# Step-by-Step Solution Let's break down the problem and solve for particle **B**'s velocity before and after the collision. --- ## **Given Data** - \( m_A = 3. \) kg - \( m_B = 1.5 \) kg ### **Before Collision** - \( v_{Ay1} = 5. \) m/s - \( v_{Ax1} = .9 \) m/s - \( v_{By1} = 6. \) m/s - \( v_{Bx1} = e = .6 \) (Assume \( v_{Bx1} \) is unknown, only \( e \) is given) ### **After Collision** - \( v_{Ax2} = -.06 \) m/s - \( v_{Ay2} \) = ? - \( v_{Bx2} \) = ? - \( v_{By2} \) = ? --- ## **Step 1: Equations of Motion** ### **Law of Conservation of Momentum** For two particles, \[ m_A v_{A1} + m_B v_{B1} = m_A v_{A2} + m_B v_{B2} \] Apply this separately for \( x \) and \( y \) directions. --- ### **X-Direction:** \[ m_A v_{Ax1} + m_B v_{Bx1} = m_A v_{Ax2} + m_B v_{Bx2} \] Plug in known values: \[ (3.)(.9) + (1.5) v_{Bx1} = (3.)(-.06) + (1.5) v_{Bx2} \] \[ 2.7 + 1.5 v_{Bx1} = -.18 + 1.5 v_{Bx2} \] --- ### **Y-Direction:** \[ m_A v_{Ay1} + m_B v_{By1} = m_A v_{Ay2} + m_B v_{By2} \] \[ (3.)(5.) + (1.5)(6.) = (3.)v_{Ay2} + (1.5)v_{By2} \] \[ 15. + 9. = 3. v_{Ay2} + 1.5 v_{By2} \] \[ 24. = 3. v_{Ay2} + 1.5 v_{By2} \] --- ## **Step 2: Coefficient of Restitution (e)** \[ e = \frac{v_{B2} - v_{A2}}{v_{A1} - v_{B1}} \] Apply **along y-direction** (since that's the axis of direct impact): \[ e = \frac{v_{By2} - v_{Ay2}}{v_{Ay1} - v_{By1}} \] \[ .6 = \frac{v_{By2} - v_{Ay2}}{5. - 6.} \] \[ .6 = \frac{v_{By2} - v_{Ay2}}{-1.} \] \[ v_{By2} - v_{Ay2} = -.6 \] \[ v_{By2} = v_{Ay2} - .6 \] --- ## **Step 3: Solve for Unknowns** ### **Y-Direction** Recall: \[ 24. = 3. v_{Ay2} + 1.5 v_{By2} \] Substitute \( v_{By2} \): \[ 24. = 3. v_{Ay2} + 1.5 (v_{Ay2} - .6) \] \[ 24. = 3. v_{Ay2} + 1.5 v_{Ay2} - .9 \] \[ 24. + .9 = 4.5 v_{Ay2} \] \[ 24.9 = 4.5 v_{Ay2} \] \[ v_{Ay2} = \frac{24.9}{4.5} = 5.53 \text{ m/s} \] Now, \[ v_{By2} = v_{Ay2} - .6 = 5.53 - .6 = 4.93 \text{ m/s} \] --- ### **X-Direction** Recall: \[ 2.7 + 1.5 v_{Bx1} = -.18 + 1.5 v_{Bx2} \] But we weren't given \( v_{Bx1} \) or \( v_{Bx2} \). Normally for head-on collisions, x-axis velocities don't change, or are both zero. If not specified, **assume \( v_{Bx1} = v_{Bx2} \) and both zero**. Thus, \[ 2.7 = -.18 \] This is not possible unless \( v_{Bx1} \) or \( v_{Bx2} \) is not zero. Since the problem doesn't provide them, we focus on the y-direction only. --- ## **Final Answers** ### **Before Collision:** - \( v_{By1} = 6. \) m/s (Given) ### **After Collision:** - \( v_{By2} = 4.93 \) m/s (Found above) --- ## **Summary Table** | Particle | Before Collision (\(v_y\)) | After Collision (\(v_y\)) | |----------|----------------------------|---------------------------| | A | 5. m/s | 5.53 m/s | | B | 6. m/s | 4.93 m/s | --- ## **Boxed Final Answer** \[ \boxed{ \begin{array}{l} \text{Particle B's y-velocity before collision: } v_{By1} = 6.\ \text{m/s} \\ \text{Particle B's y-velocity after collision: } v_{By2} = 4.93\ \text{m/s} \\ \end{array} } \] The x-velocities could not be uniquely determined with the given data. If you have further information about the x-component for B, it can be solved similarly.