Question

A boy of mass $m$ is standing at one end of a boat of mass $M$ and length $l$. The boy walks to the other end, the displacement of the

• A. CM is zero
• B. boat is $\frac{ml}{M+m}$
• C. boy is $\frac{ml}{M+m}$
• D. boat is $\frac{m}{M}l$

Answer 1

Answer: (A), (B)

CM is zero
boat is $\frac{ml}{M+m}$

Initially,

Cente of mass of the system$=\frac{m\times 0+M\times \frac{L}{2}}{m+M}=\frac{ML}{2(m+M)}$

Now let the boat displaces $x$ when boy moves from oneend to other end,

then the centre of mass about the same point will not change

as no External force will act on the boy boat system,

$\frac{ML}{2(m+M)}=\frac{m\times (L+x)+M\times (x+\frac{L}{2})}{m+M}$

$(m+M)x=-mL$

i.e (x is in opposite direction with a value of),

$x=\frac{mL}{m+M}$

x= displacement of boat $=\frac{mL}{m+M}$

displacement of boy is $=l-x=\frac{ML}{m+M}$

and also centre of mass of the system about any point is not diaplaced.

Hence, Option $\text{A,B}$ is the correct answer.