How to solve a simple elastic collision problem?

[Eclair_de_XII]

Homework Statement

"A particle with mass $m_1=1kg$ traveling at $v_{1_{0}}\frac{m}{s}$ collides with a stationary particle with mass $m_2=2kg$. How small is the speed of $m_1$ after the collision compared to before?"

Homework Equations

The Attempt at a Solution

$m_1v_{1_{0}}=m_1v_{1_{f}}+m_2v_{2_{f}}=m_1v_{1_{f}}+2m_1v_{2_{f}}=m_1(v_{1_{f}}+2v_{2_{f}})$
$v_{1_{0}}=v_{1_{f}}+2v_{2_{f}}$

Then this is where I kind of guess at things: basically, I equate the momentum of the first particle post-collision with the momentum of the second particle post-collision.

$m_1v_{1_{f}}-m_2v_{2_{f}}=0$

So now I have two equations:

$v_{1_{f}}-2v_{2_{f}}=0$
$v_{1_{f}}+2v_{2_{f}}=v_{1_{0}}$

Solving for the two variables gives me: $v_{1_{f}}=\frac{1}{2}v_{1_{0}}$ and $v_{2_{f}}=v_{1_{0}}$. I know this is wrong because suddenly the two particles have more momentum after the collision than what they started out with.

 

[ehild]

Homework Statement

"A particle with mass $m_1=1kg$ traveling at $v_{1_{0}}\frac{m}{s}$ collides with a stationary particle with mass $m_2=2kg$. How small is the speed of $m_1$ after the collision compared to before?"

Homework Equations

The Attempt at a Solution

$m_1v_{1_{0}}=m_1v_{1_{f}}+m_2v_{2_{f}}=m_1v_{1_{f}}+2m_1v_{2_{f}}=m_1(v_{1_{f}}+2v_{2_{f}})$
$v_{1_{0}}=v_{1_{f}}+2v_{2_{f}}$

Then this is where I kind of guess at things: basically, I equate the momentum of the first particle post-collision with the momentum of the second particle post-collision.

Do not guess. The collision is elastic, what does it mean for the kinetic energies of the particles?

 

[Eclair_de_XII]

It's conserved, meaning that: $\frac{1}{2}m_1v_{1_{0}}^2=\frac{1}{2}m_1v_{1_{f}}^2+\frac{1}{2}m_2v_{2_{f}}^2$, or more simply put: $m_1v_{1_{0}}^2=m_1v_{1_{f}}^2+m_2v_{2_{f}}^2$.

 

[ehild]

It's conserved, meaning that: $\frac{1}{2}m_1v_{1_{0}}^2=\frac{1}{2}m_1v_{1_{f}}^2+\frac{1}{2}m_2v_{2_{f}}^2$, or more simply put: $m_1v_{1_{0}}^2=m_1v_{1_{f}}^2+m_2v_{2_{f}}^2$.

Now you have two equation for the final velocities to solve.

 

[Eclair_de_XII]

Okay, so I have:

$m_1v_{1_{0}}^2=m_1v_{1_{f}}^2+m_2v_{2_{f}}^2$
$m_1v_{1_{0}}=m_1v_{1_{f}}+m_2v_{2_{f}}$

Multiplying the second equation by $v_{2_{f}}$ yields: $m_1v_{1_{0}}v_{2_{f}}=m_1v_{1_{f}}v_{2_{f}}+m_2v_{2_{f}}^2$. Subtracting it from the first equation then gives me:

$m_1v_{1_{0}}^2-m_1v_{1_{0}}v_{2_{f}}=m_1v_{1_{f}}^2-m_1v_{1_{f}}v_{2_{f}}$
$v_{1_{0}}^2-v_{1_{0}}v_{2_{f}}=v_{1_{f}}^2-v_{1_{f}}v_{2_{f}}$
$v_{1_{0}}^2-(v_{1_{0}}+v_{1_{f}})v_{2_{f}}-v_{1_{f}}^2=0$

Sorry, I don't really know how to proceed from there.

 

[Dirickby]

Starting from your two equations, try to eliminate one of the two unknowns (here, which variable is not asked for in the question?) so that you can solve the resulting equation for the other unknown.

 

[Eclair_de_XII]

try to eliminate one of the two unknowns (here, which variable is not asked for in the question?) so that you can solve the resulting equation for the other unknown.

So I'm guessing I express the other unknown as $v_{2_{f}}=\frac{m_1}{m_2}(v_{1_{0}}-v_{1_{f}})$? So...

$m_1v_{1_{0}}^2=m_1v_{1_{f}}^2+m_2(\frac{m_1}{m_2}(v_{1_{0}}-v_{1_{f}}))^2=m_1v_{1_{f}}^2+m_2(\frac{m_1^2}{m_2^2})(v_{1_{0}}^2-2v_{1_{f}}v_{1_{0}}+v_{1_{f}}^2))$
$v_{1_{0}}^2=v_{1_{f}}^2+(\frac{m_1}{m_2})(v_{1_{0}}^2-2v_{1_{f}}v_{1_{0}}+v_{1_{f}}^2))$
$v_{1_{0}}^2(1-\frac{m_1}{m_2})-v_{1_{f}}^2(1+\frac{m_1}{m_2})+(\frac{m_1}{m_2})2v_{1_{f}}v_{1_{0}}=0$

To be honest, I didn't feel like going through the quadratic equation to derive the momentum equation I was reluctant to use because I didn't understand it. But in the end, I just looked up the derivation and called it a day: http://www.batesville.k12.in.us/physics/APPhyNet/Dynamics/Collisions/elastic_deriv.htm

$v_{1_{f}}=\frac{m_1-m_2}{m_1+m_2}v_{1_{0}}=\frac{1kg-2kg}{3kg}v_{1_{0}}=-\frac{1}{3}v_{1_{0}}$

 

[Dirickby]

So I'm guessing I express the other unknown as $v_{2_{f}}=\frac{m_1}{m_2}(v_{1_{0}}-v_{1_{f}})$? So...

$m_1v_{1_{0}}^2=m_1v_{1_{f}}^2+m_2(\frac{m_1}{m_2}(v_{1_{0}}-v_{1_{f}}))^2=m_1v_{1_{f}}^2+m_2(\frac{m_1^2}{m_2^2})(v_{1_{0}}^2-2v_{1_{f}}v_{1_{0}}+v_{1_{f}}^2))$
$v_{1_{0}}^2=v_{1_{f}}^2+(\frac{m_1}{m_2})(v_{1_{0}}^2-2v_{1_{f}}v_{1_{0}}+v_{1_{f}}^2))$
$v_{1_{0}}^2(1-\frac{m_1}{m_2})-v_{1_{f}}^2(1+\frac{m_1}{m_2})+(\frac{m_1}{m_2})2v_{1_{f}}v_{1_{0}}=0$

To be honest, I didn't feel like going through the quadratic equation to derive the momentum equation I was reluctant to use because I didn't understand it. But in the end, I just looked up the derivation and called it a day: http://www.batesville.k12.in.us/physics/APPhyNet/Dynamics/Collisions/elastic_deriv.htm

$v_{1_{f}}=\frac{m_1-m_2}{m_1+m_2}v_{1_{0}}=\frac{1kg-2kg}{3kg}v_{1_{0}}=-\frac{1}{3}v_{1_{0}}$

Yes, the solution is right and you could indeed get it from solving the quadratic equation for $v_{1_f}$. Another (quicker) way would be to take your second line equation
$$ m_1(v_{1_0}^2-v_{1_f}^2) = \frac{m_1^2}{m_2}(v_{1_0}-v_{1_f})^2$$
One solution is obviously $v_{1_0}=v_{1_f}$ but this would be the answer only if there had been no collision.
So we can divide by $m_1(v_{1_0}-v_{1_f})$ on both sides and quickly get the desired answer by using $x^2-y^2 = (x-y)(x+y)$.

P.S. I hope I'm not breaking any rules here, please let me know if I am. My idea was not to give the answer but to show a quicker derivation as the OP had already called it a day.

 

[ehild]

So I'm guessing I express the other unknown as $v_{2_{f}}=\frac{m_1}{m_2}(v_{1_{0}}-v_{1_{f}})$? So...

$m_1v_{1_{0}}^2=m_1v_{1_{f}}^2+m_2(\frac{m_1}{m_2}(v_{1_{0}}-v_{1_{f}}))^2=m_1v_{1_{f}}^2+m_2(\frac{m_1^2}{m_2^2})(v_{1_{0}}^2-2v_{1_{f}}v_{1_{0}}+v_{1_{f}}^2))$
$v_{1_{0}}^2=v_{1_{f}}^2+(\frac{m_1}{m_2})(v_{1_{0}}^2-2v_{1_{f}}v_{1_{0}}+v_{1_{f}}^2))$
$v_{1_{0}}^2(1-\frac{m_1}{m_2})-v_{1_{f}}^2(1+\frac{m_1}{m_2})+(\frac{m_1}{m_2})2v_{1_{f}}v_{1_{0}}=0$

To be honest, I didn't feel like going through the quadratic equation to derive the momentum equation I was reluctant to use because I didn't understand it. But in the end, I just looked up the derivation and called it a day: http://www.batesville.k12.in.us/physics/APPhyNet/Dynamics/Collisions/elastic_deriv.htm

It is not one day but 1D, one dimension ...
If you substitute m1/m2=0.5, that quadratic equation is not that terrible.