Conflicting Change in Momentum Question
This is a peculiar question taken from the IBDP question bank (and hence also from an official exam). It is from the November 2016 SL paper 1.
The question asks for "the change in the magnitude of the momentum of the ball." This is equal to the final magnitude of momentum minus the initial magnitude of momentum.
$$\Delta |\vec{p}|=|\vec{p}|{\mathrm{f}}-|\vec{p}|{\mathrm{i}}=|\vec{p}{\mathrm{f}}|-|\vec{p}{\mathrm{i}}|$$
I would like to note that the question does not ask for the magnitude of the change in the momentum of the ball as that is a different quantity. I did make the mistake of $|a-b|=|a|-|b|$ but that is not valid (triangle inequality).
When I first saw the question it also was obvious to me that the change is zero. It's clear that $|\vec{p}|=m|\vec{v}|$ which, in this case, means $|\vec{p}|=mv$. $m$ and $v$ remain constant hence the change in the magnitude is zero.
The length of the scaled vector arrow representing the momenta is the same for before and after... so obviously there is no change? Yet for some reason the answer the IB gives is A from the exam and B from the question bank (for the exact same question), not even coinciding...
I believe there were two odd ways of thinking in making A/B the answer instead of D. 512
First, the 'IB physics' way to approach the problem: It is clear that the final $v_y$ equals the initial $v_y$ (vertical component of velocity), so there is no change, but $v_{x\mathrm{i}}=-v_{x\textrm{f}}$. Hence $\Delta|\vec{p}|=m\cdot v_{x\mathrm{f}}-(m\cdot-v_{x\mathrm{f}})=2m\cdot v_{x\mathrm{f}}$. Now comes the second issue (and why I assume the answers differ): the wording of the question does not match the diagram. It says "at an angle of $\theta$ to the wall" but the diagram shows an angle $\theta$ to the normal. If we use the diagram, we come to the IB result of $\Delta|\vec{p}|=2mv\cos\theta$, and we get $\sin$ if we consider the angle $\theta$ "to the wall."
I however cannot follow this logic with simply considering 1 component, so you could also formulate the above using the incorrect assumption that $|a-b|=|a|-|b|$: \begin{align} \Delta |\vec{p}|&=|\vec{p}{\mathrm{f}}|-|\vec{p}{\mathrm{i}}|\ &= |\vec{p}{\mathrm{f}}-\vec{p}{\mathrm{i}}|\ &= |m\vec{v}{\mathrm{f}}-m\vec{v}{\mathrm{i}}|\ &= \left| \begin{pmatrix} m\cdot v_{x\mathrm{f}}\m\cdot v_{y\mathrm{f}} \end{pmatrix}-\begin{pmatrix} m\cdot -v_{x\mathrm{f}}\m\cdot v_{y\mathrm{f}} \end{pmatrix}\right|\ &=\left|\begin{pmatrix} 2m\cdot v_{x\mathrm{f}}\0 \end{pmatrix}\right|\ &= \sqrt{(2m\cdot v_{x\mathrm{f}})^2}\ &= \sqrt{(-2mv\cos\theta)^2}\ &= 2mv\cos\theta. \end{align}
You can also do the above graphically by taking the final and initial momenta vector arrows and subtracting them (which gives you a vector arrow whose length is twice the horizontal component of velocity), but, again, this gives the result $|\vec{p}{\mathrm{f}}-\vec{p}{\mathrm{i}}|$ which is NOT what we are looking for!
I want to stress however that this 'property' of magnitudes is not true. 512