# When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater collision force. Is that true or false? Why?

Jun 17, 2014

In an ideal case of "head-to-head" elastic collision of material points occurring during a relatively short period of time the statement is false.

One force, acting on previously moving object, slows it down from initial velocity $V$ to a velocity equaled to zero, and the other force, equal to the first in magnitude but opposite in direction, acting on previously stationary object, accelerates it up to a velocity of the previously moving object.

In practice we have to consider many factors here. The first one is elastic or inelastic collision takes place. If it's inelastic, the law of conservation of kinetic energy is no longer applicable since part of this energy is converted into internal energy of molecules of both colliding objects and results in their heating.

The amount of energy thus converted into heat significantly affects the force causing the movement of the stationary object that depends very much on the degree of elasticity and cannot be quantified without any assumption about objects, the material they are made of, shape etc.

Let's consider a simple case of almost elastic "head-to-head" collision (there are no absolutely elastic collisions) of one object of mass $M$ that moves at speed $V$ with a stationary object of the same mass. The laws of conservation of kinetic energy and linear momentum allow to calculate exactly the velocities ${V}_{1}$ and ${V}_{2}$ of both objects after an elastic collision:
$M {V}^{2} = M {V}_{1}^{2} + M {V}_{2}^{2}$
$M V = M {V}_{1} + M {V}_{2}$

Cancelling the mass $M$, raising the second equation to a power of 2 and subtracting form the result the first equation, we get
$2 {V}_{1} {V}_{2} = 0$

Therefore, the solution to this system of two equations with two unknowns velocities ${V}_{1}$ and ${V}_{2}$ is
${V}_{1} = V$ and ${V}_{2} = 0$
The other algebraically correct solution ${V}_{1} = 0$ and ${V}_{2} = V$ should be discarded since physically it means that moving object goes through the stationary.

Since the previously moving object decelerates from $V$ to $0$ during the same time as previously stationary object accelerates from $0$ to $V$, the two forces acting on these objects are equal in magnitude and opposite in direction.