blixel]

Here is how I think it should look.
∑mvx(before)=∑mvx(after)
m₁v_1xinitial + m₂v_2xinitial = m₁v_1xfinal + m₂v_2xfinal

Then you should write a similar equation for the y direction.
Then there should be a before/after equation for kinetic energy.

But I don't think it's a good idea to try to work this problem. There is not going to be a valid solution.

So basically I can't go forward because in order to calculate m₁v_1xinitial + m₂v_2xinitial = m₁v_1xfinal + m₂v_2xfinal, I need the angle of θ₂, and the angle I calculated earlier isn't valid because the value of velocity 2 (after) is invalid to begin with?

If the problem is indeed invalid by design, that is insanely confusing to someone who has practically no understanding of what's going on in the first place.

 

[TomHart]

Even for a valid problem, your method of calculating the angle was not valid because it was based on equating m2's initial x momentum with m2's final x momentum. If you are given the initial velocities of both objects and final velocity of one of the objects, you can find the final velocity of the other object based on conservation of energy. (This only applies to a perfectly elastic collision though.) Then you will have to use conservation of momentum equations to find the angles.

 

[blixel]

This did not make sense to me. I don't really understand what you are doing. I agree that the sum of momentums in the x direction before = sum after, but the equation that you wrote to implement that didn't look right at all.

Here is how I think it should look.
∑mvx(before)=∑mvx(after)
m₁v_1xinitial + m₂v_2xinitial = m₁v_1xfinal + m₂v_2xfinal

Then you should write a similar equation for the y direction.
Then there should be a before/after equation for kinetic energy.

But I don't think it's a good idea to try to work this problem. There is not going to be a valid solution.

Well, I emailed the professor of this class and to my surprise, he replied right away. Anyway he said to use 20 m/s as the final velocity for mass 2 instead of 25 m/s.

So with that new information ... I now have...

v_f1=sqrt[((5kg)(20m/s)²+(6kg)(15m/s)²-(6kg)(20m/s)²)/5kg]=sqrt(190)≈13.784m/s ... (ignoring sig figs for now)

Does that seem valid?

 

[TomHart]

That looks right for vf1.

 

[blixel]

That looks right for vf1.

Great ... so to find the angles, would I break vf1 down into its x and y component and then use cosθ to find the missing angle?

 

[haruspex]

Great ... so to find the angles, would I break vf1 down into its x and y component and then use cosθ to find the missing angle?

As Tom posted, you need to write out the two momentum conservation equations (x and y). Both will involve both unknown angles. Then solve the pair of simultaneous equations.

 

[blixel]

As Tom posted, you need to write out the two momentum conservation equations (x and y). Both will involve both unknown angles. Then solve the pair of simultaneous equations.

For x, I'm getting that the total momentum before is (100kgm/s)*cos(45)+(90kgm/s)*cos(45)≈134.3503, that seems reasonable. (I think.)

But for y, I'm getting what seems like nonsense. (100kgm/s)*sin(-45)+(90kgm/s)*sin(45)≈-7.0711

Does it make sense that the y momentum (before) would be negative?

 

[TomHart]

Those numbers look familiar. I think they are right. And yes, it makes sense that the y momentum could be negative.

 

[blixel]

Those numbers look familiar. I think they are right. And yes, it makes sense that the y momentum could be negative.

Ok ... thanks for clearing that up, but I still have a problem with too many unknowns. (I don't know the equation editor here well enough to make this clear, so please pardon the photo equation.)

If I assume that my √190m/s (or ≈13.78m/s) velocity for mass 1 after is valid (from post #10), then I still can't solve the angles.

 

[TomHart]

Back in post #10 you solved for Vf1, so that is a known. If you look at your two equations in the form they were just before you solved them for Vf1:
On the right hand side of the x momentum equation you have 5v_f1cosθ₁
And on the right hand side of the y momentum equation you have 5v_f1sinθ₁
If you square both of those equations you can then add them together and use the trig identity sin² + cos² = 1
That will eliminate one of the variables.

BE CAREFUL of your signs though. The picture shows the final velocity of m2 in the downward direction. Is that how you are defining it? That seems to be the most straightforward.

 

[blixel]

Back in post #10 you solved for Vf1, so that is a known. If you look at your two equations in the form they were just before you solved them for Vf1:
On the right hand side of the x momentum equation you have 5v_f1cosθ₁
And on the right hand side of the y momentum equation you have 5v_f1sinθ₁
If you square both of those equations you can then add them together and use the trig identity sin² + cos² = 1
That will eliminate one of the variables.

BE CAREFUL of your signs though. The picture shows the final velocity of m2 in the downward direction. Is that how you are defining it? That seems to be the most straightforward.

Ok, I'm trying to wrap my head around this ... below is what I've done based on this new information. But I'm not seeing how this eliminates a variable. I now just have an insanely complicated equation with two unknown angles. (I'm sorry to be so dumb at this. This is my first experience with physics.)

 

[TomHart]

I did not go through your whole equations, but in your final equation (above), it looks to me like you eliminated one of the variables - namely, θ₂.
So you are left with a single equation with a single variable (θ₁). Once you manipulate all of that, I think you should end up with an equation in the form:
asinx + bcosx = c.

http://mathforum.org/kb/servlet/JiveServlet/download/206-2697091-9755226-1202710/sin-cos-eqs.pdf
Example IV in the link above shows how to address that.

Check your sign in your initial y momentum equation. If you are defining the velocity of m2 in the downward direction, that term should be negative. (Edit: Replaced the word 'expression' with the word 'term'.)

Also, do not feel dumb. Not that I am one of the experts on this forum (far from it), but I struggled quite a bit with this problem. The concept is very straightforward (equating before and after x momentum, y momentum, and kinetic energy), but the math gets quite tricky - at least for a hack like me. :)

 

[blixel]

I did not go through your whole equations, but in your final equation (above), it looks to me like you eliminated one of the variables - namely, θ₂.
So you are left with a single equation with a single variable (θ₁). Once you manipulate all of that, I think you should end up with an equation in the form:
asinx + bcosx = c.

http://mathforum.org/kb/servlet/JiveServlet/download/206-2697091-9755226-1202710/sin-cos-eqs.pdf
Example IV in the link above shows how to address that.

Check your sign in your initial y momentum equation. If you are defining the velocity of m2 in the downward direction, that term should be negative. (Edit: Replaced the word 'expression' with the word 'term'.)

Also, do not feel dumb. Not that I am one of the experts on this forum (far from it), but I struggled quite a bit with this problem. The concept is very straightforward (equating before and after x momentum, y momentum, and kinetic energy), but the math gets quite tricky - at least for a hack like me. :)

First I surely appreciate the patience and help. I will look at that link here in a moment, but before I go forward I want to make sure I am addressing the sign for m2 that you are talking about. I am assuming that when the masses collide, they will bounce off each other so mass 2 will be going in the opposite y direction from where it began. So I believe this is how the equation should look? (Now I'll continue on and get that PDF)

 

[TomHart]

I am assuming that when the masses collide, they will bounce off each other so mass 2 will be going in the opposite y direction from where it began.

I agree with that. But I also assume that both m1 and m2 will continue in the +x direction after the collision, don't you? If so, then why did you make m2's final x momentum term negative? (The very first equation underneath the word "After" in your picture.)

 

[blixel]

I agree with that. But I also assume that both m1 and m2 will continue in the +x direction after the collision, don't you? If so, then why did you make m2's final x momentum term negative? (The very first equation underneath the word "After" in your picture.)

Yes I assume that also. Setting that first equation's final x momentum negative was just a blunder out of ignorance. (I'm barely understanding this problem.) But I think I'm at a roadblock because that PDF is over my head. Turning this equation into something that can be solved for sin(theta) or cos(theta) is beyond me. I know the professor assigned this based on the assumption that it should only be moderately challenging. So I can't understand why this is turning out to be so complicated.

 

[TomHart]

You're actually not that far away. First of all, when you write those equations, eliminate all of the units (kg, m/s, etc.).
Then multiply the whole equation by the 120 on the bottom. Then multiply out the squares inside of the parentheses on the left side. If you notice, when you multiply those out, 2 of the terms will be (25)(190)cos²θ₁ and (25)(190)sin²θ₁. Those reduce to (25)(190) because of the sin² + cos² = 1 trig identity. So you should be left with an equation of the form acosx + bsinx = c.

It seems that these completely elastic collision problems are more complicated mathematically. If there is an easier way to solve this problem, I don't know what it is.

By the way, I still don't even think this problem is possible in the real world. How can a lighter object with a speed delta of +5 m/s (relative to the heavier object) increase the velocity of the heavier object by 5 m/s?

 

[blixel]

You're actually not that far away. First of all, when you write those equations, eliminate all of the units (kg, m/s, etc.).
Then multiply the whole equation by the 120 on the bottom. Then multiply out the squares inside of the parentheses on the left side. If you notice, when you multiply those out, 2 of the terms will be (25)(190)cos²θ₁ and (25)(190)sin²θ₁. Those reduce to (25)(190) because of the sin² + cos² = 1 trig identity. So you should be left with an equation of the form acosx + bsinx = c.

It seems that these completely elastic collision problems are more complicated mathematically. If there is an easier way to solve this problem, I don't know what it is.

By the way, I still don't even think this problem is possible in the real world. How can a lighter object with a speed delta of +5 m/s (relative to the heavier object) increase the velocity of the heavier object by 5 m/s?

I think I understand what you mean about the problem probably being unrealistic. The professor didn't think it through before making it up ... as was evident by the earlier problem where I was getting an imaginary number since the velocity for mass 2 (after) was so high that the equation was broken by design.

At any rate ... trying to follow what you're talking about at this step... I seem to be misunderstanding something...

 

[haruspex]

I think I understand what you mean about the problem probably being unrealistic. The professor didn't think it through before making it up ... as was evident by the earlier problem where I was getting an imaginary number since the velocity for mass 2 (after) was so high that the equation was broken by design.

At any rate ... trying to follow what you're talking about at this step... I seem to be misunderstanding something...

You should get into the habit of solving problems entirely symbolically as far as possible. Use variable names instead of the given numbers. It has many advantages.
Having found the second exit speed, you have two momentum equations. Writing just p for each mv product:
$p_{0x}=p_1\cos(\theta_1)+p_2\cos(\theta_2)$
$0=p_{0y}=p_1\sin(\theta_1)-p_2\sin(\theta_2)$
We need to come to one equation with only one angle in it. To do that, write each in the form $p_2f(\theta_2)= ...$. Squaring and adding eliminates θ₂ (and p₂).
See if you can figure out what to do from there.

 

[haruspex]

It seems that these completely elastic collision problems are more complicated mathematically

What makes it complicated is having two unknown angles each featuring via two different trig functions. That results in nonlinear simultaneous equations.

 

[blixel]

You should get into the habit of solving problems entirely symbolically as far as possible. Use variable names instead of the given numbers. It has many advantages.
Having found the second exit speed, you have two momentum equations. Writing just p for each mv product:
$p_{0x}=p_1\cos(\theta_1)+p_2\cos(\theta_2)$

Just to be sure I'm reading the notation correctly, does this mean initial x momentum is equal to final cosine (or x) momentum of object 1 + final cosine (or x) momentum of object 2?

$0=p_{0y}=p_1\sin(\theta_1)-p_2\sin(\theta_2)$

We need to come to one equation with only one angle in it. To do that, write each in the form $p_2f(\theta_2)= ...$. Squaring and adding eliminates θ₂ (and p₂).
See if you can figure out what to do from there.

What is the f in $p_2f(\theta_2)= ...$?

 

[haruspex]

Just to be sure I'm reading the notation correctly, does this mean initial x momentum is equal to final cosine (or x) momentum of object 1 + final cosine (or x) momentum of object 2?

Yes.

What is the f in p2f(θ2)=.. ...?

Some trig function, sin or cos.

 

[Ray Vickson]

You're actually not that far away. First of all, when you write those equations, eliminate all of the units (kg, m/s, etc.).
Then multiply the whole equation by the 120 on the bottom. Then multiply out the squares inside of the parentheses on the left side. If you notice, when you multiply those out, 2 of the terms will be (25)(190)cos²θ₁ and (25)(190)sin²θ₁. Those reduce to (25)(190) because of the sin² + cos² = 1 trig identity. So you should be left with an equation of the form acosx + bsinx = c.

It seems that these completely elastic collision problems are more complicated mathematically. If there is an easier way to solve this problem, I don't know what it is.

By the way, I still don't even think this problem is possible in the real world. How can a lighter object with a speed delta of +5 m/s (relative to the heavier object) increase the velocity of the heavier object by 5 m/s?

By far the easiest way to deal with such collision problems is to first transform them into the so-called center of momentum (CM) frame, look at the final velocities in that frame, and then transform those back into the original (lab) frame. I will go through most of the steps; I don't think this would volate PF policy, since you have already solved the problem.

Denote the initial lab-frame velocities of particles 1 and 2 by $\vec{v}_{i1}$ and $\vec{v}_{i2}$ and the initial lab-frame momenta by $\vec{p}_{i1}= m_1 \vec{v}_{i1}$ and $\vec{p}_{i2} = m_2 \vec{v}_{i2}$. The total initial lab-frame momentum is $\vec{p} = \vec{p}_{i1} + \vec{p}_{i2}$. Of course, this is also the total final lab-frame momentum, because of momentum conservation.

The CM frame is a new coordinate system moving at a constant velocity $\vec{V}$, so that in the CM frame the initial velocities are $\vec{u}_{i1}= \vec{v}_{i1} - \vec{V}$ and $\vec{u}_{i2} = \vec{v}_{i2} - \vec{V}$. (This just uses the standard kinematics of transforming velocities between frames in constant relative motion.) The CM frame is the one in which total momentum = 0. That means that $\vec{p} - (m_1+m_2) \vec{V} = \vec{0}$, or
$$\vec{V} = \frac{m_1 \vec{v}_{i1} + m_2 \vec{v}_{i2} }{m_1+m_2}. $$

Now comes the clincher: in the CM frame, the individual particle speeds remain the same before and after a perfectly elastic collision. (This follows from conservation of total kinetic energy.) So, in the CM frame, the final speed of particle 2 is the same as it was initially, before the collision. Also, in the CM frame the two particles are heading in exactly opposite directions. After the collision, the deflection angle can be anything from $0^o$ (for a glancing collision in which the particles just barely touch) to $180^o$ (for a head-on collision, in which each particle bounces back). If we let $\omega$ be the initial angle and $\theta$ be the deflection between $\vec{u}_{i2}$ and $\vec{u}_{f2}$ we have $\vec{u}_{f2} = |\vec{u}_{i2}| (-\cos(\omega + \theta), \sin(\omega +\theta))$. (Here we choose the direction vector to agree with the initial velocity when the deflection angle is zero, so the final angle is the same as the initial angle.)

By the way, the deflection angle is limited in the lab frame even though it is unlimited in the CM frame.

Now we can transform that back into the lab frame to get the final lab-frame velocity of particle as $\vec{v}_{f2} = \vec{u}_{f2} + \vec{V}$, and we can get the final lab-frame speed as $|\vec{v}_{f2}|$.

For your data we have $\vec{v}_{i1} = (20/\sqrt{2},-20/\sqrt{2})$, $\vec{v}_{i2} = (15/\sqrt{2},15/\sqrt{2})$. The total momentum is $\vec{p} = (95 \sqrt{2}, -5 \sqrt{2})$, and $\vec{V} = \vec{p}/(m_1+m_2) = (95\sqrt{2}/11, -5\sqrt{2}/11)$. We have
$\vec{u}_{i1} = \vec{v}_{i1}- \vec{V} = (15 \sqrt{2}/11,-105 \sqrt{2}/11)$ and $\vec{u}_{i2} = \vec{v}_{i2} - \vec{V} = (-25 \sqrt{2}/11,175 \sqrt{2}/11).$
The initial speed of particle 2 in the CM frame is $S_{CM,2} = |\vec{u}_{i2}| = 125/11$, so $\vec{u}_{f2} = (125/11)(-\cos(\omega), \sin(\omega))$. The final lab-frame velocity of particle 2 is $\vec{v}_{f2} = \vec{u}_{f2} + \vec{V}$.

We can determine $\omega$ from the direction of $\vec{u}_{i2}$: $\omega = \arctan(7)$. Substituting that into the above and expanding $\cos(\omega + \theta)$ and $\sin(\omega + \theta)$ we get the final lab-speed of particle 2 as
$$S_{LAB,2} = (5/11)\sqrt{1349 - 260 \cos(\theta) + 1320 \sin(\theta)}.$$

This has a maximum value of about 23.59 m/s, when the deflection angle in the CM frame is about $\theta = 101.143^o$.

So, indeed, in the lab frame the heavier particle can be going faster after the collision; on the other hand, the lighter particle is going much slower after the collision. You can work the details for particle 1.

So: believe it or not, this way of working things out is about as simple as it gets! With a bit of practice, it becomes almost second-nature, and it removes a lot of the uncertainty about how to deal with the effects of deflection angles and the like.