Question

derivation of elastic collision

Answer 1

Since this is an isolated system, the total momentum of the two particles is conserved:

Also, since this is an elastic collision, the total kinetic energy of the 2-particle system is conserved:

Multiplying both sides of this equation by 2 gives:

Suppose we solve equation 1 for v2:

and then substitute this result into equation 2:

Expanding and multiplying both sides by m2 in order to clear fractions gives:

Now, gather up like terms of v1:

Notice that equation 4 is a standard quadratic in v1, like Ax2 + Bx + C = 0, where:

So, we can use the quadratic formula () to solve for v1:

Inside the radical, the last term of the discriminant has factors like (a + b)(a - b) = a2 - b2, so:

Now, expand and simplify:

So, there are 2 solutions (of course...). Taking the positive sign in the numerator of equation 5 gives:

Physically, this means that no collision took place - the velocity of m1 was unchanged. That isn't the solution we have come this far to find. Taking the negative sign in the numerator of equation 5 gives:

That's it! Now, to find v2, substitute equation 6 into equation 3:

There it is! Equations 6 and 7 give the velocities of the two particles after the collision