91Ó°ÊÓ

An isolated system is one where no external forces influence the motion of the objects within the system. In the context of our problem, the interaction involves only the man and the stone. It's crucial to note that no other forces, such as friction or air resistance, act upon them. This makes it an isolated system.

When dealing with momentum conservation, ensuring the system is isolated is essential. This isolation guarantees that all momentum changes occur solely due to internal forces.

The total momentum of this system remains constant over time, which means that any action within the system (like the man pushing the stone) will not change the total momentum of the system. Instead, it merely redistributes it among the components of our system — the man and the stone.

The conservation of momentum principle is elegantly captured in the momentum equation. Mathematically, it can be expressed as:

• Before the shoving: \[ m_{man} \times v_{man, initial} + m_{stone} \times v_{stone, initial} = 0 \]

Initially, both the man and the stone are at rest, so their velocities are zero, making the total initial momentum zero.

• After the shoving: \[ m_{man} \times v_{man} + m_{stone} \times v_{stone} = 0 \]

This equation indicates that the momentum of one object is balanced by the momentum of the other, maintaining the total momentum at zero. Here, the negative sign signifies that the two momenta are in opposite directions.

As a result, we get the equation:
\[ m_{man} \times v_{man} = - m_{stone} \times v_{stone} \]
This equation shows the relationship between the masses and velocities of the two bodies post-interaction. It's essential for solving problems involving momentum conservation.

To find the man's velocity after he shoves the stone, we use the momentum equation derived above. First, we substitute the known values into the equation:

• Mass of man, \(m_{man}\) = 91 kg
• Mass of stone, \(m_{stone}\) = 0.068 kg
• Velocity of stone, \(v_{stone}\) = 4.0 m/s

Next, we rewrite the equation incorporating the given values:
\[ 91 \times v_{man} = - (0.068 \times 4.0) \]
Simplifying inside the parenthesis:
\[ 91 \times v_{man} = -0.272 \]
Now, we solve for \(v_{man}\) by dividing both sides by the mass of the man:
\[ v_{man} = - \frac{0.272}{91} \]
Performing the division gives:
\[ v_{man} = -0.00299 \text{ m/s} \]
This negative sign indicates that the man's velocity is in the opposite direction to the stone's motion. This conclusion aligns with our expectation from the conservation of momentum in an isolated system. The man's velocity is tiny compared to the stone's because of his much greater mass.