Solving Elastic Collision Problems

Two methods will be shown to solve the following problem.

Problem:  A cart with mass 0.340 kg moving on a frictionless linear air track at 1.2 m/s strikes a second cart of unknown mass at rest.  The collision between the two carts is elastic.  After the collision, the first cart continues in its original direction at 0.66 m/s.  a) What is the mass of the second cart?  b) What is the velocity of the second cart after impact?

The Brute Force Method:  Using the lab frame of reference

The problem can be solved by applying the laws of conservation of momentum and kinetic energy. The collision situation is diagrammed below, where +x is to the right.

From the lab frame of reference (e.g., a student in the lab examining a collision of two air track gliders), the equations are:

and 

Solving these two equations simultaneously yields:

Substituting given values: m2 = 98.7 g and v2f = 1.86 m/s.

An Easier Method: Center-of-Mass

This method eliminates the algebra required when using the lab frame. While it's important to understand and be able to use the method described above, you'll find the following method much quicker and easier. In this method, we look at the collision from the reference frame of the center of mass. We know that the center of mass (CM) is unaccelerated in the absence of a net, external force acting on the system. Hence, the CM serves as a valid inertial frame of reference. More than that, the collision has this simple characteristic from the CM's viewpoint:

The velocity of either object after the collision is the opposite of the velocity of that same object before the collision. Another way of saying this is that the approach velocity of either object to the CM is the opposite of that object's separation velocity from the CM.

Let's show that this works for the given problem.

Here are the initial velocities in the lab frame. Note that vcm = m1v1/(m1 + m2).

The final velocities in the lab frame are:

For object 1, the approach velocity to the CM is (1.2 - 0.93) m/s = +0.27 m/s and the separation velocity is (0.66 - 0.93) m/s = -0.27 m/s.

For object 2, the approach velocity to the CM is (0 - 0.93) m/s = -0.93 m/s and the separation velocity is (1.86 - 0.93) m/s = +0.93 m/s.

Other Examples

Here's another problem. Given the following initial and final situations:

Is this collision elastic?

Begin by calculating vcm. Then calculate the approach and separation velocities for each object. Check whether the approach velocities are the opposites of the separation velocities.

Solution

See the handout, "Totally Elastic Collisions", for other sample problems and some exercises you can try.

Additional Problems