Non-acceleration and 0 force

If a mass is moving at the rate of 30/ft. per minute for 5 minutes on a straight line and it strikes a second stationary mass and effects a change of position to this second mass, then we know from F=ma that the force is 0 since the acceleration is 0. Then why do we say because it is 0 that there is no force when in fact the first mass changed the position of the second.

• Don't forget to accept one answer by clicking on the check-mark in the left column, next to your preferred answer. – DrSheldon Oct 14 '18 at 2:41

It appears you are mixing the force at one given time with the acceleration at a different given time.

Don't feel bad; it is a common beginner mistake to think that there is "one force" and "one acceleration" in a given problem. Part of learning physics is learning how to treat values from different times, as well as values from different objects.

During the 5-minute interval, the first mass stays at the same velocity. This means its acceleration is zero. By $$F_{net}=ma$$, the net force on this object during that time interval is zero.

During this same time interval, the second mass stays at rest. Its acceleration is zero, and thus the net force on it is zero.

The collision occurs at a different time. Neither force nor acceleration care about what happened during the 5-minute interval, they only depend on what is happening now. The first object is now slowing down -- a form of acceleration -- so $$a \ne 0$$. This means the net force on the first object $$F_{net}=ma\ne 0$$. This net force comes from the second mass pushing on the first mass.

By the Third Law, during the collision the first mass pushes back on the second mass. This creates a net force on the second mass, which accelerates the second mass during the collision.

• Thank you Dr. Shelton your explanation is crystal clear and beyond superb! It is simple, non-befuddling, incomparably better than the others. They need more physics instructors like you!!!! – Robert Telarket Sep 27 '18 at 13:21

We don't say it is $$0$$, why is $$F=0$$? If the second mass was stationary, then sometime after being struck by the first mass it is in a new position, it must have undergone acceleration $$\to F\neq0$$ precisely because $$F=ma$$.

• I am not referring to the second mass. For the first mass it's force is 0 because it's acceleration is 0. Then why do we say it has no force, since from F=ma it is 0, when it effected a change in the second mass. – Robert Telarket Sep 27 '18 at 2:42
• The situation you described is impossible. Any action is balanced by an equal and opposite reaction (Newton's 3rd law). So if the second mass experienced a force from the first one, then the first one experienced the same force (in the opposite direction) from the second. It cannot be the case that 1) the first mass continues at constant velocity and 2) the second mass is pushed by the first mass. – bRost03 Sep 27 '18 at 2:49

We never say force acting is zero.We just say there is no external forces to the two masses(external force to the system is zero).This is because only if the external forces are zero(or their effect should be negligible), momentum conservation can be applied.But the force applied by one mass on the other(internal forces) is not zero.Since this is the only force acting and it acts for the same time(a short time) on both the bodies, we say the magnitude impulse on one body equals the other(opposite in direction) and thus the momentum change is also equal in magnitude but opposite in direction(since impulse equals change in momentum).

Hey when the collison occur the force will act on it this force will persist until they are in contact which is given by F =(m1v1-m1u1)/t=(m2v2-m2u2)/t=impulse/t Here v1 v2 are final velocities and u1 u2 are initial velocities t is the time upto which force acts or the time of contact Here we can see that force is not 0 it will act during collison for very short period of time .Here F is the average force acting for time t.This force will be equal and opposite for both particles and force will cause the change in momentum