Andrew K. answered • 02/08/21
Student-Athlete and Physics/Computer Science Double Major at MIT
For a perfectly elastic collision, the both total momentum of the system and total kinetic energy of the system must be conserved. Because the object is at rest to start, the initial momentum and kinetic energy of the system are both equal to the momentum and kinetic energy of the object. The momentum is 17.8 gm/s and the energy is 79.21 g*m^2/s^2. Because momentum is conserved, we have the final momentum of the system equals the initial momentum of the system, so 2 * v_p + 1 * v_o = 17.8 where v_p is the final velocity of the particle that was initially in motion and v_o is the final velocity of the object that was initially at rest. Because kinetic energy is conserved in perfectly elastic collisions, we have the final kinetic energy of the system equals the initial kinetic energy of the system, so (v_p)^2 + 1/2 (v_o)^2 = 79.21. We now have a system of equations that we can solve for the velocities, and we get v_p = 89/30 m/s and v_o = 178/15 m/s.
For b, we do the same process, but the stationary object has mass 10, so we get the system of equations 2 * v_p + 10 * v_o = 17.8 and (v_p)^2 + 5 (v_o)^2 = 79.21. Solving this we get v_p = - 89/15 m/s and v_o = 89/30 m/s. In this case, the particle that starts in motion collides with the stationary object and goes backwards with speed 89/15 (because speed is never negative).
