I want to justify Newton's second law for the linear momentum of a particle:

$$\vec F = \frac{d \vec p}{dt}$$

using really basic linear algebra. Basically, just with vector subtraction.

This is the case I am focused on:

Imagine we throw a ball against a static block (assume that the magnitude of the velocity of the ball after the collision doesn't change).

The momentum of the particle changes after the collision because the velocity of the ball changes (just in terms of direction as I stated). This means that an external force is exerted on the block.

In the following picture we expect the block to move to the right:

enter image description here

If you draw the vector $\Delta \vec p = \vec p_f - \vec p_i$ you won't get $\vec F$ because $\vec F \neq \Delta \vec p$ but $\vec F = \frac{\Delta \vec p}{\Delta t}$.

How can we explain by vector operations that $\vec F = \frac{\Delta \vec p}{\Delta t}$?



closed as unclear what you're asking by Somos, Lord Shark the Unknown, max_zorn, Dbchatto67, Leucippus Apr 19 at 5:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ ${d\vec p\over dt}\ne{\Delta\vec p\over \Delta t}$. $\endgroup$ – amd Apr 18 at 18:38
  • $\begingroup$ I agree but I did so as an approximation to avoid having to differentiate. $\endgroup$ – JD_PM Apr 18 at 18:40
  • $\begingroup$ What would you differentiate, though? $\vec p$ isn’t differentiable at the point of impact. $\endgroup$ – amd Apr 18 at 18:44
  • $\begingroup$ @amd why not??? $\endgroup$ – JD_PM Apr 18 at 18:48
  • $\begingroup$ What is your mathematical definition of $\ \vec F\ $ and $\ \vec p\ $? $\endgroup$ – Somos Apr 18 at 18:58

The ODE $$\vec F={d\vec p \over dt}$$ does not describe the experiment you are performing: We have $\vec F=\vec0$ and ${d\vec p\over dt}=0$ during $99.9\%$ of the time of the experiment and a totally unclear behavior of $t\mapsto\vec F(t)$ during the short time of the "exposure". What we, however, know is that $$\Delta\vec p=\vec p_f-\vec p_0$$ has been totally deferred to the block whereby we can say that $$\Delta\vec p=\int_0^f \vec F(t)\>dt\ ,$$ whatever the RHS means, $f\ll1$ denoting the end time of the process. If the block has mass $m$ and final velocity $\vec v_f$ then this integral is (up to sign) equal to $m\,\vec v_f$.

  • 1
    $\begingroup$ Thank you for your answer. I am still wondering if $\vec F = \frac{d \vec p}{dt}$ can be justified with vector operations. $\endgroup$ – JD_PM Apr 18 at 20:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.