Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another, as shown below, 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 3 times as far as skater 2. What is the ratio m1 / m2 of their masses?
Two ice skaters stand at rest in the center of an ice rink. When they push off against each 45-kg skater acquires a speed of 0.62 m/s. If the speed of the other skater is 0.89 m/s, what is that skater's mass?
Well, it sounds like these ice skaters have gone through quite the experience! Let's break this down and have a little fun with it.
Since skater 1 glides 3 times as far as skater 2, we can call the distance traveled by skater 2 as "x." According to the problem, skater 1 will then glide a distance of "3x."
Now, we know that the magnitudes of their accelerations are equal. Since we're dealing with frictional forces, let's just say that their accelerations can be represented by "a."
Using the distance, acceleration, and the good old kinematic equation, we can come up with a funny equation:
3x = (1/2) * a * t^2
x = (1/2) * a * t^2
Since we're not given any specific values for distance or time, we can just have fun with this equation. Why not replace "x" with "laughs" and "t^2" with "pun"? So, our equation becomes:
3laughs = (1/2) * a * pun
laughs = (1/6) * a * pun
Now let's bring in the masses. We're looking for the ratio m1 / m2. Let's assign "m1" the value of "jokes" and "m2" the value of "punchlines."
We'll represent the ratio as "jokes / punchlines."
According to Newton's third law, the forces that the skaters exert on each other are equal and opposite. So, we can say:
m1 * a = m2 * a
Well, since "a" is on both sides of the equation, we can cancel it out. Now our equation becomes:
jokes = punchlines
So, the ratio m1 / m2 is 1:1, or simply put, equal mass for our skaters.
Now, I hope this funny explanation brings a smile to your face. Remember, physics can be a ton of fun, even when it involves ice skaters and their comedic antics.
Let's denote the mass of skater 1 as m1 and the mass of skater 2 as m2.
From the given information, we know that initially, both skaters are stationary, so their initial velocities are 0.
When they push against each other, they move in opposite directions. Let's say skater 1 moves to the right with a velocity v1 and skater 2 moves to the left with a velocity v2.
Since skater 1 glides 3 times as far as skater 2 before coming to a stop, we can say that the distance covered by skater 1 (d1) is 3 times the distance covered by skater 2 (d2).
Using the equations of motion, we can relate the distances, velocities, and accelerations:
For skater 1:
d1 = (1/2) * a1 * t^2
where a1 is the acceleration and t is the time taken to come to a stop.
For skater 2:
d2 = (1/2) * a2 * t^2
where a2 is the acceleration and t is the same time taken as skater 1.
Since we are told that the magnitudes of their accelerations are equal:
|a1| = |a2| = a (let's assume positive values for simplicity)
From the equations above, we can write:
d1 = (1/2) * a * t^2
d2 = (1/2) * a * t^2
Since d1 = 3 * d2, we can substitute this into the equation:
3 * d2 = (1/2) * a * t^2
We can cancel out the t^2 and simplify the equation:
3 * d2 = (1/2) * a * t^2
6 * d2 = a * t^2
We can also relate the distances and velocities using time and acceleration:
v1 = a * t
v2 = -a * t (negative because of opposite direction)
We can write an equation for the distances in terms of velocities:
d1 = v1 * t
d2 = v2 * t
Since d1 = 3 * d2:
v1 * t = 3 * (v2 * t)
Now, let's relate the velocities and masses using momentum conservation:
The initial total momentum is 0 since both skaters are stationary:
m1 * 0 + m2 * 0 = m1 * v1 + m2 * v2
0 = m1 * v1 + m2 * v2
Now, we have two equations:
v1 * t = 3 * (v2 * t)
0 = m1 * v1 + m2 * v2
We can simplify the equations by canceling out the t term:
v1 = 3 * v2
0 = m1 * v1 + m2 * v2
Substituting v1 = 3 * v2 into the second equation:
0 = m1 * (3 * v2) + m2 * v2
0 = 3 * m1 * v2 + m2 * v2
0 = (3 * m1 + m2) * v2
Since v2 is not zero, we can divide both sides by v2:
0 = 3 * m1 + m2
Solving for m1:
3 * m1 = -m2
m1 = -m2 / 3
Therefore, the ratio of m1 to m2 is:
m1 / m2 = (-m2 / 3) / m2 = -1/3
Hence, the ratio of their masses is -1/3.
To solve this problem, let's break it down step by step:
STEP 1: Analyze the situation
- Two ice skaters initially stationary
- They push against each other and move in opposite directions
- Neglect kinetic friction while they are pushing
- After separation, kinetic friction slows them down until they come to a halt
- Magnitudes of their accelerations are equal
- Skater 1 glides three times as far as skater 2
STEP 2: Define known quantities
- Let m1 be the mass of skater 1
- Let m2 be the mass of skater 2
- Let d1 be the distance skater 1 glides
- Let d2 be the distance skater 2 glides
STEP 3: Use equations of motion
Since the magnitudes of their accelerations are equal, we can use the equations of motion to relate the distances and acceleration.
For skater 1:
- d1 = (1/2) * a * t^2 (Equation 1)
For skater 2:
- d2 = (1/2) * a * t^2 (Equation 2)
But since skater 1 glides three times as far as skater 2:
- d1 = 3 * d2 (Equation 3)
Note: We don't have to consider the initial velocities since they are initially stationary.
STEP 4: Solve for the ratio of their masses
To find the ratio of their masses (m1/m2), we need to eliminate 'a' and 't' from the equations.
From Equation 1 and Equation 2, we can equate the time and acceleration terms:
- (1/2) * a * t^2 = (1/2) * a * t^2
Since the time and acceleration terms cancel out, we can simplify Equation 3 as:
- 3 * d2 = d1
Substituting d1 = 3 * d2 into Equation 2:
- 3 * d2 = (1/2) * a * t^2
Eliminate 't' by combining Equation 1 and Equation 2:
- (1/2) * a * t^2 = (3/2) * a * t^2
Now, we have:
- 3 * d2 = (3/2) * a * t^2
Divide both sides by t^2 (assuming t^2 is non-zero):
- d2 = (1/2) * a (Equation 4)
From Equation 4, we see that d2 is directly proportional to the acceleration 'a'.
Since we know that the magnitudes of their accelerations are equal, we conclude that skater 2 (m2) must have twice the mass of skater 1 (m1).
Thus, the ratio of their masses is:
- m1 / m2 = 1 / 2 = 1:2
Their initial velocities are in a ratio V1/V2 = m2/m1
Their kinetic friction forces are in a ratio
F1/F2 = M1/M2
Gliding time t is proportional to
V/a = V m/F
Gliding distance D is proportional to
V*t = V^2*m/F
D1/D2 = 3 = [V1^2*m1/F1)][m2/(F2*V2^2)]
= (V1/V2)^2*(m1/m2)*F2/F1
= (m2/m1)^2
m2/m1 = 3^1/2 = 1.732
m1/m2 = 0.577