**Step 1 : Given Data and Conceptual Introduction** - Mass of slider, \( m_1 = 0.4 \) kg - Mass of block, \( m_2 = 2 \) kg - Initial velocity of slider, \( v_1 = 5 \) m/s - Distance on ramp, \( d = 0.5 \) m - Angle of incline, \( \theta = 20^\circ \) - Coefficient of kinetic friction, \( \mu_k = 0.30 \) - Find: Final velocity \( v' \) of the system after the slider stops relative to the block. **Explanation block:** The problem involves a mass sliding up a frictionless surface, then up an incline with friction, and finally moving together with the block. The solution uses energy and momentum conservation concepts. --- **Step 2 : Step-by-Step Calculation (Energy and Work on Incline)** 1. **Loss of Energy Due to Friction and Gain in Height:** - Work by friction: \( W_f = \mu_k m_1 g \cos\theta \cdot d \) - Change in potential energy: \( \Delta U = m_1 g d \sin\theta \) - Initial kinetic energy: \( E_{k,1} = \frac{1}{2} m_1 v_1^2 \) - Final kinetic energy before sticking: \[ E_{k,2} = E_{k,1} - W_f - \Delta U \] 2. **Calculations:** - \( W_f = 0.30 \times 0.4 \times 9.8 \times \cos(20^\circ) \times 0.5 \approx 0.55 \) J - \( \Delta U = 0.4 \times 9.8 \times \sin(20^\circ) \times 0.5 \approx 0.67 \) J - \( E_{k,1} = \frac{1}{2} \times 0.4 \times 5^2 = 5 \) J - \( E_{k,2} = 5 - 0.55 - 0.67 = 3.78 \) J - Velocity before sticking: \[ E_{k,2} = \frac{1}{2} m_1 v_2^2 \implies v_2 = \sqrt{\frac{2 \times 3.78}{0.4}} \approx 4.35 \text{ m/s} \] **Explanation block:** Kinetic energy is reduced by both friction and the gain in height, so the final speed before collision with the block is less than the initial speed. All values are substituted to find the new kinetic energy and speed. --- **Step 3 : Step-by-Step Calculation (Conservation of Momentum during Inelastic Collision)** 1. **Apply Conservation of Momentum:** - Slider \( m_1 \) (velocity \( v_2 \)), block \( m_2 \) (initially at rest). - After sticking, both move together with velocity \( v' \): \[ m_1 v_2 + m_2 \cdot 0 = (m_1 + m_2) v' \] \[ v' = \frac{m_1 v_2}{m_1 + m_2} \] 2. **Insert values:** - \( v' = \frac{0.4 \times 4.35}{0.4 + 2} = \frac{1.74}{2.4} \approx 0.73 \) m/s **Explanation block:** Momentum is conserved because the system is isolated during the collision. The final velocity is calculated by dividing the product of mass and velocity of the moving object by the total mass. --- **Final answer:** The final velocity \( v' \) of the combined system is **0.73 m/s**.