# Problem Restatement A boy jumps off canoe \(A\) with a velocity relative to the, landing in canoe \(C\ Given data: - Boy's velocity relative to canoe \(A\): \(v_{BA} = 5.\, m/s\) - Mass of each canoe: \(m_A = m_C = 40\, kg\) - Boy's mass: \(m_B = 30\, kg\) - Girl's mass in canoe \(D\): \(m_D = 25\, kg\) (not directly involved in this calculation) - Both canoes are initially at rest. **Objective:** Find the final speed of canoe \(A\) after the boy jumps. --- # Assumptions & Approach - The system is isolated with no external horizontal forces. - The boy jumps from canoe \(A\) into canoe \(C\). - Conservation of momentum applies before and after the jump. --- # Step 1: Initial Momentum Initially, both canoes and the boy are at rest: \[ \text{Initial total momentum} = \] --- # Step 2: Final Conditions - The boy lands in canoe \(C\), which starts at rest. - Canoe \(A\) moves with velocity \(v_{A}\) after the jump. - Canoe \(C\) moves with velocity \(v_{C}\) after the jump. **Note:** Since the boy jumps from \(A\) to \(C\), conservation of momentum involves the boy and the two canoes. --- # Step 3: Momentum of the Boy Relative to the Canoe The boy jumps with velocity \(v_{BA} = 5.\, m/s\) relative to canoe \(A\). - Relative velocity of boy to canoe \(A\): \[ v_{B/A} = 5.\, m/s \] - Boy's velocity relative to the ground: \[ v_B = v_A + v_{B/A} \] *Note:* The sign depends on direction. Assume the boy jumps forward, so: \[ v_B = v_A + 5.\, m/s \] --- # Step 4: Momentum Conservation Initial total momentum: \[ \] Final total momentum: \[ \text{(momentum of canoe A)} + \text{(momentum of canoe C)} + \text{(momentum of boy in C)} + \text{(momentum of girl in D, if involved)} \] Since the girl in \(D\) is not involved, focus on \(A\), \(C\), and the boy. - Canoe \(A\) after jump: \[ p_A = m_A v_A \] - Canoe \(C\) after jump: \[ p_C = m_C v_C \] - Boy after jump: \[ p_B = m_B v_B = m_B (v_A + 5.) \] Total momentum after the jump: \[ m_A v_A + m_C v_C + m_B (v_A + 5.) = \] --- # Step 5: Relation between \(v_A\) and \(v_C\) **Key assumption:** The boy lands in canoe \(C\) with no bounce or external force, and the system is isolated. - Since the boy jumps from \(A\) to \(C\), and both canoes are initially at rest, the momentum exchange occurs between the canoes and the boy. - **In the frame of the ground:** The boy's initial velocity relative to the ground: \[ v_B = v_A + 5. \] The boy lands in canoe \(C\), imparting some momentum to \(C\). Conservation of momentum gives: \[ m_A v_A + m_C v_C + m_B (v_A + 5.) = \] --- # Step 6: Simplify the momentum equation Expressed as: \[ m_A v_A + m_C v_C + m_B v_A + m_B \times 5. = \] Group terms with \(v_A\): \[ (m_A + m_B) v_A + m_C v_C + 150 = \] (Note: \(m_B \times 5. = 30 \times 5 = 150\)) --- # Step 7: Additional relation between \(v_A\) and \(v_C\) Since the boy jumps from \(A\) to \(C\), and both are initially at rest, the relative velocity of the boy to \(C\): \[ v_{B/C} = v_B - v_C \] The problem states the boy lands in \(C\), implying: \[ v_B = v_C \] But from the jumping perspective, the boy's velocity relative to \(A\): \[ v_{B/A} = v_B - v_A = 5.\, m/s \] From the above: \[ v_B = v_A + 5. \] \[ v_B = v_C \] Set equal: \[ v_A + 5. = v_C \] --- # Step 8: Solve for \(v_A\) Using: \[ v_C = v_A + 5. \] Substitute into the momentum conservation: \[ (m_A + m_B) v_A + m_C (v_A + 5.) + 150 = \] Plugging in known masses: \[ (40 + 30) v_A + 40 (v_A + 5) + 150 = \] Simplify: \[ 70 v_A + 40 v_A + 200 + 150 = \] \[ (70 + 40) v_A + 350 = \] \[ 110 v_A + 350 = \] Solve for \(v_A\): \[ v_A = - \frac{350}{110} \approx -3.18\, m/s \] --- # **Final Speed of Canoe \(A\):** \[ \boxed{ v_A \approx -3.18\, \text{m/s} } \] The negative sign indicates the canoe \(A\) moves in the opposite direction to the initial jump. --- # **Answer:** **The final speed of canoe \(A\) after the boy jumps is approximately \(\boxed{3.18\, m/s}\) in magnitude, in the direction opposite to the boy's jump.**