Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another, as in Figure 7.10, and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides twice as far as skater 2. What is the ratio m1/m2 of their masses?

Let's denote the mass of skater 1 as m1 and the mass of skater 2 as m2.

Given that skater 1 glides twice as far as skater 2, we can relate their distances traveled using the following equation:

d1 = 2 * d2

Next, we can use the equations of motion to relate the distance, acceleration, and initial velocity for each skater. Assuming they start from rest, the equations of motion are as follows:

d = (1/2) * a * t^2
v = a * t

For skater 1:
d1 = (1/2) * a1 * t1^2

For skater 2:
d2 = (1/2) * a2 * t2^2

Since the magnitudes of their accelerations are equal when gliding to a halt, we have:

a1 = a2 = a (let's denote their equal value as a)

Now, let's relate the times for each skater. Since skater 1 glides twice as far, the time taken by skater 1 will be twice that of skater 2:

t1 = 2 * t2

Now, let's substitute the time relation into the distance equations:

d1 = (1/2) * a * (2 * t2)^2
d2 = (1/2) * a * t2^2

Simplifying these equations, we have:

d1 = 4 * (1/2) * a * t2^2
d1 = 2 * a * t2^2

We can equate this to the original relation between d1 and d2:

2 * a * t2^2 = 2 * d2

Canceling out the 2 and solving for d2, we have:

a * t2^2 = d2

Finally, using the relation d1 = 2 * d2 and substituting the above equation, we have:

2 * a * t2^2 = 2 * (a * t2^2)
2 * a = a
2 = 1

However, this equation is not possible. Therefore, there must be an error in the information provided or in the reasoning used in the solution of this problem.

Please double-check the given information and the calculations to resolve the discrepancy.

To solve this problem, we can use the principle of conservation of momentum and the equations of motion.

When the two skaters push against each other, they both experience an equal and opposite force. This causes them to move in opposite directions with different speeds. Let's denote the initial velocities of skater 1 and skater 2 as v1 and v2, respectively.

According to the principle of conservation of momentum, the total momentum before and after the skaters separate should remain the same. The initial momentum is given by:

m1 * v1 + m2 * v2

After separation, the skaters eventually come to a halt. This means their final velocities are zero. The momentum after separation is zero since momentum is mass times velocity.

So we have:

0 = m1 * 0 + m2 * 0

From this, we can conclude that the initial momentum of the system is zero:

m1 * v1 + m2 * v2 = 0

Since skater 1 glides twice as far as skater 2, we can relate the distances traveled by the skaters to their initial velocities and the time taken to come to a halt. Let's denote the distances traveled by skater 1 and skater 2 as d1 and d2, respectively.

We know that the distances traveled are related to the accelerations and time taken as follows:

d = (1/2) * a * t^2

Since the magnitudes of their accelerations are equal, we can write:

(1/2) * a * t1^2 = (1/2) * a * t2^2

Since skater 1 glides twice as far as skater 2:

d1 = 2 * d2

Substituting the expression for distance travelled into the equation relating distances to accelerations and time, we get:

(1/2) * a * t1^2 = 2 * (1/2) * a * t2^2

Simplifying the equation:

t1^2 = 4 * t2^2

Taking the square root of both sides:

t1 = 2 * t2

We can now substitute the expression for time taken into the equation relating initial momentum:

m1 * v1 + m2 * v2 = 0

(m1 * a * t1) + (m2 * (-a) * t2) = 0

Substituting the value of t1:

(m1 * a * 2 * t2) + (m2 * (-a) * t2) = 0

Dividing both sides by t2:

2 * m1 - m2 = 0

Rearranging the equation:

2 * m1 = m2

Therefore, the ratio m1/m2 of their masses is 2/1 or 2.

So, the ratio of m1 to m2 is 2:1.

If they decelerate at the same rate, and skater 1 goes twice as far, she also glides for a period that is twice as long. She must have started with twice the velocity of the other skater. V1 = 2 V2

Skater 1 and skater 2 has equal momenta at pushoff.

Therefore m1*V1 = m2*V2, and

m1/m2 = V2/V1 = 1/2