Physics - Center of Mass Question with Solution | TestHub

This problem involves two stages: an inelastic collision, followed by the upward swing of the combined mass. **Stage 1: Inelastic Collision**

Just before the collision, the ball of mass $m$ has momentum $mv$, and the bob of mass $M$ is at rest. After the inelastic collision, the ball gets embedded in the bob, forming a combined mass $(M+m)$ moving with a common velocity $V$. By conservation of linear momentum:$mv+M(0)=(M+m)V\Rightarrow V=\frac{mv}{M+m}$. **Stage 2: Upward Swing**

The combined mass $(M+m)$ starts moving with velocity $V$ and rises to a maximum height $h$. During this stage, mechanical energy is conserved. The kinetic energy of the combined mass is converted into gravitational potential energy.$\frac{1}{2}(M+m)V^2=(M+m)gh$. Substitute the expression for $V$ from Stage 1 into the energy conservation equation:$\frac{1}{2}(M+m){\left(\frac{mv}{M+m}\right)}^2=(M+m)gh$.$\frac{1}{2}(M+m)\frac{m^2{v}^2}{(M+m{)}^2}=(M+m)gh$.$\frac{m^2{v}^2}{2(M+m)}=(M+m)gh$. Solving for $h$: $h=\frac{m^2{v}^2}{2(M+m{)}^{2}g}$.