Let's break down the problem step by step: ### Problem Restatement A \( 15 \)-lb block **B** starts from rest and slides down a \( 25 \)-lb wedge **A** (angle \( 30^\circ \)), which is on a smooth (frictionless) horizontal surface. **Neglecting friction**, determine the velocity of **A** as **B** reaches the bottom. --- ### Step 1: **Concepts Used** - **Conservation of momentum** (horizontal direction, since no external horizontal forces) - **Conservation of energy** (since friction is neglected) --- ### Step 2: **Let’s Define Variables** - Mass of \( B \): \( m_B = \frac{15}{32.2} = 0.466 \) slugs - Mass of \( A \): \( m_A = \frac{25}{32.2} = 0.777 \) slugs - Let \( v_A \) = velocity of wedge A (to the right) - Let \( v_B \) = velocity of block B (relative to ground) at the bottom - Block **slides a vertical height** \( h \) (not given, but will cancel out) --- ### Step 3: **Conservation of Horizontal Momentum** Initially, both are at rest, so total initial horizontal momentum = 0. Let’s say **positive x** is to the right. At the bottom: - Block B has horizontal velocity component \( v_{Bx} \) - Wedge A has velocity \( v_A \) to the right \[ m_B v_{Bx} + m_A v_A = 0 \] \[ v_{Bx} = -\frac{m_A}{m_B} v_A \] --- ### Step 4: **Relative Velocity** Block **B** moves **down the incline** with respect to wedge **A**. Let \( v_{B/A} \) = velocity of B **relative to** A (along the incline). The **absolute velocity** of block B (relative to ground): \[ \vec{v}_B = \vec{v}_A + \vec{v}_{B/A} \] Decompose \( \vec{v}_{B/A} \) into x and y components: - \( v_{B/A,x} = v_{B/A} \cos(30^\circ) \) (to the right) - \( v_{B/A,y} = -v_{B/A} \sin(30^\circ) \) (downward) So, \[ v_{Bx} = v_A + v_{B/A}\cos(30^\circ) \] --- ### Step 5: **Conservation of Energy** The total mechanical energy is conserved: - Initial potential energy of B: \( m_B g h \) - At bottom: all energy becomes kinetic (both objects) \[ m_B g h = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B (v_{Bx}^2 + v_{By}^2) \] --- ### Step 6: **Express All Velocities in Terms of \( v_A \)** From earlier: \[ v_{Bx} = v_A + v_{B/A} \cos(30^\circ) \] But from momentum: \[ v_{Bx} = -\frac{m_A}{m_B} v_A \] So: \[ -\frac{m_A}{m_B} v_A = v_A + v_{B/A} \cos(30^\circ) \] \[ v_{B/A} \cos(30^\circ) = -\frac{m_A}{m_B} v_A - v_A = -\left(\frac{m_A}{m_B} + 1\right) v_A \] \[ v_{B/A} = -\frac{\left(\frac{m_A}{m_B} + 1\right)v_A}{\cos(30^\circ)} \] For vertical component: \[ v_{By} = -v_{B/A} \sin(30^\circ) \] --- ### Step 7: **Plug Into Energy Equation** \[ m_B g h = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B \left[ v_{Bx}^2 + v_{By}^2 \right] \] But the block slides down the incline a **vertical height** \( h \). Let’s assume the incline length is such that the block drops a height \( h \). Let’s express everything in terms of \( v_A \): 1. \( v_{B/A} \) as above 2. \( v_{Bx} = -\frac{m_A}{m_B} v_A \) 3. \( v_{By} = -v_{B/A} \sin(30^\circ) \) Calculate \( v_{B/A} \): \[ \frac{m_A}{m_B} = \frac{0.777}{0.466} = 1.668 \] \[ v_{B/A} = -\frac{1.668 + 1}{\cos(30^\circ)} v_A = -\frac{2.668}{0.866} v_A = -3.08 v_A \] So, \[ v_{By} = -(-3.08 v_A)\cdot 0.5 = 1.54 v_A \] \[ v_{Bx} = -1.668 v_A \] Plug into energy: \[ m_B g h = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B \left[ (-1.668 v_A)^2 + (1.54 v_A)^2 \right] \] \[ (-1.668)^2 = 2.784 \] \[ (1.54)^2 = 2.371 \] \[ 2.784 + 2.371 = 5.155 \] So, \[ m_B g h = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B \cdot 5.155 v_A^2 \] \[ m_B g h = \frac{1}{2} v_A^2 (m_A + 5.155 m_B) \] \[ v_A^2 = \frac{2 m_B g h}{m_A + 5.155 m_B} \] --- ### Step 8: **Final Answer (In Terms of h)** \[ \boxed{v_A = \sqrt{ \frac{2 m_B g h}{m_A + 5.155 m_B} } } \] If you know the vertical height \( h \) that block B descends, **plug in \( h \) and calculate the final numerical answer**. --- #### **If Block B Slides Entire Height of Wedge** If you want the answer in terms of the wedge height \( h \), just plug in the values: - \( m_B = 0.466 \) slugs - \( m_A = 0.777 \) slugs - \( g = 32.2 \) ft/s² \[ v_A = \sqrt{ \frac{2 \cdot 0.466 \cdot 32.2 \cdot h}{0.777 + 5.155 \cdot 0.466} } \] \[ 0.777 + 2.404 = 3.181 \] \[ v_A = \sqrt{ \frac{29.997 h}{3.181} } \] \[ v_A = \sqrt{9.44 h} \] --- ### **Summary** - The velocity of A when B reaches the bottom is \[ \boxed{v_A = \sqrt{9.44 h} \text{ ft/s}} \] where \( h \) is the **vertical height** block B descends. If you are given a specific height, plug it in for \( h \) to get the final number!