Derivation of the final velocity formulas for elastic collisions + physical checks on the solution.
00:00 Given an initially moving cart and a stationary "target cart", we perform a derivation of the perfectly elastic collision general formula; i.e., we solve for the final velocity of each cart in terms of the two masses and the initial velocity of the first cart. The problem is broken into parts to guide us through the algebra, then we perform two physical checks on the solution to make sure the solution behaves correctly.
00:50 Equations for momentum and kinetic energy: in the first part of the general elastic collision problem, we are asked to write down the equations expressing conservation of momentum and conservation of energy for the collision. This gives us m_1*v_0=m_1*v_1+m_2*v_2 and 1/2*m_1*v_0^2=1/2*m_1*v_1^2+1/2*m_2*v_2^2. This is a nonlinear system of two equations and two unknowns, and our goal is to solve for the unknown final velocities in terms of the two masses and the initial velocity of the initially moving cart.
02:00 Solve for one velocity in terms of the other: in the next part of the problem, we are asked to use the momentum equation to solve for v_1 in terms of v_2. Because the momentum equation is linear, this is simple to do, and we quickly obtain our substitution.
02:40 Substitute into the kinetic energy equation: next, we take our substitution for the first final velocity and plug it into the kinetic energy equation. We start simplifying by cancelling the factor of 1/2 on every term, then we have to square a binomial term, clean things up and combine like terms. We find that we are left with a factorable quadratic equation in v_2. We throw out the non-physical solution for v_2 (it cannot be zero!), and solve for v_2 in terms of m_1, m_2 and v_0, obtaining the general final velocity for the elastic collision: v_2=2m_1*v_0/(m_1+m_2).
07:05 Use v_2 to find v_1: now we sub our solution for v_2 back into the substitution we derived from the momentum equation for v_1 in terms of v_2. After combining like terms and simplifying, we obtain a nice symmetric solution for v_1: v_1=(m_1-m_2)/(m_1+m_2)*v_0. We're done with the derivation of the final velocity formulas for elastic collisions!
09:07 Physical checks on the solution: as with any general solution of a physics problem, we can check to see if our answer is reasonable by thinking about some limiting cases and special cases where we know what physical behavior to expect from the solution. Here, we look at the limiting case of the second mass growing large (tending to infinity). We expect that v_2 should be zero and v_1 should be -v_0, and both of these limits work in the elastic collision general solution. Next, we try the special case m_1=m_2, where we remember that an elastic collision with equal masses results in the first mass coming to rest and the second mass leaving with a velocity equal to the original velocity of the first mass. This solution works as well, and we're done!
