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

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In the discussion about solving the collision problem involving a 10g and a 15g object, participants focus on applying the conservation of momentum and kinetic energy equations to find the final velocities after an elastic collision. The equations used include momentum conservation and kinetic energy conservation, leading to two equations with two unknowns. A suggested approach is to express one final velocity in terms of the other and substitute it into the second equation to simplify the problem. Participants emphasize the importance of considering both positive and negative solutions when taking square roots in the final calculations. The conversation highlights common pitfalls in algebraic manipulation and the need for careful attention to detail in solving physics problems.
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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?
 
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What is conserved in an elastic collision? (momentum and...)
 
hunter151 said:
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).
 
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.
 
hunter151 said:
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?
 
When you took the square root of Vf1, you forgot that there are two solutions, + and -. ;)
 
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