Physics Solution: Boat Displacement with Water Resistance

1. Physics Principles

To solve this problem, we must analyze the dynamics of the system consisting of the Boy and the Boat. The key concepts involved are:

• System Definition: We treat the Boy + Boat as a single system.
• Newton’s Second Law (Momentum form): The net external force on a system is equal to the rate of change of its total linear momentum ($F_{ext} = \frac{dP}{dt}$).
• Impulse-Momentum Theorem: The integral of force over time equals the change in momentum.
• Viscous Drag: The external force provided by the water is proportional to the velocity of the boat ($F = -kv$).

2. Visual Representation

3. Mathematical Derivation

Let us define the variables:

• $v_{boat}$: Velocity of the boat relative to the stationary water.
• $P_{system}$: Total linear momentum of the (Boy + Boat) system.
• $F_{ext}$: External force acting on the system.

The problem states that the water resistance is proportional to velocity. This is the only external horizontal force acting on the system (gravity and buoyancy cancel out vertically).

$$ F_{ext} = -k v_{boat} $$

From Newton’s Second Law for the system:

$$ F_{ext} = \frac{dP_{system}}{dt} $$

Substituting the force expression:

$$ -k v_{boat} = \frac{dP_{system}}{dt} $$

We know that velocity $v_{boat} = \frac{dx}{dt}$, where $x$ is the position of the boat.

$$ -k \frac{dx}{dt} = \frac{dP_{system}}{dt} $$

Now, we integrate this equation with respect to time, covering the entire duration of the event (from the moment the boy starts walking until the entire system comes to a halt).

$$ \int_{t=0}^{t_{end}} -k \frac{dx}{dt} dt = \int_{t=0}^{t_{end}} \frac{dP_{system}}{dt} dt $$

This simplifies to integrals over position ($x$) and momentum ($P$):

$$ -k \int_{x_i}^{x_f} dx = \int_{P_i}^{P_f} dP $$ $$ -k [x]_{initial}^{final} = [P_{system}]_{initial}^{final} $$

Let $s$ be the net displacement of the boat, so $s = x_f – x_i$.

$$ -k (s) = P_{final} – P_{initial} $$

Substituting these values into our equation:

$$ -k (s) = 0 – 0 $$ $$ -k s = 0 $$

Since $k$ (the proportionality constant) is non-zero, the displacement must be:

$$ s = 0 $$

4. Intuitive Explanation

Normally, if there were no friction, the boat would displace in the opposite direction of the boy’s walk to keep the Center of Mass stationary (resulting in a displacement of roughly 3.0 m).

However, in this specific case involving a resistive force proportional to velocity ($F \propto v$), the total impulse (change in momentum) imparted by the water resistance is directly proportional to the total displacement. Since the system starts at rest and ends at rest, the net change in momentum is zero. Therefore, the net impulse from the external force must be zero, which mathematically implies the net displacement is zero. The boat moves back while the boy accelerates, and forward while he decelerates, eventually returning to its original position.