Let's break down the solution step by step. --- ## **Step 1: Data and Variables** Given: - Mass of catapult, \( M = 30\, \text{kg} \) - Mass of projectile, \( m = 1.2\, \text{kg} \) - Spring constant, \( k = 350\, \text{N/m} \) - Initial height, \( h_i = .25\, \text{m} \) - Final height, \( h_f = 1\, \text{m} \) - Initial angle, \( \theta = 40^\circ \) - Gravity, \( g = 9.8\, \text{m/s}^2 \) --- ## **Step 2: Conservation of Energy** **Initial State (I):** - Projectile and catapult at rest (total kinetic energy = ) - Spring compressed (potential energy in spring) - Projectile at height \( h_i \) **Final State (II):** - Spring at rest length (no spring energy) - Projectile and catapult moving (kinetic energy) - Projectile at height \( h_f \) **Energy Conservation Equation:** \[ E_{\text{initial}} = E_{\text{final}} \] \[ \frac{1}{2} k x^2 + mgh_i = \frac{1}{2} m v_p^2 + \frac{1}{2} M v_c^2 + mgh_f \] where: - \( x \) is the spring compression, - \( v_p \) is the velocity of the projectile, - \( v_c \) is the velocity of the catapult. --- ## **Step 3: Conservation of Momentum** Since there is **no external force horizontally**: \[ M v_{c,x} + m v_{p,x} = \] Let’s define positive \( x \) as the direction along the incline. --- ## **Step 4: Geometry and Kinematics** The spring launches the projectile at angle \( \theta \), so: - Projectile velocity: \( v_p \) at angle \( \theta \) above horizontal. - Catapult velocity: \( v_c \) (horizontal since it moves on wheels). ### Velocity Components: - Projectile: \( v_{p,x} = v_p \cos\theta \), \( v_{p,y} = v_p \sin\theta \) - Catapult: \( v_{c,x} \) (horizontal), \( v_{c,y} = \) --- ## **Step 5: Find Spring Compression, \( x \)** From the geometry: - The spring is compressed along the incline: the vertical change is \( h_f - h_i = 1 - .25 = .75\, \text{m} \). - Along the incline, the compression \( x = \frac{.75}{\sin\theta} \). \[ x = \frac{.75}{\sin 40^\circ} = \frac{.75}{.6428} \approx 1.167\, \text{m} \] --- ## **Step 6: Solve Conservation Equations** ### **Conservation of Energy:** \[ \frac{1}{2} k x^2 + mgh_i = \frac{1}{2} m v_p^2 + \frac{1}{2} M v_c^2 + mgh_f \] Calculate each term: - \( \frac{1}{2} k x^2 = .5 \times 350 \times (1.167)^2 \approx 2383\, \text{J} \) - \( mgh_i = 1.2 \times 9.8 \times .25 = 2.94\, \text{J} \) - \( mgh_f = 1.2 \times 9.8 \times 1 = 11.76\, \text{J} \) Plug in: \[ 2383 + 2.94 = .5 \times 1.2 \times v_p^2 + .5 \times 30 \times v_c^2 + 11.76 \] \[ 2385.94 - 11.76 = .6 v_p^2 + 15 v_c^2 \] \[ 2374.18 = .6 v_p^2 + 15 v_c^2 \tag{1} \] --- ### **Conservation of Momentum (Horizontal):** \[ M v_{c,x} + m v_{p,x} = \] \[ 30 v_c + 1.2 v_p \cos 40^\circ = \] \[ v_c = -\frac{1.2}{30} v_p \cos 40^\circ = -.04 v_p \times .766 = -.03064 v_p \] --- ### **Substitute \( v_c \) into Energy Equation:** \[ 2374.18 = .6 v_p^2 + 15 (.03064 v_p)^2 \] \[ 2374.18 = .6 v_p^2 + 15 \times .000939 v_p^2 \] \[ 2374.18 = .6 v_p^2 + .014085 v_p^2 \] \[ 2374.18 = .614085 v_p^2 \] \[ v_p^2 = \frac{2374.18}{.614085} \approx 3865.6 \] \[ v_p \approx \sqrt{3865.6} \approx 62.18\, \text{m/s} \] --- ### **Find \( v_c \):** \[ v_c = -.03064 \times 62.18 \approx -1.906\, \text{m/s} \] --- ## **Step 7: Final Answers** ### **Projectile:** - **Magnitude:** \( v_p \approx 62.2\, \text{m/s} \) - **Direction:** \( 40^\circ \) above the horizontal ### **Catapult:** - **Magnitude:** \( v_c \approx 1.91\, \text{m/s} \) - **Direction:** Opposite to projectile's horizontal component (to the left, if the projectile goes right) --- ## **Summary Table** | Object | Velocity Magnitude (m/s) | Direction | |-------------|-------------------------|-----------------------------------| | Projectile | 62.2 | \( 40^\circ \) above horizontal | | Catapult | 1.91 | Horizontal, opposite to projectile| --- ### **Boxed Final Answers** \[ \boxed{ \begin{array}{l} \text{Projectile Velocity: } 62.2\, \text{m/s} \text{ at } 40^\circ \text{ above horizontal} \\ \text{Catapult Velocity: } 1.91\, \text{m/s} \text{ in opposite horizontal direction} \end{array} } \]