

G91c. Solving Onedimensional Elastic Collisions Introduction Consider this problem: A glider with mass m and speed v_{1}_{i} moving along a frictionless, horizontal air track collides with a stationary glider having a mass m/3. What are the velocities of both gliders after the collision? We can apply conservation of momentum to the solution, since the net, external force on the system of gliders is 0. That would give us mv_{1}_{i} + 0 = mv_{1}_{f} + (m/3)v_{2f}. This equation has two unknowns and, hence, does not have a unique solution. Mathematically, there are an infinite number of pairs of values for the final velocities that satisfy the equation. Nevertheless, most people would express the belief that if the collision were carried out, there would be only one possible pair of values for the final velocities. We have faith that there is a unique outcome in nature; that is, nature is predictable. In order to resolve this dilemma, we look for a second equation, independent of the one above, that contains the same two unknown velocities and can be solved simultaneously with the equation above in order to determine unique values for the final velocities. In the case of an elastic collision, that relationship is conservation of kinetic energy. We'll see below how to set up the problem and solve the equations for momentum and kinetic energy conservation in order to determine the final velocities.
What follows is the solution of the general problem of finding the final velocities for an elastic collision of two objects in one dimension.
Setting up the Problem The problem involves two blocks of mass m_{1} and m_{2} that collide elastically on a frictionless surface. The initial and final states are shown below. In addition to the masses, we’re given the initial velocities v_{1i} and v_{2i}. The goal of the problem is to determine the final velocities of the blocks in terms of the masses and the initial velocities. The system is comprised of the two blocks. The external forces of weight and normal add to 0 for each block; hence, F_{net,ext} = 0 for the system. A diagram is shown below with initial and final states indicated. The direction of +x is chosen to the right. _{} Solving the Problem Two independent equations are required to solve this problem, since there are two unknowns. The equations we will use are those for conservation of momentum and kinetic energy. Applying conservation of momentum: (1) Applying conservation of kinetic energy: (2) With some algebraic simplification and rearranging, Equation (2) can be rewritten as: (3) Factoring the binomials gives: (4) Equation (1) can be rewritten as: (5) Dividing Equation (4) by equation (5) yields a simple result: (6) Equation (6) can be rewritten as: (7) Equation (1) can be rewritten as: (8) Equation (7) is substituted into Equation (8) and rearranged to get all terms with v_{1f} on the lefthand side of the equation: The velocity terms are factored to obtain: The final result for v_{1f} is: (9) Now we substitute Equation (9) into Equation (7) to obtain: Grouping the velocity terms and putting all the mass terms over the common denominator of (m_{1} + m_{2}) gives:
Final simplification yields the result for v_{2f}: (10) Equations (9) and (10) are the solution of the problem. Checking the Results Now we apply checks to the results. Suppose m_{1} = m_{2}. The final velocities reduce to: (11) In effect, the blocks exchange velocities as a result of the collision. This result is expected for elastic collisions of equal mass objects. Suppose block 2 is initially at rest: v_{2i} = 0. The final velocities reduce to: (12) This is the result given in the textbook without proof as equation 912 on p. 259. Now you have proof. Equations (12), by the way, apply to the problem situation that introduced this page. For that problem, m_{1} = m, m_{2} = m/3, and v_{2}_{i} = 0. With those substitutions, v_{1}_{f} = (1/2)v_{1}_{i} v_{2f}_{ }= (3/2)v_{1}_{i}


© North Carolina School of Science and Mathematics, All Rights Reserved. These materials may not be reproduced without permission of NCSSM. 