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Q: A $0.150\text{ kg}$ glider is moving to the right ($+x$) on a frictionless, horizontal air track with a speed of $0.80\text{ m/s}$. It has an elastic collision with a $0.300\text{ kg}$ glider moving to the left ($-x$) with a speed of $2.20\text{ m/s}$.

a.) What is the initial momentum of each glider? Express the momentum in terms of unit vectors.

b.) Use the relative velocity formula to find $v_{2f}$ in terms of $v_{1f}$.

c.) Use the relative velocity result to solve conservation of momentum to find the velocity (magnitude and direction) of each glider after the collision.

I've figured part (a) using the definition of momentum: $p=mv$:

1st glider: $$p_1 = m_1 v_1 = (0.15\text{ kg})(+0.8\hat{x}\text{ m/s}) = +0.12\hat{x}\text{ kg m/s}$$

2nd glider: $$p_2 = m_2 v_2 = (0.3\text{ kg})(-2.20\hat{x}\text{ m/s}) = -0.66\hat{x}\text{ kg m/s}$$

Parts (b) and (c) are what have me confused at the moment. I'm not positive I have the equations for relative velocity right nor how to solve for $v_{2f}$ in terms of $v_{1f}$. My book lists this an equation that can be gotten from manipulation of a kinetic energy equation: $v_{1i} – v_{2i} = -(v_{1f} - v_{2f})$. Is this the relative velocity formula? would just isolating $v_{2f}$ in this equation be solving for $v_{2f}$ in terms of $v_{1f}$?

Any help appreciated. thanks!

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No problem :-) I wanted to add this to our list of good homework questions on meta so I figured it would help to make it look pretty. Also I just noticed you had a mistake with the units: momentum is in kg m/s but you had written it in Newtons. – David Z Oct 25 '11 at 21:03
up vote 3 down vote accepted

I have no clue whatsoever as to what the "relative velocity formula" is, but I think that the idea here is to find a relation between the $v_i^f$ (assuming $f$ means final).

That comes straight out of momentum conservation (using $s$ for starting and $f$ for final and that the problem is in one dimension): $$ p^f = p^s \implies m_1 v_1^f + m_2 v_2^f = m_1 v_1^s + m_2 v_2^s $$ which has only two unknowns, namely the $v_i^f$.

(Kinetic) energy conservation would lead to: $$E^f = E^s \implies m_1 (v_1^f)^2 + m_2 (v_2^f)^2 = m_1 (v_1^s)^2 + m_2 (v_2^s)^2$$ which again has the two same unknowns.

So, two equations with two unknowns...

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Do not forget velocities have signs (so, also do momenta). – adavid Oct 25 '11 at 19:06
Hmm, I am still not seeing how I will find v2f in terms of v1f. My book says v1i – v2i = -(v1f - v2f), along with conservation of momentum, can be used to solve for the two unknowns. I believe I am supposed to end up with these equations for the "relative velocity result": – Matt Oct 25 '11 at 21:07
Am I just missing some algebra we can do with the equations we are currently looking at to get those resultant equations? – Matt Oct 25 '11 at 21:09
Well @Matt, can you see how v2f comes out of the equation you just pasted in the comment above? I mean, there are 4 symbols in a linear combination, 2 have known values, so the other 2 can be expressed in terms of each other. (This is all b is asking for.) – adavid Oct 25 '11 at 21:12
How v2f comes out of: v1i – v2i = -(v1f - v2f) or the equation from the imgur link? – Matt Oct 25 '11 at 21:28

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