# Problem Breakdown - Boy \( B \) jumps off canoe \( A \) with a velocity relative to canoe \( A \): \( U_{B/A} = 4.5\, \text{m/s} \) - Mass of each canoe: \( m_A = m_C = 40\, \text{kg} \) - Mass of boy: \( m_B = 30\, \text{kg} \) - Initial state: both canoes at rest - Goal: Find the final velocity of canoe \( A \) after the boy jumps off. --- # Assumptions and Principles - **Conservation of Momentum**: Since no external horizontal forces are acting, total momentum before and after jumping is conserved. - **Relative Velocity**: The boy's velocity relative to canoe \( A \) is given. --- # Step 1: Define Initial Conditions Initial total momentum: \[ P_{initial} = \] since everything is at rest. --- # Step 2: Determine the Boy's Velocity relative to ground The boy jumps with velocity \( U_{B/A} = 4.5\, \text{m/s} \) relative to canoe \( A \). Let: - \( v_A \) = velocity of canoe \( A \) after the boy jumps. - \( v_B \) = velocity of the boy relative to ground after jumping. Since the boy's velocity relative to canoe \( A \) is \( 4.5\, \text{m/s} \), and the canoe's velocity after the jump is \( v_A \): \[ v_B = v_A + U_{B/A} \] or \[ v_B = v_A + 4.5 \] --- # Step 3: Apply Conservation of Momentum Initial total momentum: \[ \] Final total momentum: \[ m_A v_A + m_B v_B + m_C v_C \] but since the boy jumps into canoe \( C \), which is initially at rest, and the problem seems to involve only the initial jump, let's clarify: - The boy jumps **off** canoe \( A \) into **canoe \( C \)**. - Canoe \( C \) is initially at rest and catches the boy. Thus, **after the jump**: - Canoe \( A \) has velocity \( v_A \) - Canoe \( C \) has velocity \( v_C \) - The boy has velocity \( v_B \) --- # Step 4: Momentum conservation equations Initial total momentum: \[ \] Final total momentum: \[ m_A v_A + m_C v_C + m_B v_B \] Since no external forces: \[ = m_A v_A + m_C v_C + m_B v_B \] --- # Step 5: Relationship between velocities - The boy jumps **from** canoe \( A \), with \( U_{B/A} = 4.5\, \text{m/s} \). - The boy lands in canoe \( C \). Assuming the boy's velocity relative to ground: \[ v_B = v_A + 4.5 \] - Canoe \( C \) is initially at rest, and after catching the boy, it moves with velocity \( v_C \). If the boy lands in canoe \( C \): \[ v_B = v_C \] since the boy and canoe \( C \) move together after the landing. --- # Step 6: Write the momentum conservation in terms of \( v_A \) and \( v_C \) \[ = m_A v_A + m_C v_C + m_B v_B \] but \( v_B = v_C \), so: \[ = m_A v_A + m_C v_C + m_B v_C \] \[ = m_A v_A + (m_C + m_B) v_C \] \[ m_A v_A = - (m_C + m_B) v_C \] --- # Step 7: Find \( v_C \) in terms of \( v_A \) \[ v_C = - \frac{m_A v_A}{m_C + m_B} \] Using known masses: \[ m_A = 40\, \text{kg} \] \[ m_B = 30\, \text{kg} \] \[ m_C = 40\, \text{kg} \] Plugging in: \[ v_C = - \frac{40 v_A}{40 + 30} = - \frac{40 v_A}{70} = - \frac{4 v_A}{7} \] --- # Step 8: Express \( v_B \) in terms of \( v_A \) From earlier: \[ v_B = v_A + 4.5 \] But after landing, \( v_B = v_C \): \[ v_C = v_B = v_A + 4.5 \] From the previous step: \[ v_C = - \frac{4 v_A}{7} \] Set equal: \[ v_A + 4.5 = - \frac{4 v_A}{7} \] --- # Step 9: Solve for \( v_A \) Rearranged: \[ v_A + \frac{4 v_A}{7} = -4.5 \] Express \( v_A \) terms over common denominator: \[ \frac{7 v_A}{7} + \frac{4 v_A}{7} = -4.5 \] \[ \frac{11 v_A}{7} = -4.5 \] Multiply both sides by 7: \[ 11 v_A = -4.5 \times 7 \] \[ 11 v_A = -31.5 \] Finally: \[ v_A = - \frac{31.5}{11} \approx -2.864\, \text{m/s} \] --- # **Final Answer:** \[ \boxed{ \text{The final velocity of canoe \( A \)} \approx \mathbf{-2.86\, \text{m/s}} } \] The negative sign indicates that canoe \( A \) moves in the opposite direction to the initial jump relative to the ground.