A man of mass M stands at one end of a plank of length L which is at rest on a frictionless horizontal surface. The man walks to the other end of the plank. If mass of the plank is M/3, the distance that the man moves relative to ground is? Explain how pls...

The answer is L/4

initially the center of mass is at rest.

therefore the center of mass does not move in horizontal direction.

it is L/8 distance away from the end of the side of the man.

let x be the dispacement of the plank after the man walks to the other side.

then (L-x) will be the displacement of man.

let  initially the center of mass be at origin i.e x_i=0.

as we said earlier ,the center of mass will not move in horizontal direction therefore the center of mass remains stationary,

x_i=x_f

L/8=(m(L-x)+m/3(L/2-x))/(m+m/3)    ( )

thus we get x=3L/4,

therefore the man moved L-x=L/4.

Approved

Lets consider that the man moved "x" distance in ground reference ------1Consider man of mass "m" at the origin ---2If man wasent there the com of log would b at L/2 -------3Now by 2 and 3 we can conclude that the mass of the system is at L/8 ---------4 (by the formula of "m1L / m1 + m2" ...... Putting values {m/3 X L/2}÷ m+ m/3)Put the values in the formula of conservation of com i.e. com before = com after Com = {mx + m/3 (L-x)} ÷ m + m/3L/8 = {mx + m/3 (L-x)} ÷ 4m/3L/4 = x That`s the answer

Lets consider that the man moved "x" distance in ground reference ------1Consider man of mass "m" at the origin ---2If man wasent there the com of log would b at L/2 -------3Now by 2 and 3 we can conclude that the mass of the system is at L/8 ---------4 (by the formula of "m1L / m1 + m2" ...... Putting values {m/3 X L/2}÷ m+ m/3)Put the values in the formula of conservation of com i.e. com before = com after Com = {mx + m/3 (L-x)} ÷ m + m/3L/8 = {mx + m/3 (L-x)} ÷ 4m/3L/4 = x That`s the answer........

Dear Student,
Please find below the solution to your problem.

Let distance moved by the planck be x, then wrt ground the man will move by a distance L−x
Now the net displacement of the centre of mass will be zero in ground frame
M(L−x)=M∗x/3
x=3∗L/4
(L−x)=L/4

Thanks and Regards