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:
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 $\vec F = \frac{\Delta \vec p}{\Delta t}$?
Thanks
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 $m\,\vec v_f$.