Here's a bit of a fun physics paradox, which I will pose, and then answer below.
(These ideas were inspired by Grant Sanderson's fascinating video on how the digits of $\pi$ are hidden within elastic collisions.)
An object is travelling to the right, encounters an immovable barrier, undergoes a perfectly elastic collision, and returns to the left at an equal but opposite velocity.
Since the collision is perfectly elastic, kinetic energy and momentum are both conserved.
But the momentum after the collision is not equal to the momentum after the collision, because $p \ne -p$. The net change, $\Delta p$ should be zero, but instead:
$$\Delta p = p_f - p_i = -p - p = -2p \ne 0$$
What's up with that?
The paradox is created by assuming the "immovable" absorbs no momentum at all.
Let $m$ and $M$ be the masses of the moving object and the barrier, respectively.
Let $v_i$ be the velocity of the moving object prior to the collision. The initial conditions are:
$$p_{total}= mv_i$$ $$E_k{_{total}}= \frac{1}{2}m{v_i}^2$$
Let $v_f$ be the velocity of the moving object after the collision. And let $V$ be the velocity of the "immovable" barrier after the collision.
$$p_{total}= mv_f + MV$$ $$E_k{_{total}}= \frac{1}{2}m{v_f}^2+\frac{1}{2}M{V}^2$$
Assign values as follows:
$m = 1$, $v_i = 1$ and $M=10^{10}$
Units are irrelevant. The purpose of this exercis is to show comparative values. The mass, $M$ is immensely larger than the mass, $m$. Eventually, it will be shown that the velocity, $V$ of the "immovable" barrier is obscenely small compared to $v_i$ and $v_f$.
Now solve for the two unknowns, $v_f$ and $V$, using substitution:
$$p_{total}= mv_i = (1)(1) = 1 = mv_f + MV = (1)v_f + (10^{10})V$$ $$v_f = -(10^{10})V+1 \approx -(10^{10})V$$
Note that if we choose larger values for $M$, the ratio $\frac {v_f}{V}$ becomes larger and negative. The "immovable" barrier recoils from the collision at an infinitesimal rate, in the opposite direction of $v_f$.
$$E_k{_{total}}= \frac{1}{2}m{v_i}^2 = \frac{1}{2}(1)(1)^2 = \frac{1}{2}$$ $$\frac{1}{2}= \frac{1}{2}m{v_f}^2+\frac{1}{2}M{V}^2 = \frac{1}{2}(1)(-(10^{10})V+1)^2+\frac{1}{2}(10^{10})V^2$$ Double everything to get rid of the fractions, then expand: $$1=(10^{20})V^2−2(10^{10})V+1+(10^{10})V^2 $$ $$0=(10^{20}+10^{10})V^2−2(10^{10})V$$ $$0=V((10^{20}+10^{10})V−2(10^{10}))$$
Solving for $V$ yields two solutions:
$$V=0$$ $$ \text{(The velocity before the collision)}$$ $$V=\frac{2(10^{10})}{10^{20}+10^{10}}$$ $$ \text{(The itty-bitty velocity after the collision)}$$
Applying what was calculated earlier: $$v_f = -(10^{10})V+1 = \frac{-2(10^{20})}{10^{20}+10^{10}}+1[[\approx -1]]$$
The total momentum after the collision:
$$p_{total}= mv_f + MV$$ $$= (1)(\frac{-2(10^{20})}{10^{20}+10^{10}}+1) + (10^{10})(\frac{2(10^{10})}{10^{20}+10^{10}})=1$$ $$p_f=p_i=1 \\ \therefore \Delta p=0$$
To see more intuitively the distribution of momentum after the collision, approximate each of the figures, by eliminating the (relatively) insignificant $10^{10}$ from each denominator.
$$mv_f=(1)(\frac{-2(10^{20})}{10^{20}+10^{10}}+1) \approx \frac{-2(10^{20})}{10^{20}+10^{10}}+1 =\frac{-2(10^{20})}{10^{20}}+1 = -2 + 1 = 1$$ $$MV = (10^{10})(\frac{2(10^{10})}{10^{20}+10^{10}})=\frac{2(10^{20})}{10^{20}+10^{10}}\approx \frac{2(10^{20})}{10^{20}} = 2$$
The "immovable" barrier actually carries a momentum that has twice the magnitude of the moving object. However, its velocity is mindbogglingly small, and approaches zero as the mass of the barrier approaches infinity. We could show this more rigorously with painful limits and epsilons and such, but the intuition shown here should suffice.
Finally, to correct the diagram in the question: