Question

a boat of mass 4m and length l is at rest on water surface. two men A and B of mass M and 2M are standing on the opposite ends of a boat now A travels l/4 relative to boat towards the center and B moves distance 3 l/4 to meat A find the displacement of boat on water when A and B meet

Answer 1

Answer:

The displacement of the boat will be $\frac{5l}{28}$

Explanation:

Mass of the boat = 4M

Mass of the person A = M

Mass of person B = 2M

Length of the boat = l

A moved l/4 towards the centre and B moves 3l/4 towards the centre w.r.t. the boat

If we take the position of A as origin then w.r.t. the origin, the displacement of the centre of mass will be zero

Initially the centre of mass of A is at 0 and B is at l

centre of mass of the system

$x_{cm}=\frac{M\times 0+2M\times l+4M\times (l/2)}{M+2M+4M}$

$\implies x_{cm}=4l/7$

Let the displacement of com of boat w.r.t. the ground be x

When A moves l/4 towards the centre then

displacement of A w.r.t. the ground from the origin = $x+l/4$

Similarly when B moved towards the centre then

Displacement of B w.r.t. the ground from the origin = $l-3l/4+x=l/4+x$

Distance of com of Boat from the origin = $l/2+x$

The com of the system will be at the same position

Hence

$\frac{4l}{7}=\frac{M(x+l/4)+2M(l/4+x)+4M(l/2+x)}{7M}$

$\implies 4l=x+l/4+2l/4+2x+2l+4x$

$\implies 4l=7x+3l/4+2l$

$\implies 7x=4l-\frac{3l}{4}-2l$

$\implies 7x=\frac{5l}{4}$

$\implies x=\frac{5l}{28}$

Therefore, the displacement of the boat will be $\frac{5l}{28}$

Hope this helps.