# Confusion with non-head-on elastic collisions

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1. Dec 2, 2014

### Jazz

1. The problem statement, all variables and given/known data

Two pucks of equal masses collide in a non-head-on collision. Puck 1 with a velocitiy $u_1 = 2.50\ m/s$ impacts puck 2 at rest, in a 28-degree angle. Assuming no friction between the pucks, find their final velocities.

Data:

$m_1 = 5.00\ kg$

$u_1 = 2.50\ m/s$

$\theta_i = 28.0º$

$m_2 = 8.00\ kg$

$u_2 = 0$

$v_1 = \ ?$

$v_2 = \ ?$

$\theta_f =\ ?$

2. Relevant equations

$\Delta p_1 + \Delta p_2 = constant$

$\Delta KE_1 + \Delta KE_2 = constant$

$\theta_f = \arcsin\left(\frac{u_1sin(\theta_i)}{v_1}\right)\hspace{35pt}(1)$

3. The attempt at a solution

This problem may seem similar to the one I posted before, but I think now I've been able to summarize my thoughts about what I'm not understanding. I would like to analyze two situations while solving the problem I've written above.

The system I’ve been trying to analyze in different cases (changing masses, angles and velocities) is one like this:

I’m having trouble trying to interpret a non-head-on elastic collision of two objects ($m_1 = m_2$ and $m_1 \neq m_2$), either when one is at rest or they’re approaching one another. I’m able to find their final velocities and comprobe that Momentum and KE are conserved, but there is something I still don’t understand.

**First case: $m_2 < m_1$ (the current problem):

According to my predictions about what happens after the collision, $m_2$ will remove a little of $u_1\cos(\theta)$. Since $u_1\sin(\theta)$ is not affected by the impact, it remains the same. Then, $v_1 < u_1$ and it will also have a stepper direction from the x-axis compare to that of $u_1$.

After finding the velocities, $v_1$ would be $-0.58\ m/s$, smaller than the magnitude of $u_1$ as predicted, but I’m not able to calculate its direction since the numerator in equation $(1)$ is greater than the denominator.

I interpret the negative sign as moving in diagonal pointing in this direction: $\nwarrow$. Does it mean my prediction of the direction was wrong?

The same happen if I consider $m_2$:

• Moving toward the origin of my coordinate system; unless its speed is large enough as to make $m_1$ move in an angle of $90º$ from the x-axis after the collision.
• Being at rest; unless it has a mass large enough as to make $m_1$ move in a $90º$ as well.

I have no idea about what I’m doing wrong there.

**Second Case: $m_2 = m_1$

According to my understanding, in a head-on collision of two objects of equal masses, the change is momentum can be viewed as an interchange of velocities. If one of them is at rest, then after the collision $m_1$ is brought to rest and the other one ends up with a velocity $v_2 = u_1$.

I tried to use this approach with a non-head-on collision of equal masses:

By setting the x-axis along the line of impact, the horizontal component of $u_1$ is ‘’given’’ to $m_2$ and the vertical component doesn’t change, ‘’remaining’’ with $m_1$. Then $u_1\cos(\theta)$ becomes $v_2$ and $u_1\sin(\theta)$ becomes $v_1$.

But by the CoM equation I notice that this can’t be right, because $m_1v_1 + m_2v_2$ will not be equal to the initial momentum of the system, unless I treat the final momentum of each ball as the sides of a right triangle and try to solve it by the Pythagorean Theorem.

On the other hand if I try to apply Newton’s Experimental Law (NEL), it will give me $v_2 = u_1$, which means that the incoming ball was left at rest. But this can’t be right either, since there is neither velocity along the line of impact with that magnitude nor force removing $u_1\sin(\theta)$.

By doing the math, it seems that NEL will always treat a collision of equal masses as a head-on one, either $m_2$ being at rest or approaching $m_1$, since:

$m_1 = m_2 = m;$

$v_2= \frac{2m_1u_1+u_2(m_2-m_1)}{m_1+m_2} = \frac{2mu_1+u_2(0)}{2m} = \frac{2mu_1}{2m} = u_1$

And I’m really puzzled here :). The fact that having to calculate the CoM of the system as a vector addition by the Pythagorean Theorem (PT) confuses me a little bit, because in previous examples with unequal masses, I’ve just needed to add the final momentum of each object without worrying about their final directions (as long as the calculation was correct and taking into account their signs though).

Mathematically it makes perfect sense (doing that of the PT), but it seems odd having to do it instead of ''admiting'' that in a non-head-on collision of objects with equal masses, the angle between particles of equal mass will be $90º$ and the momentum will not be conserved.

Thanks !!

2. Dec 2, 2014

### haruspex

This is fully elastic, right?
If m2 < m1 then m1 should continue in the positive x direction. If equal it should move off in the y direction only. If m2 > m1 then m1 will recoil.
How else would you add two vectors at right angles?

3. Dec 3, 2014

### Jazz

Well, my problem was to be dealing with two pucks of equal masses, but in the data $m_2 > m_1$ :), and so the first case should have been $m_2 > m_1$ with $m_1$ scattering away from the wall formed by the y-axis.

Yep, fully elastic. I got it :).

I've noticed that my mistake all the time was having treated momentum as a conserved scalar quantity and never as a vector one. By formulas on paper I found it to be conserved but never worry about how the final momentum ''fits'' with the initial one in a vector form. That's why the issue of the angles never made sense to me. I see that my proposal of a non-conserved momentum for a non-head collision of equal masses doesn't make sense either (for a while I thought I'd discovered something new).

I always drew diagrams for velocities but never one for momentum:

It's not hard to see that this can be solved in a couples of ways using trig formulas.

The last time I found $v'_2$ by NEL, and then use the Law of Cosine to find $p'_1$ which turns out to be $\small{7.30\ kg \cdot m/s}$. This corresponds to a velocity of $\frac{p'_1}{m_1} = \frac{7.30\ kg \cdot m/s}{5.00\ kg} = \small{1.46\ m/s}$. I can find $\beta$ without any problem in this way. On the other hand, if I try to solve for $v_1'$ by NEL instead, it's found to be $\small{-0.58\ m/s}$ (which was the velocity that does not allow me to work out the angle $\beta$).

Where is this difference coming from?

4. Dec 3, 2014

### haruspex

You seem to have changed or added notation. What is $v_1'$? Is that $v_1\cos(\beta)$?
In equation 1, the denominator is the magnitude of the total velocity, not just the x component. Is that your problem?
By the way, I get -0.51 m/s for the x-component of m1 after collision. Please check that your numbers do result in momentum and energy being conserved.

5. Dec 4, 2014

### Jazz

Sorry. $u$'s for initial velocities and just $v$'s (without the prime symbol) for final ones.

Yep. I've checked them and aren't conserved /:. Fortunately, I realized where I started to mess things up.

In Newton's Experimental Law, I was considering the four velocities as the total ones. I was carrying out calculations as if it were a head-on collision, when I knew by the diagram that that was not the case.

Because of that, I get a wrong magnitude $(-0.58\ m/s)$ and I was treating it as the overall $v_1$, when it should be $v_1\cos(\beta)$. The answer I got now for the x component of $v_1$ $(-0.509\ m/s)$ agrees with that you've indicated. Now the negative sign does make sense, since it goes to the negative x direction of my coordinate system.

The Law of Cosine is not necessary.

So finally I got the following values, with which Momentum and KE are conserved:

$p'_1 = 6.397\ kg \cdot m/s$

$v_1 = 1.279\ m/s$

$\beta = 66.54º$ (North of West).

$p'_2 = 13.584\ kg \cdot m/s$

$v_2 = 1.698\ m/s$

$KE_i = 15.625\ J$

$KE_f = 15.622\ J$ (it should be due to rounding).

It seems they're correct.

Last edited: Dec 4, 2014