Question

Two people, one twice as massive as the other, try to play tug of war on a nearly frictionless, level, icy surface. When they begin to tug, they are 12 m apart. They tug by pulling the rope hand over hand, until their hands meet along the rope.

How far does the heavier person slide?

Answer 1

Using conservation of momentum, the heavier person slides 4 meters towards the lighter person on a frictionless surface in a tug of war scenario.

Step-by-step explanation:

The question involves the concept of momentum conservation in physics. Since the icy surface is nearly frictionless, the forces exerted by the two people on each other through the rope are internal to the system, and the total momentum of the system must remain constant.

Let's denote the mass of the lighter person as m and that of the heavier as 2m. Assuming no external forces, the conservation of momentum dictates that the momentum before and after they start pulling must be equal.

Initially, both individuals are at rest, so the total momentum is zero. When they start to pull, if the lighter individual moves x meters towards the heavier one, conservation of momentum (momentum of heavier person = - momentum of lighter person ) gives us: 2m × (12 - x) = m × x.

Solving for x yields x = 8 meters. Since they start 12 meters apart, the heavier person would slide 12 - x = 4 meters before their hands meet. Therefore, the heavier person slides 4 meters towards the lighter person.