Two people,one twice as massive as the other,attempt a tug-of-war with 12 meters of massless rope on frictionless ice.After a brief time,they meet.The heavier person slides a distance of
A)3 m.
B)4 m.
C)6 m.
D)0 m.

Question

Quizplus · Accepted Answer

Since there is no friction, the heavier person will continue to slide until the tension in the rope completely disappears. Let the acceleration of the heavier person be a and the acceleration of the lighter person be 2a (since the heavier person has twice the mass).

Using Newton's second law, F = ma, we can set up equations for the forces on each person: 
For the heavier person: 
T - mg = ma 
For the lighter person: 
mg - T = 2ma

To find when the tension in the rope goes to zero, we can set these two equations equal to each other: 
ma + mg = 2ma 
mg = ma 
a = g

So the acceleration of the heavier person is equal to the acceleration due to gravity. We can then use the equation vf^2 = vi^2 + 2ad to find how far the heavier person slides: 
0 = v^2 + 2ad 
d = v^2/2a

Since the heavier person starts with zero velocity, we just need to find the final velocity. Using the equation v = at, we get: 
v = at = (g)(t)

Now we just need to find the time it takes for the rope to become taut. This is the time it takes for the lighter person to move the entire length of the rope, which is 12m. Using the equation x = 1/2at^2, we get: 
12 = 1/2(2a)t^2 
t = sqrt(6/g)

Plugging in this value for t, we can find the distance the heavier person slides: 
d = v^2/2a = (g)(6/g)/2(g) = 3m

Therefore, the correct answer is choice B.

Two People,one Twice as Massive as the Other,attempt a Tug-Of-War