Let's carefully analyze the problem step-by-step. --- ### **Given Data:** | Parameter | Value | |--------------------------------|-----------------------------------------| | Length of bar, \( L \) | 1 m | | Mass of bar, \( M \) | 8 kg | | Mass of ball, \( m \) | 3 kg | | Initial speed of ball, \( v_0 \) | 10 m/s | | Impact point along bar, \( r \) | \( 34L = 0.34 \) m | | Impact angle, \( \beta \) | 60° | | Coefficient of restitution, \( e \) | 0.7 | | Initial angular velocity, \( \omega_0 \) | 0 (at rest) | --- ### **Step 1: Determine the velocity components of the ball at impact** The ball strikes the bar at a point \( r = 0.34 \) m from the pivot O, with speed \( v_0 = 10 \text{ m/s} \) at an angle \( \beta = 60^\circ \). - The velocity components relative to horizontal: \[ v_{x} = v_0 \cos \beta = 10 \times \cos 60^\circ = 10 \times 0.5 = 5\, \text{m/s} \] \[ v_{y} = v_0 \sin \beta = 10 \times \sin 60^\circ \approx 10 \times 0.866 = 8.66\, \text{m/s} \] --- ### **Step 2: Find the velocity of the ball **just before impact** Since impact is in the horizontal plane, and the initial velocity vector is at \( 60^\circ \) to the horizontal, the velocity components are: \[ \boxed{ \mathbf{v}_{b} = (v_{x}, v_{y}) = (5, 8.66) \text{ m/s} } \] --- ### **Step 3: Determine the velocity of the ball relative to the bar at impact** The bar is initially at rest, so the initial velocity of the bar is zero. - The **tangential velocity** of the ball at the impact point relative to the pivot is: \[ \mathbf{v}_{b,\text{impact}} = \mathbf{v}_{b} = (5, 8.66) \text{ m/s} \] - The **position vector** from the pivot to impact point: \[ \mathbf{r} = r \hat{\mathbf{r}} \quad \text{where} \quad r = 0.34\, \text{m} \] Assuming impact occurs at angle \( \theta = 60^\circ \) from the horizontal: \[ \hat{\mathbf{r}} = (\cos 60^\circ, \sin 60^\circ) = (0.5, 0.866) \] Thus, \[ \mathbf{r} = 0.34 \times (0.5, 0.866) \approx (0.17, 0.294) \] --- ### **Step 4: Compute the tangential velocity component of the ball** The tangential velocity component relative to the pivot: \[ v_{t} = \mathbf{v}_{b} \cdot \hat{\mathbf{r}} = (5)(0.5) + (8.66)(0.866) \] Calculate: \[ 5 \times 0.5 = 2.5 \] \[ 8.66 \times 0.866 \approx 7.5 \] Sum: \[ v_{t} \approx 2.5 + 7.5 = 10\, \text{m/s} \] **Note:** The magnitude of the relative tangential velocity component along the impact direction is approximately 10 m/s. --- ### **Step 5: Apply the coefficient of restitution** The impact affects the **normal component** of the relative velocity at the contact point. - Since the impact is at an angle, decompose the velocity into components **normal** and **tangential** to the contact surface. Assuming the impact normal makes an angle \( \beta = 60^\circ \) with horizontal: - The normal component of the ball's velocity: \[ v_{n} = v_0 \sin \beta = 8.66\, \text{m/s} \] - The tangential component: \[ v_{t} = v_0 \cos \beta = 5\, \text{m/s} \] Because the problem specifies the impact at angle \( \beta \), and the impact is characterized by coefficient \( e \), the **normal component** of the relative velocity after impact**: \[ v_{n}^{\text{after}} = - e \times v_{n}^{\text{before}} \] \[ v_{n}^{\text{after}} = - 0.7 \times 8.66 \approx -6.06\, \text{m/s} \] The tangential component remains unchanged (assuming no friction): \[ v_{t}^{\text{after}} = v_{t}^{\text{before}} = 5\, \text{m/s} \] --- ### **Step 6: Find the change in the ball's velocity** - The **relative velocity** after impact: \[ \mathbf{v}_{b,\text{after}} = (v_{t}, v_{n}^{\text{after}}) \] Expressed in components: \[ v_{x,\text{after}} = v_{t} \cos \beta - v_{n}^{\text{after}} \sin \beta \] \[ v_{y,\text{after}} = v_{t} \sin \beta + v_{n}^{\text{after}} \cos \beta \] But more straightforwardly, since the impact is at angle \( \beta \): \[ v_{x,\text{after}} = v_{t} \cos \beta - e v_{n} \sin \beta \] \[ v_{y,\text{after}} = v_{t} \sin \beta + e v_{n} \cos \beta \] Calculate: \[ v_{x,\text{after}} = 5 \times 0.5 - 6.06 \times 0.866 \approx 2.5 - 5.25 = -2.75\, \text{m/s} \] \[ v_{y,\text{after}} = 5 \times 0.866 + 6.06 \times 0.5 \approx 4.33 + 3.03 = 7.36\, \text{m/s} \] --- ### **Step 7: Find the change in angular momentum** The change in the **ball's velocity** at impact affects the angular momentum imparted to the bar. - The **initial** velocity at impact: \[ \mathbf{v}_b = (5, 8.66)\, \text{m/s} \] - The **final** velocity at impact: \[ \mathbf{v}_b' = (-2.75, 7.36)\, \text{m/s} \] - The **change in velocity**: \[ \Delta \mathbf{v}_b = \mathbf{v}_b' - \mathbf{v}_b = (-2.75 - 5, 7.36 - 8.66) = (-7.75, -1.3) \] - The **change in angular momentum** of the ball about the pivot: \[ \Delta L_b = m (\mathbf{r} \times \Delta \mathbf{v}_b) \] \[ = m (x \Delta v_y - y \Delta v_x) \] where \( x = 0.17\, \text{m} \), \( y= 0.294\, \text{m} \). Calculate: \[ \Delta L_b = 3 \times (0.17 \times -1.3 - 0.294 \times -7.75) \] \[ = 3 \times (-0.221 + 2.278) = 3 \times 2.057 \approx 6.17\, \text{kg m}^2/\text{s} \] This is the **angular impulse** imparted to the bar. --- ### **Step 8: Apply conservation of angular momentum** The **initial angular momentum** of the system is zero (both at rest). The **final angular momentum** of the system: \[ I_{total} \omega = \text{impulse} \] where: - \( I_{total} \) is the total moment of inertia of the system - \( \omega \) is the angular velocity of the bar after impact The **moment of inertia**: \[ I_{bar} = \frac{1}{3} M L^2 = \frac{1}{3} \times 8 \times 1^2 = \frac{8}{3} \approx 2.666\, \text{kg m}^2 \] - The **ball's contribution** to the angular momentum: \[ I_{ball} = m r^2 = 3 \times (0.34)^2 = 3 \times 0.1156 \approx 0.347\, \text{kg m}^2 \] **Total moment of inertia:** \[ I_{total} = I_{bar} + I_{ball} = 2.666 + 0.347 \approx 3.013\, \text{kg m}^2 \] --- ### **Step 9: Compute the angular velocity \( \omega \)** Using angular impulse: \[ I_{total} \times \omega = \Delta L_b \implies \omega = \frac{\Delta L_b}{I_{total}} \approx \frac{6.17}{3.013} \approx 2.05\, \text{rad/sec} \] --- ## **Final Answer:** \[ \boxed{ \boxed{ \omega \approx \mathbf{2.05\, \text{rad/sec}} } } \] The bar's angular speed immediately after impact is approximately **2.05 radians per second**.