Let's break down and solve the problem step by step: --- **Given:** - Ball mass (\(m\)) = 0.5 kg - Inclined plane angle (\(\theta\)) = 30° - Magnitude of impulse (\(\Delta J\)) = 7.3 N·s - Acceleration due to gravity (\(g\)) = 9.81 m/s² --- ### **a) Find the magnitude of the initial velocity of the ball.** #### **Step 1: Understanding the Impulse** Impulse is the change in momentum: \[ \Delta J = m \Delta v \] where \(\Delta v\) is the change in velocity (in the direction of impulse). #### **Step 2: Analyzing the Collision** - The ball hits the plane with a velocity just before impact, \(v\). - Only the perpendicular component of velocity is changed by the impulse; the parallel component remains the same. - The impulse acts perpendicular to the plane, reversing the perpendicular component. Let the velocity just before collision be \(v\). - Perpendicular component: \(v_{\perp} = v \sin\theta\) - Parallel component: \(v_{\parallel} = v \cos\theta\) After collision: - \(v_{\parallel}\) stays the same. - \(v_{\perp}\) reverses and changes due to the impulse. #### **Step 3: Calculating Change in Perpendicular Velocity** Before: \(v_{\perp, \text{before}} = -v \sin\theta\) (downward, into the plane) After: \(v_{\perp, \text{after}} = +v \sin\theta\) (upward, away from the plane) So, change in perpendicular velocity: \[ \Delta v_{\perp} = v_{\perp, \text{after}} - v_{\perp, \text{before}} = v \sin\theta - (-v \sin\theta) = 2v \sin\theta \] #### **Step 4: Applying Impulse-Momentum Theorem** \[ \Delta J = m \Delta v_{\perp} \] \[ 7.3 = 0.5 \times 2v \sin(30^\circ) \] \[ 7.3 = 0.5 \times 2v \times 0.5 \] \[ 7.3 = 0.5v \] \[ v = \frac{7.3}{0.5} = 14.6~\text{m/s} \] --- **Final Answer (a):** \[ \boxed{14.6~\text{m/s}} \] --- ### **b) Calculate the time between the collision and the ball reaching the highest point in its trajectory after the impact.** #### **Step 1: Velocity After the Collision** After the collision: - Perpendicular component: \(v_{\perp, \text{after}} = +v \sin\theta\) - Parallel component: \(v_{\parallel} = v \cos\theta\) The highest point is reached when the vertical component of velocity becomes zero. #### **Step 2: Calculating the Vertical Component After Collision** Resolve velocity after impact into vertical (\(y\)) direction: - The vertical axis makes an angle \(\theta\) with the perpendicular to the slope, so: \[ v_{y, \text{after}} = v_{\perp, \text{after}} \cos\theta + v_{\parallel} \sin\theta \] \[ = (v \sin\theta) \cos\theta + (v \cos\theta) \sin\theta \] \[ = v \sin\theta \cos\theta + v \cos\theta \sin\theta \] \[ = 2v \sin\theta \cos\theta \] \[ = v \sin(2\theta) \] Where we used the trigonometric identity \(\sin(2\theta) = 2 \sin\theta \cos\theta\). #### **Step 3: Maximum Height Condition** At the highest point, the vertical component of velocity becomes zero: \[ v_{y, \text{after}} - g t = 0 \] \[ t = \frac{v_{y, \text{after}}}{g} \] \[ t = \frac{v \sin(2\theta)}{g} \] #### **Step 4: Substitute Values** Given: - \(v = 14.6~\text{m/s}\) - \(\theta = 30^\circ\) - \(g = 9.81~\text{m/s}^2\) \[ \sin(2\theta) = \sin(60^\circ) = 0.866 \] \[ t = \frac{14.6 \times 0.866}{9.81} \] \[ t = \frac{12.6436}{9.81} \] \[ t \approx 1.29~\text{seconds} \] --- **Final Answer (b):** \[ \boxed{1.29~\text{seconds}} \] --- ### **Summary of Answers** **(a) Magnitude of initial velocity:** \[ \boxed{14.6\ \text{m/s}} \] **(b) Time to highest point after collision:** \[ \boxed{1.29\ \text{seconds}} \]