# 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?

## X m.=Distance the heavier person moves

2x m.= Lighter person distance.

x + 2x = 12
3x = 12
X = 4 m.

## Well, the answer to this question is going to feel a little slippery. The force exerted by each person in a tug of war depends on their mass, so let's call the mass of the lighter person M, and the mass of the heavier person 2M.

Now, because the surface is icy and frictionless, it means that the only force acting on the heavier person will be... gravity! That's right, gravity will be trying its best to give them a one-way ticket to the Winter Olympics bobsled team.

So, in this icy tug of war, the heavier person won't slide at all! Or, to be more precise, they will slide a grand total of zero meters. They'll be in the same spot, as if they were doing some sort of chilly shuffling dance move.

But don't worry, the lighter person won't have to do all the work. They can show off their strength by pulling the heavier person closer to them, while sharing some laughs and trying not to accidentally turn the tug of war into an impromptu ice-skating competition.

## To find out how far the heavier person slides, we need to use the principle of conservation of momentum.

Since the surface is nearly frictionless, we can assume that there are no external forces acting on the system. This means that the total momentum of the system before the tug of war is equal to the total momentum after the hands meet along the rope.

Let's assume the mass of the lighter person is m, and the mass of the heavier person is 2m.

At the start, the lighter person is at one end of the 12 m distance, while the heavier person is at the other end. The total initial momentum is given by:

initial momentum = (mass of lighter person) × (velocity of lighter person) + (mass of heavier person) × (velocity of heavier person)

Since both persons are initially at rest, the initial velocity of both persons is 0. Therefore, the initial momentum can be simplified to:

initial momentum = 0 × 0 + (2m) × 0 = 0

Now, let's consider the situation after their hands meet. At this point, the lighter person has slid a certain distance, while the heavier person has not moved. So, the total final momentum is given by:

final momentum = (mass of lighter person) × (velocity of lighter person) + (mass of heavier person) × (velocity of heavier person)

Since the heavier person does not move, their velocity is 0. Therefore, the final momentum can be simplified to:

final momentum = (m) × (velocity of lighter person) + (2m) × 0 = m × (velocity of lighter person)

Since the total momentum is conserved, the initial momentum must be equal to the final momentum:

initial momentum = final momentum
0 = m × (velocity of lighter person)

This implies that the velocity of the lighter person is 0. In other words, the lighter person does not slide at all. Therefore, the heavier person slides the entire 12 m distance.

## To find out how far the heavier person slides, we need to understand the concept of center of mass and conservation of momentum.

Let's denote the mass of the lighter person as m1 and the mass of the heavier person as m2. Given that the heavier person is twice as massive as the lighter person, we can write the mass relationship as m2 = 2m1.

Considering a nearly frictionless, level, icy surface, we can assume that there is no external force acting on the system once they start pulling the rope. This means the total momentum before and after they start tugging remains constant.

Initially, the two people are 12 meters apart. As they pull the rope hand over hand, they move towards each other. At some point, their hands meet along the rope. Let's call the distance covered by the heavier person as x meters.

Since there is no external force, the total momentum before and after remains the same. The total momentum of a system of particles is defined as the sum of the individual momenta. Mathematically, this can be expressed as:
(mass1 x velocity1) + (mass2 x velocity2) = (mass1 x final_velocity1) + (mass2 x final_velocity2)

In this situation, since the surface is nearly frictionless, the velocity of both individuals will be zero initially and final_velocity2 will be zero because the heavier person stops once their hands meet. Hence, we can simplify the equation to:
mass1 x 0 + mass2 x 0 = mass1 x 0 + mass2 x final_velocity2

Now, substituting mass2 = 2mass1 and the fact that initial velocities are zero, we get:
2mass1 x 0 = mass1 x 0 + 2mass1 x final_velocity2

This further simplifies to:
0 = 0 + 2mass1 x final_velocity2

Since mass1 is non-zero, the only way for the equation to hold true is for final_velocity2 to be zero. This means the heavier person stops as soon as their hands meet.

Therefore, the heavier person does not slide at all.