finding "sweet spot" "a ball of mass m makes an elastic collision with rod of mass M and length L. the ball hits a distance y from the bottom of the rod. find the value of y corresponding to the "sweet spot" for the collision, where the bottom of the rod remains instantaneously at rest after the collision." condition for hitting sweet spot: I (moment of inertia) for rod is instantaneously 1/3ML^2 (for rotation about stationary axis). angular momentum is conserved if we consider the system to be the rod and the ball. therefore: L_i = L_f; m*(v_i)*y = 1/3(ML^2)wf (omega final, final angular velocity of rod after collision) at this point, unknowns are y and wf. to find wf, we consider the fact that energy and linear momentum are conserved: E_i = E_f 1/2*m*v_i^2 = 1/2*I*wf^2 + 1/2*m*v_f^2 eliminate 1/2, use linear momentum to eliminate v_f p_i = p_f m*v_i = M*v_CM + m*v_f we know that the center of mass will be rotating, so we can use the relation v_CM = r*w, where r is the location of the center of mass (L/2). therefore, solving for v_f, we obtain v_f = v_i - M/2m*(L*wf) this is where i run into problems. my strategy was to use this v_f and solve for wf in the energy conservation equation, and then consequently solve for y in the angular momentum conservation equation. however, upon subsituting the above formula for v_f into the energy conservation equation, it becomes an extremely convoluted quadratic equation. what have i done wrong? is there an easier way to do this (preferably, without using impulses?) thanks so much for any help.