Question

A wooden plank of length L rests on a frictionless floor. A boy of mass M now runs over the plank starting from its one end. If
M
mass of wooden plank is the distance covered by the boy relative to the ground will be​

Answer 1

Given:

Length of the wooden plank = L

Mass of the boy = M

Mass of wooden plank = M

To Find:

Distance covered by the boy over the wooden plank relative to the ground.

Solution:

Let, initially the boy stands at position x=0, length of plank be x=L

In the system,

CM of boy lies in x=0

CM of the wooden plank = $\frac{L}{2}$

∴ Initially center of mass of the system = $\frac{M\times 0+M\times\frac{L}{2} }{M+M}= \frac{L}{4}$

Now, let the boy moves distance s, then for this system

position of center of mass of the boy = s

and CM of plank will shift (L-s)

CM of the plank is= $\frac{L}{2}-(L-s) = s-\frac{L}{2}$

∴CM of this system = $\frac{M\times s+M\times (s-\frac{L}{2}) }{M+M}=\frac{2s-\frac{L}{2} }{2} = \frac{4s-L}{2}$

Here, no external force is present, hence CM will not change

∴$\frac{L}{4}=\frac{4s-L}{2}$

$=>L = 8s-2L$

$=>s=\frac{3L}{8}$

∴ Distance covered by the boy relative to the ground w.r.t. ground will be, $\bold{\frac{3L}{8}}$