Let's analyze the problem step-by-step. --- # Problem Summary: - Boy \( B \) jumps from canoe \( A \) to canoe \( C \). - Initial velocity of \( B \) relative to canoe \( A \): \( v_{B/A} = 5\, \text{m/s} \). - The angle of jump: \(30^\circ\). - Mass of boy \( B \): \( 30\, \text{kg} \). - Mass of girl \( D \): \( 25\, \text{kg} \). - Both canoes are initially at rest. - The goal: Find the final speeds of both canoes after the boy jumps. --- # Assumptions: - The canoes are only affected by the boy's jump (no external horizontal forces). - The canoes are free to move horizontally. - The jump is instantaneous, and momentum is conserved during the jump. --- # Step 1: Understand the initial conditions - Before the jump: - Both canoes are at rest. - Total initial momentum: zero. - During the jump: - The boy exerts a force on canoe \( A \) to propel himself. - Since the system is isolated (no external horizontal forces), total momentum is conserved. --- # Step 2: Find the velocity of the boy relative to ground - The boy's velocity relative to canoe \( A \): \( 5\, \text{m/s} \) at \(30^\circ\). - To find the velocity in ground frame \( v_{B} \): \[ v_{B} = v_{B/A} + v_{A} \] But initially, the canoe \( A \) is at rest, so \( v_A = \). - The velocity of the boy relative to ground: \[ v_{B} = 5\, \text{m/s} \quad \text{at} \quad 30^\circ \] - Components: \[ v_{B_x} = 5 \cos 30^\circ = 5 \times \frac{\sqrt{3}}{2} \approx 4.33\, \text{m/s} \] \[ v_{B_y} = 5 \sin 30^\circ = 5 \times .5 = 2.5\, \text{m/s} \] Since the jump is in the horizontal direction, we focus on the horizontal component \( v_{B_x} \). --- # Step 3: Apply conservation of momentum - Before the jump: \[ \text{Total momentum} = \] - After the jump: \[ m_B v_{B_x} + m_A v_{A_{f}} + m_D v_{D_{f}} = \] where: - \( m_B = 30\, \text{kg} \), - \( v_{B_x} \approx 4.33\, \text{m/s} \), - \( v_{A_{f}} \) = final velocity of canoe \( A \), - \( m_A \) = mass of canoe \( A \). (Assuming canoe's mass is negligible or can be considered negligible, as it's not given. Alternatively, if the problem suggests the canoes are massless, the momentum transfer is only between the boy and the canoes.) Similarly, for the girl \( D \): - The girl remains in canoe \( D \), which is initially at rest. - When the boy jumps to canoe \( C \), the canoes move in response. --- # Step 4: Find the final velocities - The key is that the boy jumps from canoe \( A \) to \( C \). - The final velocities of the canoes can be found considering the momentum transfer. --- # Final step: **Conclusion based on the problem's typical approach** Given the nature of the problem, the standard approach involves: - Recognizing that the boy's horizontal velocity relative to the ground is \( 4.33\, \text{m/s} \). - The total horizontal momentum before is zero, so after the jump, the total must still be zero: \[ m_A v_{A_{f}} + m_B v_{B} + m_D v_{D_{f}} = \] - Since the girl \( D \) is initially at rest and remains in her canoe, and assuming the canoes are massless or have negligible mass, the key is: \[ m_B v_{B_x} + m_A v_{A_{f}} = \] which yields: \[ v_{A_{f}} = - \frac{m_B}{m_A} v_{B_x} \] But without the mass of canoe \( A \), we typically assume the canoes are massless or that the system is simplified to the boy's motion. --- # **Summary of the solution:** - **Horizontal velocity of the boy relative to ground**: \[ v_{B_x} \approx 4.33\, \text{m/s} \] - **Final velocity of canoe \( A \)**: \[ v_{A} = - \frac{m_B}{m_A} v_{B_x} \] - **Final velocity of canoe \( C \)**: - Since the boy lands on canoe \( C \), conservation of momentum applies again when he lands, and the same principles are used. --- # **Final answer:** The **final speeds** of both canoes depend on their masses and the detailed momentum exchange, but with the given data, the primary horizontal velocity of the boy after jump is approximately **4.33 m/s** in the original direction. --- **Note:** To get precise final velocities, more data about the mass of the canoes or additional assumptions are needed.

# Theory Behind the Problem ## Conservation of Momentum The principle of conservation of momentum is fundamental in physics, especially in analyzing collisions and interactions in closed systems. It states that the total momentum of an isolated system remains constant if no external forces act upon it. ### Key Concepts: 1. **Momentum**: Defined as the product of mass and velocity (\( p = mv \)). 2. **Isolated System**: A system with no external forces acting on it, allowing the momentum before an event (like a jump) to equal the momentum after. ## Application in the Problem In this problem: - The boy jumps off from one canoe to another. - Both canoes are initially at rest, meaning their initial momentum is zero. - When the boy jumps, he exerts a force on the first canoe (canoe \( A \)), causing it to move in the opposite direction to conserve momentum. ### Steps Involved: 1. **Identify Initial Conditions**: Both canoes and the boy are at rest. 2. **Calculate the Boy's Velocity**: Decompose the boy's jump into x (horizontal) and y (vertical) components to understand his motion. 3. **Apply Conservation of Momentum**: Use the total momentum before and after the jump to find the final velocities of both canoes. ### Equations Used: - The horizontal component of the boy's velocity: \[ v_{B_x} = v_{B/A} \cos(\theta) \] - The conservation of momentum equation: \[ m_B v_{B_x} + m_A v_{A_{f}} + m_D v_{D_{f}} = 0 \] where \( m_A \) and \( m_D \) are the masses of the canoes and the girl, respectively. ## Implications - The direction of the boy's jump influences the velocities of both canoes. - The final speeds of the canoes can be determined based on their masses and the velocity of the boy. - The problem illustrates the practical application of conservation laws in dynamics. ### Conclusion Understanding the underlying principles of momentum conservation allows for solving complex motion problems in physics. By analyzing forces and motions separately, one can predict the outcomes of interactions in a system.