Distance a platform moves when a person walks on it

[songoku]

Homework Statement

A metal platform of mass M and length L is placed on a frictionless floor. A man of mass m walks on the platform from the left end to the right end of the platform. Let left end of platform denoted as point O. How much does the platform move on the floor with respect to O?

Homework Equations

conservation of momentum

d = v.t

The Attempt at a Solution

Using conservation of momentum:
total momentum of system before the man walks = total momentum of system after the man walks
0 = m.v₁ + M.v₂

v₂ = m.v₁ / M

When the person walks to the right, the platform will move to the left so the time needed for the man to reach the right end:
t = d / v = L / (v₁ + v₂)

Distance moved by the platform with respect to O
= v₂ . t
= m.v₁ / M . L / (v₁ + v₂)
= m.v₁ / M . L / (v₁ + m.v₁ / M)
= m/(m + M) . L

Is this correct? Thanks

 

[tnich]

Homework Statement

A metal platform of mass M and length L is placed on a frictionless floor. A man of mass m walks on the platform from the left end to the right end of the platform. Let left end of platform denoted as point O. How much does the platform move on the floor with respect to O?

Homework Equations

conservation of momentum

d = v.t

The Attempt at a Solution

Using conservation of momentum:
total momentum of system before the man walks = total momentum of system after the man walks
0 = m.v₁ + M.v₂

v₂ = m.v₁ / M

When the person walks to the right, the platform will move to the left so the time needed for the man to reach the right end:
t = d / v = L / (v₁ + v₂)

Distance moved by the platform with respect to O
= v₂ . t
= m.v₁ / M . L / (v₁ + v₂)
= m.v₁ / M . L / (v₁ + m.v₁ / M)
= m/(m + M) . L

Is this correct? Thanks

That looks right.

 

[Orodruin]

The question has failed to state explicitly the crucial assumption that the man and platform are initially at rest.

An alternative solution, which relies on the same principle is that in the absence of external forces, the centre of mass will be in rectilinear motion (and therefore at rest in this case). From this you directly obtain $m x + My = 0$, where $x$ is the distance covered by the man and $y$ the distance covered by the platform. The fact that the man should move a distance $L$ relative to the platform gives you $x = L+y$. Thus
$$
m(L+y) + My = 0 \quad \Longrightarrow \quad y = -\frac{mL}{M+m}.
$$
(Note that I put the positive direction to be the direction in which the man walks.)

This method completely avoids your assumption of constant velocity, it is not needed.

 

[songoku]

The question has failed to state explicitly the crucial assumption that the man and platform are initially at rest.

An alternative solution, which relies on the same principle is that in the absence of external forces, the centre of mass will be in rectilinear motion (and therefore at rest in this case). From this you directly obtain $m x + My = 0$, where $x$ is the distance covered by the man and $y$ the distance covered by the platform. The fact that the man should move a distance $L$ relative to the platform gives you $x = L+y$. Thus
$$
m(L+y) + My = 0 \quad \Longrightarrow \quad y = -\frac{mL}{M+m}.
$$
(Note that I put the positive direction to be the direction in which the man walks.)

This method completely avoids your assumption of constant velocity, it is not needed.

Wow I never learned this thing before. In what chapter should I learn about this? Or maybe you have suggestion where I can learn about this ?

Thanks