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 $v_{1f}$?
Answer
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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