Two-Dimensional Elastic Collision of Equal Masses

[flamespirit919]

Homework Statement

Show that if an elastic collision between a mass and a stationary target of equal mass is not head-on that the projectile and target final velocities are perpendicular. (Hint: Square the conservation of momentum equation, using $p^2=p\cdot p$, and compare the resulting equation with the energy conservation equation.)

Homework Equations

$p=mv$
$K=\frac{1}{2}mv^2$

The Attempt at a Solution

So I followed the hint and got $$\overrightarrow{p_1^2}=\overrightarrow{p_1^{'2}}+\overrightarrow{p_2^{'2}}+2\left(\overrightarrow{p_1}\cdot \overrightarrow{p_2}\right)$$ Plugging in values I got $$m^2\overrightarrow{v_1^2}=m^2\overrightarrow{v_1^{'2}}+m^2\overrightarrow{v_2^{'2}}+2\left(m\overrightarrow{v}_1\cdot m\overrightarrow{v}_2\right)$$ $$\overrightarrow{v_1^2}=\overrightarrow{v_1^{'2}}+\overrightarrow{v_2^{'2}}$$ For the x- and y-components I got the following assuming the target traveled along the x-axis after the collision $$v_1^2cos^2(\theta)=v_1^{'2}cos^2(\theta ^{'})+v_2^{'2}$$ $$v_1^2sin^2(\theta)=v_1^{'2}sin^2(\theta ^{'})$$ This is where I got stuck and wasn't sure how to solve for $v_1^{'}$ or $v_2^{'}$. I'm not entirely sure when or how to apply the equation for kinetic energy. Am I in the right direction or did I do something wrong?

 

[kuruman]

You did something wrong. The last two equations are dimensionally incorrect. Review their derivation.

 

[flamespirit919]

You did something wrong. The last two equations are dimensionally incorrect. Review their derivation.

I forgot to square the initial velocity. Were there any other mistakes?

 

[flamespirit919]

I think I got it.

So the momentum in the x and y-direction is $$v_1=v_1^{'}cos(\theta _1)+v_2^{'}cos(\theta _2)$$ $$0=v_1^{'}sin(\theta _1)-v_2^{'}sin(\theta _2)$$ I then used the following to find the magnitude of the momentum $$p^2=p_x^2+p_y^2$$ Doing this I got $$v_1=v_1^{'2}(cos^2(\theta _1)+sin^2(\theta _1))+v_2^{'2}(cos^2(\theta _2)+sin^2(\theta _2))+2v_1^{'}v_2^{'}(cos(\theta _1)cos(\theta _2)-sin(\theta _1)sin(\theta _2))$$ Using trig identities and the law of cosines, I simplified it to $$v_1^2=v_1^{'2}+v_2^{'2}+2v_1^{'}v_2^{'}cos(\theta _1+\theta _2)$$ Knowing that kinetic energy is conserved in an elastic collision I used the equation for kinetic energy and simplified $$\frac{1}{2}mv_1^2=\frac{1}{2}mv_1^{'2}+\frac{1}{2}mv_2^{'2}$$ $$v_1^2=v_1^{'2}+v_2^{'2}$$ With this formula, $v_1^2$ can be substituted for $v_1^{'2}+v_2^{'2}$ After substituting the equation simplifies to $$0=2v_1^{'}v_2^{'}cos(\theta _1+\theta _2)$$ Solving results in $$v_1^{'}=0$$ $$v_2^{'}=0$$ $$cos(\theta _1+\theta _2)=0$$ Solving for the last equation results in $$\theta _1+\theta _2 = 90°\text{ or }\frac{\pi}{2}$$ Thus, the two velocities must be perpendicular.

 

[haruspex]

Solving results in $$v_1^{'}=0$$ $$v_2^{'}=0$$ $$cos(\theta _1+\theta _2)=0$$

it results in at least one of those three. How do you rule out the first two?

 

[flamespirit919]

it results in at least one of those three. How do you rule out the first two?

The question asks to prove that the two velocities will be perpendicular, so I assumed the velocities to be non-zero because otherwise there would be no answer as the angle between the two velocities would not exist.

 

[haruspex]

The question asks to prove that the two velocities will be perpendicular, so I assumed the velocities to be non-zero because otherwise there would be no answer as the angle between the two velocities would not exist.

You are given that the collision is not head-on. From that you can show neither velocity is zero.

 

[flamespirit919]

You are given that the collision is not head-on. From that you can show neither velocity is zero.

That makes sense, thank you very much.

 

[kuruman]

Here is a quick geometric proof, not that your algebraic proof is incorrect.
Momentum conservation for a two equal-mass collision requires that $\vec{v}_1={\vec{v_1}'}+{\vec{v_2}'}$.
Since the collision is not one-dimensional, if you draw a vector addition diagram, you get a closed triangle with the magnitudes of the three vectors as its sides.
Energy conservation for a two equal-mass collision requires that $v_1^2= {v'_1}^2+{v'_1}^2$.
Therefore, said triangle obeys the Pythagorean theorem with $v_1$ as the hypotenuse and ${v'_1}$ and ${v'_2}$ as the two right sides.

 

[flamespirit919]

Here is a quick geometric proof, not that your algebraic proof is incorrect.
Momentum conservation for a two equal-mass collision requires that $\vec{v}_1={\vec{v_1}'}+{\vec{v_2}'}$.
Since the collision is not one-dimensional, if you draw a vector addition diagram, you get a closed triangle with the magnitudes of the three vectors as its sides.
Energy conservation for a two equal-mass collision requires that $v_1^2= {v'_1}^2+{v'_1}^2$.
Therefore, said triangle obeys the Pythagorean theorem with $v_1$ as the hypotenuse and ${v'_1}$ and ${v'_2}$ as the two right sides.

Thank you! I wish I had seen that sooner.