Question

The man of mass m1 and the woman of mass m2 are

standing on opposite ends of the platform of mass m0 which
moves with negligible friction and is initially at rest with s = 0.
The man and woman begin to approach each other. Derive
an expression for the displacement s of the platform when
the two meet in terms of the displacement x1 of the man
relative to the platform.

Answer 1

To find the displacement of the platform when the man and woman meet, we can use the principle of conservation of momentum. We can express their momenta as m1v1 and -m2v2, where v1 is the velocity of the man and v2 is the velocity of the woman. By equating their momenta, we can solve for the displacement of the platform.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of momentum. The initial momentum of the system is zero since the platform is at rest. When the man and woman start moving towards each other, their momenta are equal in magnitude but opposite in direction.

We can express their momenta as m1v1 and -m2v2, where v1 is the velocity of the man and v2 is the velocity of the woman. According to the principle of conservation of momentum, the total initial momentum must be equal to the total final momentum, which is zero since the platform is at rest again. Therefore, we have:

(m1v1) + (-m2v2) = 0

Simplifying this equation, we get:

m1v1 = m2v2

Since the displacement s of the platform is equal to the combined displacement of the man and woman, we can write:

s = x1 - x2

where x1 is the displacement of the man relative to the platform and x2 is the displacement of the woman relative to the platform.