Solving Collision Problem: 10g & 15g Objects Moving in Opposite Directions

[the7joker7]

Homework Statement

1) A 10.0-g object moving to the right at 20 cm/s makes an elastic head-on collision with a 15.0-g object moving in the opposite direction at 30.0 cm/s. Find the velocity of each object after collision.

Homework Equations

m$$_{1}$$v$$_{i1}$$ + m$$_{2}$$v$$_{i2}$$ = m$$_{1}$$v$$_{f1}$$ + m$$_{2}$$v$$_{f2}$$

(.5)m$$_{1}$$v$$_{1i}$$$$^{2}$$ + .(5)m$$_{2}$$v$$_{2i}$$$$^{2}$$ = (.5)m$$_{1}$$v$$_{1f}$$$$^{2}$$ + .(5)m$$_{2}$$v$$_{2f}$$$$^{2}$$

The Attempt at a Solution

I plugged values into the equations and got...

(.01)(.2) + (.015)(-.3) = (.01)(v$$_{f1}$$) + (.015)(v$$_{f2}$$)

(.5)(.01)(.2^2) + (.5)(.015)(-.3^2) = (.5)(.01)(v$$_{f1}$$$$^{2}$$) + (.5)(.015)(v$$_{f2}$$$$^{2}$$)

But I'm still stuck with two variables. How do I solve one of them?

 

[hunter151]

What is conserved in an elastic collision? (momentum and...)

 

[the7joker7]

What is conserved in an elastic collision? (momentum and...)

Energy.

I was under the impression I had the equation for conservation of energy there (the 2nd one).

 

[hunter151]

Oh wow, I am an idiot. You are correct on the equations. Now you're faced with two equations and two unknowns; therefore, you can solve. One way of doing things is to solve one of the equations for one variable, then plug this result into the second equation, to be left with a single equation with one unknown variable.

For example, you could find Vf1 in terms of Vf2 from the first equation, and plug this Vf1 into the second equation. Now you are faced with an equation of just Vf2 as unknown.

 

[the7joker7]

Oh wow, I am an idiot. You are correct on the equations. Now you're faced with two equations and two unknowns; therefore, you can solve. One way of doing things is to solve one of the equations for one variable, then plug this result into the second equation, to be left with a single equation with one unknown variable.

For example, you could find Vf1 in terms of Vf2 from the first equation, and plug this Vf1 into the second equation. Now you are faced with an equation of just Vf2 as unknown.

Thanks.

I'm trying to do that now, and...

.0002 - .000675 = (.005)vf1^2 + (.0075)vf2^2

.005vf1^2 = -.000475 - .0075vf2^2

vf1^2 = .095 - 1.5vf2^2

vf1 = .3082 - 1.5vf2

-.0025 = (.01)(.3082 - 1.5vf2) + (.015)(vf2)

-.005582 = -(.015vf2) + (.015vf2)

Well, my vf2s cancel out :/

Did I make a math error?

 

[hunter151]

When you took the square root of Vf1, you forgot that there are two solutions, + and -. ;)