Question

(i) In a boat of mass 4 $M$ and length $l$ on a frictionless water surface. Two men $A$ (mass = $M$) and $B$ (mass $2M$) are standing on the two opposite ends. Now $A$ travels a distance $l/4$ relative to boat towards its centre and $B$ moves a distance $3l/4$ relative to boat and meet $A$. Find the distance travelled by the boat on water till $A$ and $B$ meet.

• A. The distance travelled by the boat on water is $5l/28$.
• B. The distance travelled by the boat on water is $l/4$.
• C. The distance travelled by the boat on water is $3l/4$.
• D. The distance travelled by the boat on water is $l/7$.

Answer 1

Answer: (A)

The distance travelled by the boat on water is $\frac{5l}{28}$.

Answer 2

The problem can be solved using the principle of conservation of the center of mass. Since the water surface is frictionless, there are no external horizontal forces acting on the system (boat + men). Therefore, the center of mass of the system remains stationary.

Let the total mass of the system be $M_{total} = 4M (\text{boat}) + M (\text{man A}) + 2M (\text{man B}) = 7M$.

We set up a coordinate system. Let the initial position of the left end of the boat be $x=0$. The boat has length $l$. Initial positions:

• Boat's center of mass: $x_{boat,i} = l/2$.
• Man A's initial position: $x_{A,i} = 0$ (assuming A is at the left end).
• Man B's initial position: $x_{B,i} = l$ (assuming B is at the right end).

The initial position of the center of mass of the system ($X_{COM,i}$) is: $X_{COM,i} = \frac{(4M)(l/2) + (M)(0) + (2M)(l)}{4M + M + 2M} = \frac{2Ml + 0 + 2Ml}{7M} = \frac{4Ml}{7M} = \frac{4l}{7}$

Now, let the boat move a distance $d$ to the right. The final position of the boat's center of mass will be $x_{boat,f} = l/2 + d$.

Man A travels a distance $l/4$ relative to the boat towards its center. If A starts at the left end (position 0 relative to the boat's left end), his new position relative to the boat's left end is $l/4$. The absolute position of Man A will be: $x_{A,f} = (\text{final position of boat's left end}) + (\text{A's position relative to boat's left end}) = d + l/4$.

Man B moves a distance $3l/4$ relative to the boat and meets A. This means Man B also ends up at the same position relative to the boat's left end as Man A, which is $l/4$. The absolute position of Man B will be: $x_{B,f} = d + l/4$.

The final position of the center of mass of the system ($X_{COM,f}$) is: $X_{COM,f} = \frac{(4M)(l/2 + d) + (M)(d + l/4) + (2M)(d + l/4)}{7M}$ $X_{COM,f} = \frac{2Ml + 4Md + Md + Ml/4 + 2Md + 2Ml/4}{7M}$ $X_{COM,f} = \frac{2Ml + 4Md + Md + Ml/4 + 2Md + Ml/2}{7M}$ Combine terms: $X_{COM,f} = \frac{(2Ml + Ml/4 + Ml/2) + (4Md + Md + 2Md)}{7M}$ $X_{COM,f} = \frac{(8Ml/4 + Ml/4 + 2Ml/4) + 7Md}{7M}$ $X_{COM,f} = \frac{11Ml/4 + 7Md}{7M}$ $X_{COM,f} = \frac{11l/4 + 7d}{7} = \frac{11l}{28} + d$

Since the center of mass of the system remains stationary ($X_{COM,i} = X_{COM,f}$): $\frac{4l}{7} = \frac{11l}{28} + d$

Solving for $d$: $d = \frac{4l}{7} - \frac{11l}{28}$ $d = \frac{16l}{28} - \frac{11l}{28}$ $d = \frac{5l}{28}$

The distance travelled by the boat on water is $d = \frac{5l}{28}$.