two figure skaters, one weighting 625N and the other 725N,push off against each other on frictionless ice. a) if the heavest skater travels at 1.50m/s, how fast will the lighter one travel? b)how much kinetic energy is created during the skaters manauver, and where does this energy come from?
a) The net momentum will remain zero. They will leave in opposite drections.
(725/g)kg* 1.50m/s + (625/g)V = 0
V is the speed of the ligther skater. Solve for it. The acceleration of gravity (g) can be cancelled out.
V = -(725/625)*1.50 m/s
The minus sign denotes going in the opposite direction.
a) To solve this problem, we can use the principle of conservation of momentum. The initial momentum of the system, before the skaters push off, is zero because there is no motion. The final momentum of the system, after the skaters push off, is also zero because the skaters are moving in opposite directions.
According to the conservation of momentum, the total momentum before the push-off is equal to the total momentum after the push-off. Therefore, the product of the mass and velocity of the heavier skater is equal to the product of the mass and velocity of the lighter skater.
Mathematically, we can express this as:
(mass of heavier skater)(velocity of heavier skater) = (mass of lighter skater)(velocity of lighter skater)
Let's denote the mass of the heavier skater as m1, the mass of the lighter skater as m2, the velocity of the heavier skater as v1, and the velocity of the lighter skater as v2. We are given the values of m1 (725 N), v1 (1.50 m/s), and we need to find v2.
Using the above equation, we can solve for v2:
(m1)(v1) = (m2)(v2)
(725 N)(1.50 m/s) = (625 N)(v2)
v2 = (725 N)(1.50 m/s) / (625 N)
v2 = 1.74 m/s
Therefore, the lighter skater will travel at a speed of 1.74 m/s.
b) To calculate the kinetic energy created during the skaters' maneuver, we can use the formula:
Kinetic energy = 0.5 * mass * velocity^2
The total kinetic energy created during the maneuver is equal to the sum of the kinetic energies of the two skaters. Let's denote the mass of the heavier skater as m1, the mass of the lighter skater as m2, the velocity of the heavier skater as v1, and the velocity of the lighter skater as v2. We are given the values of m1 (725 N), v1 (1.50 m/s), m2 (625 N), and v2 (1.74 m/s).
The kinetic energy of the heavier skater is given by:
Kinetic energy 1 = 0.5 * m1 * v1^2
The kinetic energy of the lighter skater is given by:
Kinetic energy 2 = 0.5 * m2 * v2^2
The total kinetic energy is:
Total kinetic energy = Kinetic energy 1 + Kinetic energy 2
Plugging in the given values, we can calculate the total kinetic energy.
Please provide the mass of the heavier skater (m1).
To solve this problem, we can use the concept of conservation of momentum. The total momentum before the push-off is equal to the total momentum after the push-off.
a) We can calculate the speed of the lighter skater using the principle of conservation of momentum. The formula for momentum (p) is given by p = m * v, where m is the mass of the object and v is its velocity.
Given:
Mass of the heavier skater (m1) = 625 N
Mass of the lighter skater (m2) = 725 N
Velocity of the heavier skater (v1) = 1.50 m/s
The total initial momentum is given by:
Initial momentum = m1 * v1 + m2 * v2
Let v2 be the final velocity of the lighter skater.
Now, we can equate the initial momentum to the final momentum:
m1 * v1 + m2 * v2 = total momentum
Substituting the given values:
625 N * 1.50 m/s + 725 N * v2 = total momentum
Simplifying the equation, we get:
937.5 N * m/s + 725 N * v2 = total momentum
To solve for v2, isolate v2 on one side of the equation:
725 N * v2 = total momentum - 937.5 N * m/s
v2 = (total momentum - 937.5 N * m/s) / 725 N
b) The kinetic energy created during the skaters' maneuver is a result of the work done on the skaters' bodies. In this case, the work is done against the friction between the skaters' skates and the ice. Since the ice is frictionless, no work is done, and thus no energy is created during the maneuver.
Therefore, the kinetic energy created during the skaters' maneuver is zero, and the energy does not come from any external source.