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F=ma with constant acceleration

  1. Feb 21, 2014 #1
    I'm new to physics, so I'm sure the explanation is very simple, but I cannot wrap my head around Newton's 2nd law F=ma. It is very easy to understand until I plug 0 into a. Wouldn't this make the force of the reaction disappear?

    For instance if a 100kg rock traveling east at constant velocity of 25m/s2 hits a 20kg rock traveling west at constant velocity of 25m/s2 at the point of impact neither object is accelerating (due to constant velocities). Wouldn't the total acceleration be 0 and nullify the force? Where am I going wrong? What am I forgetting?

    Thank you!
  2. jcsd
  3. Feb 21, 2014 #2


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    If neither object accelerates then they will go right through one another. Normally when two rocks collide, they stop or bounce off each other. In either case, their velocities are not constant during the collision.
  4. Feb 21, 2014 #3


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    During the collision, the rocks experience a repulsive force from each other which acts to accelerate them and change their velocities. Thus, the velocity is no longer constant but is changing through the duration of the collision.
  5. Feb 21, 2014 #4
    Ah, OK I think I see. So at the instantaneous time of impact each rock a=0 but the next time frame the collision changes the a's for each rock generating force? This force builds upon itself (over the next several milliseconds) as each rock adjusts to the force applied by the other due to their masses and their (now changing) a?

    This example above would clarify my problem with a=0, but is it accurate?
  6. Feb 22, 2014 #5
    It seems to me you're confusing velocity with acceleration. m/s^2 is a unit of acceleration, not velocity which is measured with the unit: m/s.

    That example is very simple. During the collision the two objects produce a force on each other causing them to accelerate changing their velocities and trajectories.
  7. Feb 22, 2014 #6


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    Nothing is totally incompressible. Perhaps it would help to replace the rocks with something not so rigid so that it is easier to see what's going on....

    Consider a car impacting a wall. The car has a crumple zone at the front so the bulk of the car does not stop the instant the bumper/fender hits the wall. The bulk of the car keeps going for a distance "s" and short time "t" while it comes to a stop. During that time the car decelerates from it's impact velocity "u" to it's final velocity "v" which in this case is zero.

    If you assume the deceleration is constant you can work out the deceleration using..


    but since v=0 then

    a = -u/t

    It's negative because the acceleration is in the opposite direction to the impact velocity. From that you can work out the force f that exists for the duration of the impact.

    In the real world you may not know the duration of the impact t but you might know the length of the crumple zone and hence measure the stopping distance "s". In which case you might be able apply other SUVAT equations such as ..


    If v=0 this rearranges to

    a = -U2/2s

    This shows that the acceleration "a" (and therefore the impact forces) are inversely dependant on the stopping distance. This is where your rocks come in. Rock isn't easily compressed so the stopping distance for two rocks colliding (or a coffee cup hitting a concrete floor or similar) is going to be short and that makes the acceleration and the forces very high.
  8. Jun 7, 2014 #7


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    The point of F=ma is that when a force, F, is applied to mass, m, the mass is accelerated at acceleration, a.

    At constant velocity implies zero acceleration, so the net force is zero. That works in a vacuum with no gravity. In most cases, there is some force due to friction, wind resistance, and gravity.

    Neglecting wind resistance, two blocks traveling at constant velocity do not have a ne force applied. However, at the moment they collide, they apply a force to each other. If the collision is elastic, then they recoil. Otherwise, the collision is inelastic, and the forces are manifest in the compressive and shear forces in the bodies, which can be stored as elastic energy, but more likely may result in internal deformations and fracture.

    In collisions, one must consider kinetic energy and momentum.


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