1. Derivation of Initial Relative Velocity ($v_{rel,0}$)

Let $V$ be the velocity of the wedge and $v_{rel}$ be the velocity of the disc relative to the wedge immediately after entering the incline. The velocity of the disc relative to the ground is $\vec{v}_{d} = \vec{V} + \vec{v}_{rel}$.

Conservation of Momentum (x-axis):

$$ m(V + v_{rel}\cos\theta) + MV = mu $$ Given $M=m$, this simplifies to: $$ 2V + v_{rel}\cos\theta = u \implies V = \frac{u – v_{rel}\cos\theta}{2} $$

Conservation of Energy:

$$ \frac{1}{2}m |\vec{v}_{d}|^2 + \frac{1}{2}MV^2 = \frac{1}{2}mu^2 $$ Expanding $|\vec{v}_{d}|^2 = (V + v_{rel}\cos\theta)^2 + (v_{rel}\sin\theta)^2$: $$ m(V^2 + v_{rel}^2 + 2V v_{rel}\cos\theta) + MV^2 = mu^2 $$ Substitute $M=m$ and divide by $m$: $$ 2V^2 + v_{rel}^2 + 2V v_{rel}\cos\theta = u^2 $$ $$ 2V(V + v_{rel}\cos\theta) + v_{rel}^2 = u^2 $$ Substitute $2V = u – v_{rel}\cos\theta$: $$ (u – v_{rel}\cos\theta) \left( \frac{u – v_{rel}\cos\theta}{2} + v_{rel}\cos\theta \right) + v_{rel}^2 = u^2 $$ $$ (u – v_{rel}\cos\theta) \left( \frac{u + v_{rel}\cos\theta}{2} \right) + v_{rel}^2 = u^2 $$ $$ \frac{u^2 – v_{rel}^2 \cos^2\theta}{2} + v_{rel}^2 = u^2 $$ $$ u^2 – v_{rel}^2 \cos^2\theta + 2v_{rel}^2 = 2u^2 \implies v_{rel}^2 (2 – \cos^2\theta) = u^2 $$ Using $2 – \cos^2\theta = 1 + \sin^2\theta$: $$ v_{rel,0} = \frac{u}{\sqrt{1 + \sin^2\theta}} $$

2. Derivation of Relative Acceleration ($a_{rel}$)

Consider the forces in the non-inertial frame of the wedge (moving with acceleration $A$ to the right).
Forces on disc ($m$): Gravity $mg$ (down), Normal $N$, Pseudo-force $mA$ (left).

Equation for Wedge ($M$): The normal force $N$ from the disc pushes the wedge right. $$ N\sin\theta = MA $$ Equation for Disc perpendicular to incline: $$ N + mA\sin\theta = mg\cos\theta \implies N = m(g\cos\theta – A\sin\theta) $$ Substitute $N$ into the wedge equation: $$ m(g\cos\theta – A\sin\theta)\sin\theta = MA $$ $$ mg\sin\theta\cos\theta = A(M + m\sin^2\theta) $$ $$ A = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta} $$ Now, finding $a_{rel}$ (acceleration of disc down the incline relative to wedge): $$ ma_{rel} = mg\sin\theta + mA\cos\theta $$ $$ a_{rel} = g\sin\theta + A\cos\theta = g\sin\theta + \frac{mg\sin\theta\cos^2\theta}{M + m\sin^2\theta} $$ $$ a_{rel} = g\sin\theta \left( 1 + \frac{m\cos^2\theta}{M + m\sin^2\theta} \right) = g\sin\theta \left( \frac{M + m(\sin^2\theta + \cos^2\theta)}{M + m\sin^2\theta} \right) $$ $$ a_{rel} = \frac{g\sin\theta(M+m)}{M + m\sin^2\theta} $$ For $M=m$: $$ a_{rel} = \frac{2g\sin\theta}{1 + \sin^2\theta} $$

3. Calculation of Displacements

Time to stop ($t$): Using $v = u – at$ with final velocity 0: $$ t = \frac{v_{rel,0}}{a_{rel}} = \frac{u/\sqrt{1+\sin^2\theta}}{2g\sin\theta / (1+\sin^2\theta)} = \frac{u\sqrt{1+\sin^2\theta}}{2g\sin\theta} $$

Horizontal Relative Displacement ($\Delta x_{rel}$): Distance up incline $L = v_{rel,0}^2 / (2a_{rel})$. $$ L = \frac{u^2 / (1+\sin^2\theta)}{4g\sin\theta / (1+\sin^2\theta)} = \frac{u^2}{4g\sin\theta} $$ Horizontal component relative to wedge: $$ \Delta x_{rel} = L\cos\theta = \frac{u^2\cot\theta}{4g} $$

Wedge Displacement ($x_w$): Using Center of Mass relation ($mx_d + Mx_w = (m+M)x_{cm}$): $$ x_w + (x_w + \Delta x_{rel}) = 2 v_{cm} t $$ With $v_{cm} = u/2$: $$ 2x_w = ut – \Delta x_{rel} \implies x_w = \frac{ut}{2} – \frac{\Delta x_{rel}}{2} $$ Substitute $t$ and $\Delta x_{rel}$: $$ x_w = \frac{u}{2} \left( \frac{u\sqrt{1+\sin^2\theta}}{2g\sin\theta} \right) – \frac{u^2\cos\theta}{8g\sin\theta} $$