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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...


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...


Grade:12th Pass

6 Answers

Swapnil Saxena
102 Points
9 years ago

The answer is L/4

G Narayana Raju
51 Points
9 years ago

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 xi=0.

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

xi=xf

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.

JATIN
11 Points
4 years ago
Its simple....
just apply formula,
 
Let the one point be origin (0,0)
so,
m1x1+m2x2/m1+m2
= 0+ML/3/4M/3
L/4
 
Udbhav Sinha
28 Points
4 years ago
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
Udbhav Sinha
28 Points
4 years ago
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........
Rishi Sharma
askIITians Faculty 646 Points
one year ago
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

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