Physics - Center of Mass Question with Solution | TestHub

Since there are no external horizontal forces, the horizontal momentum of the (block + wedge) system is conserved. Initially, the system is at rest, so the total initial momentum is zero.

Let $V$ be the velocity of the wedge to the left, and $v$ be the velocity of the block relative to the wedge downwards along the incline. The horizontal velocity of the block relative to the ground is $v\cos \theta -V$. Conservation of horizontal momentum: $m(v\cos \theta -V)-MV=0\Rightarrow mv\cos \theta =(M+m)V\Rightarrow V=\frac{mv\cos \theta }{M+m}$. By conservation of mechanical energy: $mgh=\frac{1}{2}M{V}^2+\frac{1}{2}m(v^2+V^2-2vV\cos \theta )$.

Substitute $V$ and solve for $v$, then substitute $v$ back into the expression for $V$. A simpler approach is to use the kinetic energy of the block relative to the ground: $mgh=\frac{1}{2}M{V}^2+\frac{1}{2}m(v_x^2+v_y^2)$, where $v_x=v^{\prime }\cos \theta$ and $v_y=v^{\prime }\sin \theta$ are the components of the block's velocity relative to the ground. The block's velocity relative to ground has horizontal component $v_x=v_r\cos \theta -V$ and vertical component $v_y=v_r\sin \theta$. Using the relation $V=\frac{m{v}_r\cos \theta }{M+m}$ and energy conservation: $mgh=\frac{1}{2}M{V}^2+\frac{1}{2}m((v_r\cos \theta -V{)}^2+(v_r\sin \theta {)}^2)$.

After algebraic manipulation and substitution, we get $V=\sqrt{\frac{2m^{2}gh{\sin }^2\theta }{M(M+m{\sin }^2\theta )}}$. This comes from $mgh=\frac{1}{2}M{V}^2+\frac{1}{2}m{v}_{rel}^2$, where $v_{rel}$ is the velocity of block relative to wedge. The correct relation for energy is $mgh=\frac{1}{2}M{V}^2+\frac{1}{2}m{v}_{block,ground}^2$. Horizontal momentum conservation gives $MV=m(v_{block,x})$. Also, $v_{block,x}=v_{block,rel,x}-V$. The block's velocity relative to the wedge along the incline is $v_{rel}$. Its horizontal component is $v_{rel}\cos \theta$. So $MV=m(v_{rel}\cos \theta -V)$. Energy conservation: $mgh=\frac{1}{2}M{V}^2+\frac{1}{2}m((v_{rel}\cos \theta -V{)}^2+(v_{rel}\sin \theta {)}^2)$.

Solving these simultaneously yields $V=\sqrt{\frac{2m^{2}gh{\sin }^2\theta }{M(M+m{\sin }^2\theta )}}$. Note that the vertical component of the block's velocity relative to the ground is $v_{rel}\sin \theta$, not $v_{rel}$ itself. The $v_{rel}$ here is the speed along the incline. The correct form of the solution is $V=\sqrt{\frac{2m^{2}gh{\sin }^2\theta }{M(M+m{\sin }^2\theta )}}$. The term ${\sin }^2\theta$ arises because the vertical drop $h$ is related to the distance travelled along the incline, and the horizontal component of the relative velocity is $v_{rel}\cos \theta$, while the vertical component is $v_{rel}\sin \theta$. The energy equation can be written as $mgh=\frac{1}{2}M{V}^2+\frac{1}{2}m{v}_{block,ground}^2$. We know $v_{block,ground}^2=(v_{rel}\cos \theta -V{)}^2+(v_{rel}\sin \theta {)}^2$. From horizontal momentum conservation $MV=m(v_{rel}\cos \theta -V)$. Substituting this into the energy equation and solving for $V$ gives the result.