Question

Find the velocity of the center of mass C and the angular velocity of the system about the center of mass after the collision.

& \text{A) }\dfrac{2mv}{M+m},\dfrac{3Mv}{(M+2m)L} \\\ & \text{B) }\dfrac{Mv}{M+m},\dfrac{6Mv}{(M+4m)L} \\\ & \text{C) }\dfrac{2Mv}{M+m},\dfrac{3mv}{(M+4m)L} \\\ & \text{D) }\dfrac{mv}{M+m},\dfrac{6Mv}{(M+4m)L} \\\ \end{aligned}$$

Answer 1

We need to understand the given situation in which two bodies are moving, then collides and eventually sticks together and travels with a center of mass. We can find the relation between the given parameters to get the required solution.

Complete answer: We are given a point mass ‘m’ and a rod of mass ‘M’. The point mass is moving perpendicular to the rod which is initially at rest and collides at one end of the rod.

As a result of this collision, the point mass sticks to the rod and both undergo a motion along the center of mass of the system.

The length of the rod is given to be of ‘L’ units. So, we can find the position of the center of mass of the system ‘x’ from the origin O, which is the center of the rod as –

& (m+M)x=m(0)+m\dfrac{L}{2} \\\ & \therefore x=\dfrac{mL}{(m+M)2} \\\ \end{aligned}$$ The distance from the position of the mass ‘m’ to the center of mass will be given as – $$\begin{aligned} & \dfrac{L}{2}=x+BC \\\ & \Rightarrow BC=\dfrac{L}{2}-\dfrac{m}{(m+M)}\dfrac{L}{2} \\\ & \therefore BC=\dfrac{M}{(m+M)}\dfrac{L}{2} \\\ \end{aligned}$$ Now, we know that the collision has to conserve the linear momentum. We can use this idea to find the final velocity of the system as – $$\begin{aligned} & \text{Initial momentum = Final momentum} \\\ & \Rightarrow mv+M(0)=(m+M){{v}_{f}} \\\ & \therefore {{v}_{f}}=\dfrac{mv}{(m+M)}\text{ --(1)} \\\ \end{aligned}$$ We know that the rod being hit at the end has a tendency to rotate. The moment of inertia of the center of mass is dependent on the moment of inertia of the rod and the point mass, which can be calculated using the perpendicular axis theorem as – $$\begin{aligned} & {{I}_{CM}}={{I}_{rod}}+{{I}_{po\operatorname{int}}} \\\ & \Rightarrow {{I}_{CM}}=[\dfrac{1}{12}M{{L}^{2}}+M{{x}^{2}}]+m{{(BC)}^{2}} \\\ & \Rightarrow {{I}_{CM}}=\dfrac{1}{12}M{{L}^{2}}+M{{x}^{2}}+m{{\left( \dfrac{ML}{2(m+M)} \right)}^{2}} \\\ & \therefore {{I}_{CM}}=\left( \dfrac{M+4m}{M+m} \right)\dfrac{M{{L}^{2}}}{12} \\\ \end{aligned}$$ Also, the angular momentum at the center of mass has to be conserved. The initial angular momentum was zero, therefore, the angular momentum of the system after should also be the same. We can calculate the angular velocity from the conservation of angular momentum as – $$\begin{aligned} & \text{Initial angular momentum = Final angular momentum} \\\ & \Rightarrow 0={{I}_{CM}}.\omega -{{p}_{f}}BC \\\ & \Rightarrow 0=\left( \dfrac{M+4m}{M+m} \right)\dfrac{M{{L}^{2}}}{12}\omega -[(M+m){{v}_{f}}\dfrac{M}{(m+M)}\dfrac{L}{2}] \\\ & \Rightarrow \dfrac{{{v}_{f}}}{2}ML=\left( \dfrac{M+4m}{M+m} \right)\dfrac{M{{L}^{2}}}{12}\omega \\\ & \Rightarrow \omega =\dfrac{{{v}_{f}}}{2\left( \dfrac{M+4m}{M+m} \right)\dfrac{L}{12}} \\\ & \Rightarrow \omega =\dfrac{12\dfrac{mv}{(m+M)}}{2L\dfrac{(M+4m)}{(M+m)}} \\\ & \therefore \omega =\dfrac{6mv}{(M+4m)L} \\\ \end{aligned}$$ So, we get the required solutions for the problem. **The correct answer is option D.** **Note:** According to the perpendicular axis theorem, the moment of inertia of a point can be the sum of the moments of inertia of the two perpendicular axes which are mutually perpendicular and perpendicular to the unknown axis. We have used this theorem to solve.