When to use relative velocity in centre of mass problems? I am confused about when we should use relative velocity of an object placed on a movable platform like a wedge while solving center of mass related dynamics problems. For example, in the first problem below consisting of two wedges we have not used relative velocity for block $B$. However, in the second problem when the ball is in the semi-circular gap of the smooth wedge at the point $B$, we are using the relative velocity of the ball w.r.t wedge in order to solve the problem.
Kindly enlighten me as to when we should use the relative velocity and when we should use separate ground reference frame variables in the dynamical equations.



 A: The OP is a good exercise in kinematics and dynamics.
The solution to the first problem is accurate and clear since all velocities are measured relative to an absolute or the ground frame. Clearly, there are typographical errors in the solution to the second problem. Let us first correct the analysis. Interpreting the question (in reverse) from the solution the ball of mass $m$ is moving within the semicircle surface in a wedge placed on a smooth surface which has the mass $m$ as well. The velocity of the ball w.r.t. the wedge when it is located at the top of it's semicircular trajectory is $0$ and at point $B$ is $0<v_1$ directed at an angle $-45^\circ$ to the horizontal, while the wedge is has an initially velocity of $0 $ and the speed $0<v$ directed horizontally to the left of the figure at the time instant at which the ball is located at point $B$. Finally, it is assumed that the ball does not lose contact with the semicircular surface of the wedge in the duration of the motion being analyzed.
To solve the second problem, the equation 1 is written correctly as $g\frac{r}{\sqrt{2}}=\frac{v^2}{2}+\frac{1}{2}(\frac{v_1^2}{2}-2\frac{v_1v}{\sqrt{2}}+v^2) + \frac{1}{2}\frac{v_1^2}{2}$ wherein the right hand side is the mass normalized sum of the kinetic energies of the wedge $\frac{1}{2}m(-v)^2$ and the ball $\frac{1}{2}m((v_1\cos45^\circ-v)^2+(-v_1\sin45^\circ)^2)$. The equation after the phrase 'put in equation 1' is written as correctly $3v^2=g\frac{r}{\sqrt{2}}$ which indeed results from substituting the expression of $v_1$ in terms of $v$ obtained from the horizontal linear momentum conservation equation $0=m(-v)+m(v_1\cos45^\circ-v)$ $\equiv v_1=2\sqrt{2}v$ into the equation 1, which is the mechanical energy conservation law.

The relative velocity $v_1$ of the ball w.r.t. the wedge is required in the analysis in the solution to the second problem because this is the quantity which enables us to relate the initial momentum of the ball when it is located at the top of the wedge surface to the momentum when it is located at point $B$. In particular, the absolute or ground reference frame velocity of the ball when located at point $B$ written as coordinates is ($(v_1\cos45^\circ-v),-v_1\sin45^\circ$) that is, $(v_1\cos45^\circ-v)$ horizontally to the right and $v_1\sin45^\circ$ vertically downwards. The underlying reason that we require this manipulation is that while the relative velocity has a direct relationship with the geometry of the wedge (tangential to the semicircular surface), the velocity measured w.r.t. the ground does not.
Alternately, we can choose to denote the ground reference frame velocity of the ball at point $B$ as ($\tilde{v}_{1x},\tilde{v}_{1y}$) so that the horizontal linear momentum conservation equation $0=-mv+m\tilde{v}_{1x}$ yields $v=\tilde{v}_{1x}$ and the energy conservation equation is $mg\frac{r}{\sqrt{2}}=\frac{1}{2}m(-v)^2+\frac{1}{2}m\tilde{v}_{1x}^2 + \frac{1}{2}m\tilde{v}_{1y}^2$ yields $g\frac{r}{\sqrt{2}}=\frac{2v^2+\tilde{v}_{1y}^2}{2}$. Therefore, in order to express $v$ in terms of the given parameters $g, \;r$ we now need to express $\tilde{v}_{1y}$ in terms of $v$, which requires the relative velocity relationship $v_\text{ball}^\text{ground}=v_\text{wedge}^\text{ground}+v_\text{ball}^\text{wedge}$ or $(\tilde{v}_{1x},\tilde{v}_{1y})=(-v,0)+(v_1\cos45^\circ,-v_1\sin45^\circ)$. The underlying reason that we require this manipulation is that while the relative velocity has a direct relationship with the geometry of the wedge (tangential to the semicircular surface), the velocity measured w.r.t. the ground does not.
A: I suggest you that always solving it w.r.t ground. It  is easier and also avoids confusions .you can directly conserve momentum and energy which is not possible if u solve using relative motion.
