This question can be explained using Newton's Third Law of Motion and the principle of conservation of momentum. When a man jumps from a boat, he exerts a force on the boat, and the boat exerts an equal and opposite force on him.
Identify the system: Consider the man and the boat as a combined system initially at rest. The total momentum of this system is zero.
\[ P_{\text{initial}} = 0 \]Apply Newton's Third Law: When the man jumps towards the shore, he pushes the boat backward (away from the shore) with a certain force. According to Newton's Third Law, the boat simultaneously pushes the man forward (towards the shore) with an equal and opposite force.
\[ F_{\text{man on boat}} = -F_{\text{boat on man}} \]Apply Conservation of Momentum: Since there are no external horizontal forces acting on the man-boat system (neglecting water resistance and air resistance for simplicity), the total momentum of the system must remain conserved. As the initial momentum is zero, the final momentum must also be zero.
\[ P_{\text{final}} = P_{\text{man}} + P_{\text{boat}} = 0 \]If the man moves towards the shore with a momentum \( P_{\text{man}} = m_{\text{man}} v_{\text{man}} \), then the boat must move in the opposite direction with an equal and opposite momentum \( P_{\text{boat}} = m_{\text{boat}} v_{\text{boat}} \).
\[ m_{\text{man}} v_{\text{man}} + m_{\text{boat}} v_{\text{boat}} = 0 \] \[ m_{\text{boat}} v_{\text{boat}} = -m_{\text{man}} v_{\text{man}} \]The negative sign indicates that the velocity of the boat (\(v_{\text{boat}}\)) is in the opposite direction to the velocity of the man (\(v_{\text{man}}\)). Since the man jumps towards the shore, the boat moves away from the shore.
Correct Option: B) Moves away from the shore
When the man jumps towards the shore, he applies a force on the boat in the opposite direction (away from the shore). Due to Newton's Third Law and the conservation of momentum, the boat will move backward, i.e., away from the shore.
This question tests understanding of Newton's Third Law of Motion, which describes the nature of action and reaction forces. We need to recall the key characteristics of these force pairs.
Correct Option: C) Act on different bodies
This is a fundamental aspect of Newton's Third Law. If you push a wall, your hand exerts a force on the wall (action), and the wall exerts an equal and opposite force on your hand (reaction). The action is on the wall, and the reaction is on your hand.
This problem can be solved using the principle of conservation of momentum. In an isolated system (man + boat), the total momentum before the man jumps is equal to the total momentum after he jumps. Since the man and boat are initially at rest, the initial momentum is zero.
The total momentum before the jump is equal to the total momentum after the jump.
\[ (m_m + m_b)u = m_m v_m + m_b v_b \]Since the initial velocity \(u = 0\), the left side of the equation becomes zero.
\[ (60 \text{ kg} + 240 \text{ kg}) \times 0 \text{ m/s} = (60 \text{ kg} \times 4 \text{ m/s}) + (240 \text{ kg} \times v_b) \]\[ 0 = 240 \text{ kg} \cdot \text{m/s} + 240 \text{ kg} \cdot v_b \]Rearrange the equation to find \(v_b\).
\[ -240 \text{ kg} \cdot \text{m/s} = 240 \text{ kg} \cdot v_b \]\[ v_b = \frac{-240 \text{ kg} \cdot \text{m/s}}{240 \text{ kg}} \]\[ v_b = -1 \text{ m/s} \]The negative sign indicates that the boat moves in the opposite direction to the man's jump, which is expected for recoil. The magnitude of the recoil velocity is \(1 \text{ m/s}\).Correct Option: B) 1 m/s
This problem can be solved using the principle of conservation of momentum. According to this principle, in an isolated system, the total momentum before an event (like firing a bullet) is equal to the total momentum after the event. Since the gun and bullet system is initially at rest, the total initial momentum is zero. Therefore, the total momentum after firing must also be zero.
The mass of the bullet is given in grams, which needs to be converted to kilograms to match the unit of the gun's mass.
\[ m = 25 \text{ g} = \frac{25}{1000} \text{ kg} = 0.025 \text{ kg} \]Let \(V\) be the recoil velocity of the gun. The initial momentum of the gun-bullet system is zero (as both are at rest). After firing, the total momentum is the sum of the momentum of the gun and the momentum of the bullet.
\[ M \times V + m \times v = 0 \]Rearrange the equation to find \(V\):
\[ M \times V = - m \times v \]\[ V = - \frac{m \times v}{M} \]D) 2.5 m/s is the correct magnitude of the recoil velocity of the gun, calculated using the conservation of momentum.
This problem tests the understanding of Newton's third law of motion and the concept of forces within a closed system. When a bird flies, it exerts a downward force on the air, and the air, in turn, exerts an upward force on the bird. The key is to consider the entire cage-bird-air system.
Correct Option: C) Remain the same. The total mass within the closed system (cage + bird + air) does not change. When the bird flies, the force it exerts on the air to stay aloft is transferred to the cage through the air, effectively replacing the direct contact force it had when sitting. Thus, the total weight measured by an external scale remains constant.
To find the momentum of a body, we use the fundamental formula that defines momentum as the product of its mass and velocity. We are given the mass and velocity, so we will directly apply this formula.
Momentum (\(p\)) is defined as the product of mass (\(m\)) and velocity (\(v\)).
\[ p = m \times v \]The unit of momentum is the product of the unit of mass (kilogram, kg) and the unit of velocity (meter per second, m/s), which is kg·m/s.
D) 10 kg·m/s — The calculation \(2 \text{ kg} \times 5 \text{ m/s} = 10 \text{ kg} \cdot \text{m/s}\) correctly yields this result.