Part A

The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, the two skaters pushing off against each other is an isolated system.

The momentum before the push is zero because they are stationary. After the push, the total momentum should still be zero.

The momentum of an object is given by the product of its mass and velocity (p=mv).

Let's denote:
m1 = mass of the heavier skater = 765 N / 9.8 m/s^2 = 78.06 kg (we divide by 9.8 to convert weight into mass)
v1 = velocity of the heavier skater = 1.30 m/s

m2 = mass of the lighter skater = 625 N / 9.8 m/s^2 = 63.78 kg
v2 = velocity of the lighter skater (this is what we're trying to find)

Since the total momentum before and after the push should be the same and equal to zero, we have:

m1*v1 = m2*v2

Substituting the known values:

78.06 kg * 1.30 m/s = 63.78 kg * v2

Solving for v2 gives:

v2 = (78.06 kg * 1.30 m/s) / 63.78 kg = 1.59 m/s

So, the lighter skater will travel at 1.59 m/s.

Part B

The kinetic energy (K) of an object is given by the formula K = 1/2 * m * v^2. The total kinetic energy of the system after the push is the sum of the kinetic energies of the two skaters.

K = 1/2 * m1 * v1^2 + 1/2 * m2 * v2^2
= 1/2 * 78.06 kg * (1.30 m/s)^2 + 1/2 * 63.78 kg * (1.59 m/s)^2
= 65.95 J + 80.67 J
= 146.62 J

So, 146.62 joules of kinetic energy is "created" during the skaters' maneuver.

Part C

The kinetic energy of the system was produced by the work the two skaters do on each other. When they push off against each other, they do work against the force of friction, which is converted into kinetic energy.

Question

Part A

The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, the two skaters pushing off against each other is an isolated system.

The momentum before the push is zero because they are stationary. After the push, the total momentum should still be zero.

The momentum of an object is given by the product of its mass and velocity (p=mv).

Let's denote:
m1 = mass of the heavier skater = 765 N / 9.8 m/s^2 = 78.06 kg (we divide by 9.8 to convert weight into mass)
v1 = velocity of the heavier skater = 1.30 m/s

m2 = mass of the lighter skater = 625 N / 9.8 m/s^2 = 63.78 kg
v2 = velocity of the lighter skater (this is what we're trying to find)

Since the total momentum before and after the push should be the same and equal to zero, we have:

m1*v1 = m2*v2

Substituting the known values:

78.06 kg * 1.30 m/s = 63.78 kg * v2

Solving for v2 gives:

v2 = (78.06 kg * 1.30 m/s) / 63.78 kg = 1.59 m/s

So, the lighter skater will travel at 1.59 m/s.

Part B

The kinetic energy (K) of an object is given by the formula K = 1/2 * m * v^2. The total kinetic energy of the system after the push is the sum of the kinetic energies of the two skaters.

K = 1/2 * m1 * v1^2 + 1/2 * m2 * v2^2
= 1/2 * 78.06 kg * (1.30 m/s)^2 + 1/2 * 63.78 kg * (1.59 m/s)^2
= 65.95 J + 80.67 J
= 146.62 J

So, 146.62 joules of kinetic energy is "created" during the skaters' maneuver.

Part C

The kinetic energy of the system was produced by the work the two skaters do on each other. When they push off against each other, they do work against the force of friction, which is converted into kinetic energy.

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