## Step 1: Find the Velocity of the Center of Mass After Collision **Conservation of Linear Momentum:** Since the collision happens on a frictionless table and the clay sticks to the rod, we use conservation of linear momentum in the \(x\)-direction (since the clay only moves in the \(x\)-direction before the collision). - **Initial total momentum (in x):** - Rod and attached mass are at rest: \(\) - Clay: \(m v_1\) \[ p_{x,\text{initial}} = m v_1 \] - **Total mass after collision:** - Mass at end of rod: \(2m\) - Clay: \(m\) - Total: \(3m\) Let \(V_{\text{cm}}\) be the velocity of the center of mass after collision. \[ m v_1 = (3m) V_{\text{cm},x} \implies V_{\text{cm},x} = \frac{v_1}{3} \] **Explanation:** The clay hits and sticks, so the total linear momentum is shared among all masses, resulting in a lower center-of-mass velocity. --- ## Step 2: Find the Angular (Rotation) Rate After Collision **Conservation of Angular Momentum:** Angular momentum is conserved about the center of mass of the system since no external torques act. - **Let the rod have length \(L\).** - The mass \(2m\) is at one end, clay \(m\) at the other. ### Find Initial Angular Momentum (about the center of mass after collision) - **Location of center of mass (CM) after collision:** - Place origin at the rod's center for calculation. - Let the rod be oriented at \(60^\circ\) to the \(x\)-axis. - The two masses are at ends, a distance \(L\) apart. Let’s call the two ends \(A\) (with mass \(2m\)) and \(B\) (clay, mass \(m\)). - Coordinates of \(A\): \((,)\) - Coordinates of \(B\): \((L\cos\beta, L\sin\beta)\), where \(\beta = 60^\circ\). **Center of mass:** \[ x_\text{CM} = \frac{2m \times + m \times L\cos\beta}{3m} = \frac{L\cos\beta}{3} \] \[ y_\text{CM} = \frac{2m \times + m \times L\sin\beta}{3m} = \frac{L\sin\beta}{3} \] - **Initial angular momentum (about CM):** - Only the clay has initial momentum: \(m v_1\). - Its position vector from CM: \(\vec{r}_B - \vec{r}_{CM}\). - Its velocity: \(v_1\) in \(+\hat{x}\). The angular momentum is \(L = \vec{r} \times m\vec{v}\). \[ \vec{r}_B - \vec{r}_{CM} = (L\cos\beta - \frac{L\cos\beta}{3},\ L\sin\beta - \frac{L\sin\beta}{3}) = (\frac{2L\cos\beta}{3}, \frac{2L\sin\beta}{3}) \] The cross product for angular momentum (z-component): \[ L_{z,\text{initial}} = m \left[ \frac{2L\cos\beta}{3} \cdot - \frac{2L\sin\beta}{3} \cdot v_1 \right] = -\frac{2mL\sin\beta}{3} v_1 \] - **After collision:** All mass rotates about CM. Let \(\omega\) be the angular velocity after collision. The moment of inertia about CM: - For \(2m\) at distance \(d_1\): \(d_1 = |\vec{r}_{CM}|\) - For \(m\) at distance \(d_2\): \(d_2 = |\vec{r}_B - \vec{r}_{CM}|\) Both are \(r = \sqrt{(2L\cos\beta/3)^2 + (2L\sin\beta/3)^2} = \frac{2L}{3}\). \[ I_{CM} = 2m \left(\frac{L}{3}\right)^2 + m \left(\frac{2L}{3}\right)^2 = 2m \frac{L^2}{9} + m \frac{4L^2}{9} = \left( \frac{2mL^2}{9} + \frac{4mL^2}{9} \right) = \frac{6mL^2}{9} = \frac{2mL^2}{3} \] **Setting angular momenta equal:** \[ -\frac{2mL\sin\beta}{3} v_1 = I_{CM} \omega \implies \omega = -\frac{2mL\sin\beta v_1}{3 \cdot \frac{2mL^2}{3}} = -\frac{\sin\beta v_1}{L} \] For \(\beta = 60^\circ\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\): \[ \omega = -\frac{\sqrt{3}}{2} \frac{v_1}{L} \] **Explanation:** The negative sign shows the direction of rotation (using the right-hand rule, consistent with the clay's impact direction). --- ## **Final Answers** ### **Path of Center of Mass** - **Moves in \(+\hat{x}\) direction** with velocity \(V_{\text{cm},x} = \frac{v_1}{3}\). ### **Velocity (including direction)** - \[ \boxed{V_{\text{cm}} = \frac{v_1}{3}~\hat{x}} \] ### **Rotation Rate of the Dumbbell** - \[ \boxed{\omega = -\frac{\sqrt{3}}{2} \frac{v_1}{L}} \] (Negative sign indicates rotation direction according to the right-hand rule.)