I am trying an experiment with Electromagnets and looking for help to calculate force required. Please note I am not asking about electromagnets though since that would be electrical question.

So the problem is simple as depicted in image below. Lets say an object of M1 mass is moving in one direction (left to right or green arrow). It passes through the Cylinder M2. Now that cylinder applies a pull force on M1 in order to stop it first and then reverse it's direction with exact same velocity V1.

My assumption is the force required to reverse direction of M1 would be twice of the momentum M1 has. Meaning it will take 2(V1*M1). I say this because to stop the M1 Cylinder will have to apply same force V1*M1. Then to pull it back it will be another force of V1*M1.

My question is, am I thinking this correct or am I way off in how this mechanism works?

Note: Please ignore things like friction, gravity, energy gain by cylinder etc. I am just trying to understand basic physics to validate a concept.


enter image description here

  • 1
    $\begingroup$ Force is rate of momentum change. $\endgroup$
    – joseph h
    Commented Oct 18, 2023 at 0:35

1 Answer 1


There are a few issues in your post. As joseph said, force is the rate of momentum change, $\vec{F} = \frac{d\vec{p}}{dt}$. You have labeled forces but written $mv$ terms, which are momentum terms. If we just look at $M_1$ and you truly meant force, then the green arrow describes a leftward $F_1$ on the mass, which is $F_1 = M_1a$. In order to "reverse" the mass such that the acceleration of the new force is exactly equal and opposite to the original force, we need to apply a force $F_2 = 2F_1 = 2M_1a$.

  • $\begingroup$ Semantics aside you did understand and answered my question. Thanks for validating my assumption. I will close this post as its answered. $\endgroup$
    – Rob
    Commented Oct 18, 2023 at 1:18
  • $\begingroup$ If it's answered, you should accept the answer, not close! @Rob $\endgroup$ Commented Oct 18, 2023 at 1:19
  • 1
    $\begingroup$ Yup did that. Didn't know there is no option for close! Thanks again. $\endgroup$
    – Rob
    Commented Oct 18, 2023 at 1:20
  • $\begingroup$ @Rob This problem is better analysed focusing upon momentum rather than on forces and accelerations. This is because the momentum change is very clearly $2M_1v_1$ here, but the forces would be $\frac{2M_1v_1}t$ i.e. can be any value you like provided you can change time, and in fact, this doesnt need to be constant and thus can be really much more varied. $\endgroup$ Commented Oct 18, 2023 at 3:19

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