Two skaters are in the exact center of a circular frozen pond. Skater 1 pushes skater 2 off with a force of 100 N for 1.4 seconds. If skater 1 has a mass of 30 kg and skater 2 has a mass of 74 kg, a) what is the relative velocity (v1 - v2) after the push to the nearest hundredth of a m/s? b) After reaching the other shore, how fast, to the nearest tenth of a m/s, must skater 1 run around the lake to meet skater 2 at the opposite shore?
To answer these questions, we need to apply the principles of conservation of momentum and Newton's second law of motion.
a) To find the relative velocity (v1 - v2) after the push, we will use the conservation of momentum principle.
The momentum before the push is given by the sum of the individual momentum of skater 1 and skater 2, which is equal to zero since they are at rest:
momentum_before = m1 * v1 + m2 * v2 = 0
The momentum after the push is again the sum of their individual momentum:
momentum_after = m1 * v1' + m2 * v2'
Where v1' and v2' are the velocities of skater 1 and skater 2 after the push, respectively.
According to the conservation of momentum, momentum_before = momentum_after.
Since skater 1 pushes skater 2 off with a force of 100 N for 1.4 seconds, the impulse exerted on skater 2 is equal to the force multiplied by the time:
impulse = force * time = 100 N * 1.4 s
From Newton's second law of motion, we know that impulse is also equal to the change in momentum:
impulse = momentum_after - momentum_before
Substituting the values:
m1 * v1' + m2 * v2' = 100 N * 1.4 s
Now we need to solve the equations for v1' and v2'. From the given information, we know the masses of both skaters:
m1 = 30 kg
m2 = 74 kg
Rearranging the equation, we have:
30 kg * v1' + 74 kg * v2' = 140 kg·m/s
We can simplify this equation by dividing each term by 2 kg:
15 v1' + 37 v2' = 70
Now we have a system of equations. To solve them, we can use substitution. Since skater 1 pushes skater 2 off the center, skater 2 should have the opposite velocity to skater 1 (v2' = -v1'):
15 v1' + 37 (-v1') = 70
15 v1' - 37 v1' = 70
-22 v1' = 70
v1' = -70 / 22
Calculating this, we get:
v1' = -3.18 m/s (rounded to the nearest hundredth)
Since v2' = -v1', we have:
v2' = 3.18 m/s (rounded to the nearest hundredth)
Therefore, the relative velocity (v1 - v2) after the push is:
v1 - v2 = v1' - v2'
v1 - v2 = -3.18 m/s - 3.18 m/s = -6.36 m/s (rounded to the nearest hundredth)
b) To find out how fast skater 1 must run around the lake to meet skater 2 at the opposite shore, we need to apply the conservation of momentum again.
Since the momentum is conserved, the total momentum of the system after the push should be zero. Skater 1's momentum is equal to their mass (m1) multiplied by their velocity (v1'). Skater 2 is standing still, so their momentum is zero.
Therefore:
momentum_after = m1 * v1' + m2 * v2'
0 = 30 kg * v1' + 74 kg * 0
0 = 30 kg * v1'
Solving for v1':
30 kg * v1' = 0
v1' = 0 / 30 kg
v1' = 0 m/s
This means that skater 1 does not have to run around the lake to meet skater 2 at the opposite shore. They can stay still.
So the speed at which skater 1 must run around the lake is 0 m/s (rounded to the nearest tenth).