Let’s break down and solve the problem step by step: --- ## **Problem Restatement** - **Truck mass, \( m_t \):** \( 6000 \) kg - **Block mass, \( m_b \):** \( 1500 \) kg - The truck bed is raised to \( 30^\circ \). - **Frictionless** surface between block and bed; truck can roll freely. - Block slides **2 m** down the bed before hitting the tailgate. - Find **speed of truck and block:** - **(a)** *Immediately before* the block hits the tailgate. - **(b)** *Immediately after* the block hits the tailgate (*plastic impact*). --- ## **Step 1: Initial Setup and Diagrams** Let’s assign variables: - Let \( v_t \): velocity of the truck (relative to ground, right positive) - Let \( v_b \): velocity of the block (relative to ground, right positive) ### **1.1. Before block slides** - Both at rest: \( v_{t,0} = v_{b,0} = 0 \) --- ## **Step 2: Block Slides Down the Bed** ### **2.1. Block's Acceleration Down the Bed** On the **inclined bed**: - Distance, \( d = 2 \) m - Angle, \( \theta = 30^\circ \) - Gravity component down the incline: \( g \sin\theta \) - **Neglecting friction** By **conservation of momentum** (the truck can roll freely): Let’s define the system as: - Block slides left *relative to the truck* (but rightward in ground frame). - The truck will roll *backwards* (left) to conserve momentum. --- ### **2.2. Relative Motion Setup** Let: - \( v_{b/t} \): velocity of block **relative to truck** - \( v_t \): velocity of truck **relative to ground** (left negative) - \( v_b = v_t + v_{b/t} \): velocity of block relative to ground **Initial momentum (system at rest):** \[ (m_t + m_b) \times 0 = 0 \] **During sliding, conservation of momentum:** \[ m_t v_t + m_b v_b = 0 \implies m_t v_t + m_b (v_t + v_{b/t}) = 0 \] \[ (m_t + m_b)v_t + m_b v_{b/t} = 0 \] \[ v_t = -\frac{m_b}{m_t + m_b}v_{b/t} \] --- ### **2.3. Find \( v_{b/t} \) (Block's speed relative to truck)** Use **energy conservation** (block slides down 2 m): - Loss in potential energy \( = \) gain in kinetic energy (since the bed/truck is moving, both gain KE): \[ \Delta PE = \text{Total KE gained} \] \[ m_b g h = \frac{1}{2} m_t v_t^2 + \frac{1}{2} m_b v_b^2 \] Where \( h = d \sin\theta = 2 \times 0.5 = 1 \) m So: \[ m_b g (2 \sin 30^\circ) = \frac{1}{2} m_t v_t^2 + \frac{1}{2} m_b v_b^2 \] Recall: \( v_b = v_t + v_{b/t} \) and \( v_t = -\frac{m_b}{m_t + m_b} v_{b/t} \) So, \[ v_b = v_t + v_{b/t} = -\frac{m_b}{m_t + m_b} v_{b/t} + v_{b/t} = \left(1 - \frac{m_b}{m_t + m_b}\right) v_{b/t} = \frac{m_t}{m_t + m_b} v_{b/t} \] --- Plug these into the **energy equation**: \[ m_b g (2 \sin 30^\circ) = \frac{1}{2} m_t \left(-\frac{m_b}{m_t + m_b} v_{b/t}\right)^2 + \frac{1}{2} m_b \left(\frac{m_t}{m_t + m_b} v_{b/t}\right)^2 \] \[ = \frac{1}{2} m_t \frac{m_b^2}{(m_t + m_b)^2} v_{b/t}^2 + \frac{1}{2} m_b \frac{m_t^2}{(m_t + m_b)^2} v_{b/t}^2 \] \[ = \frac{1}{2} v_{b/t}^2 \left[ \frac{m_t m_b^2 + m_b m_t^2}{(m_t + m_b)^2} \right] \] \[ = \frac{1}{2} v_{b/t}^2 \left[ \frac{m_t m_b (m_t + m_b)}{(m_t + m_b)^2} \right] \] \[ = \frac{1}{2} v_{b/t}^2 \left[ \frac{m_t m_b}{(m_t + m_b)} \right] \] --- Now plug in values: - \( m_t = 6000 \) kg - \( m_b = 1500 \) kg - \( g = 9.81 \) m/s² - \( \sin 30^\circ = 0.5 \) \[ m_b g (2 \times 0.5) = \frac{1}{2} v_{b/t}^2 \left( \frac{m_t m_b}{m_t + m_b} \right) \] \[ 1500 \times 9.81 \times 1 = \frac{1}{2} v_{b/t}^2 \left( \frac{6000 \times 1500}{7500} \right) \] \[ 1500 \times 9.81 = \frac{1}{2} v_{b/t}^2 \times 1200 \] \[ 14715 = 600 v_{b/t}^2 \] \[ v_{b/t}^2 = \frac{14715}{600} \] \[ v_{b/t}^2 = 24.525 \] \[ v_{b/t} = \sqrt{24.525} \approx 4.95 \text{ m/s} \] --- ### **2.4. Find \( v_t \) and \( v_b \) (relative to ground)** Recall: \[ v_t = -\frac{m_b}{m_t + m_b} v_{b/t} = -\frac{1500}{7500} \times 4.95 = -0.2 \times 4.95 = -0.99 \text{ m/s} \] (The minus sign means *leftward*.) \[ v_b = v_t + v_{b/t} = -0.99 + 4.95 = 3.96 \text{ m/s} \] --- ## **Step 3: After Plastic Impact with Tailgate** - **Plastic impact:** Block and truck move together after collision. Let the **final velocity** (after impact) be \( v_f \). **Conservation of momentum:** \[ m_t v_t + m_b v_b = (m_t + m_b) v_f \] Plug in values: \[ 6000 \times (-0.99) + 1500 \times 3.96 = (7500) v_f \] \[ -5940 + 5940 = 7500 v_f \] \[ 0 = 7500 v_f \] \[ v_f = 0 \text{ m/s} \] So, **after the plastic impact, both come to rest**! --- ## **Final Answers** ### **(a) Immediately before the block hits the tailgate:** - **Truck speed:** \( \boxed{-0.99 \text{ m/s}} \) (leftward) - **Block speed:** \( \boxed{3.96 \text{ m/s}} \) (rightward) ### **(b) Immediately after the block hits the tailgate (plastic impact):** - **Both (truck and block):** \( \boxed{0 \text{ m/s}} \) --- ### **Summary Table** | Situation | Truck (\(v_t\), m/s) | Block (\(v_b\), m/s) | |----------------------------------------------|----------------------|----------------------| | Immediately before block hits tailgate | -0.99 (left) | +3.96 (right) | | Immediately after (plastic collision) | 0 | 0 | Let me know if you need any clarifications or further explanation!