# Collision of two objects

Suppose there are two objects, denoted by A and B. When they collide with each other with the condition that A is stationary and B is in motion, B exerts an action force on A due to which A exerts a reaction force on B. Thus the net force exerted on A is $F_{{action}_{AB}}$ and the net force exerted on B is $F_{{reaction}_{BA}}$. Same is the case when A is in motion and B is at rest. But when A and B collide when they are in motion, the net force exerted on A is $F_{{action}_{AB}}$ + $F_{{reaction}_{AB}}$ and the net force exerted on B is $F_{{action}_{BA}}$ + $F_{{reaction}_{BA}}$.

Question: Is my reasoning correct for the case when A and B collide when they are in motion? Is the net force exerted on A and B when they collide when they are in motion greater in magnitude than the net force exerted on A and B when they collide when only one of them is in motion?

If I understand correctly your reasoning, you are using incorrectly the Newton's third law. The force of A is not exerted over A itself, and the same for B.

Yes, that is correct. When only one object is in motion for the collision, it exerts a force on the other object and feels a reaction force simultaneously.

When both objects are in motion, A exerts a force on B and still feels an equal reaction force in response. But now, A also feels the force exerted from B.

The magnitude of the total net force is greater. But when you include direction (which forces have since they're vectors) then the forces and reaction forces cancel much of each other out.

• How can the action and reaction forces cancel out? They are in the same direction and act on the same body. Don't they?
– MrAP
Mar 25, 2017 at 5:14
• What I meant was when you find the net force, forces that act in the negative direction are subtracted from forces that act in the positive direction. But in a free body diagram, you wouldn't really draw reaction forces. Mar 25, 2017 at 6:14
• @InertialIgnorance "When only one object is in motion for the collision, it exerts a force on the other object and feels a reaction force simultaneously" Why the moving object is the object that exerts the force? Or why also the non-moving object exers a force other than the reaction force? May 28, 2020 at 10:39

Your reasoning is correct and with Newton's third law you also have that $\mathbf{F}_{action_{AB}}=-\mathbf{F}_{reaction_{BA}}$ and $\mathbf{F}_{action_{BA}}=-\mathbf{F}_{reaction_{AB}}$.

The three cases you consider are in fact all identical (in the non-relativistic case considered here) and are essentially just a choice of the frame of reference. Let's say you have A fixed and B approaching at velocity $v$. Now imagine you are sitting on B. What you see is that A is approaching at the same velocity $v$ (opposite direction), while B is fixed. Similarly you can be in a frame of reference where both A and B are moving towards each other at velocity $v/2$. In all cases the relative velocity $\mathbf{v}_A-\mathbf{v}_B$ (note vectors) is all that matters and has in all three cases the magnitude $v$.

All you need to do is to apply Newton's third law correctly.

In all instances there are only two forces acting in the system of two balls, one on each of the balls.
One is the force on ball $A$ due to ball $B$ and the other is the force on ball $B$ due to ball $A$.
These force are equal in magnitude and opposite in direction.

The words reaction and action are sometimes used to differentiate between these two forces but the choice which force you call action and which you call reaction is totally arbitrary and is (perhaps) not necessary in your example particularly if such labels cause confusion.