Two ice skaters start out motionless in the center of an ice rink. They then push each other apart. The
first skater (mass = 80 kg) moves to the right with a velocity of 2.5 m/s. The second skater (mass = 58 kg)
moves to the left at some unknown speed. If we assume that the ice is frictionless, then this is a case where
the total momentum would be conserved. Given that assumption, what is the velocity of the second skater?
P=(m +m)v
=(80kg +58kg)(2.5)
=345 kg. M/s
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant if no external forces act on it. In this case, the ice is assumed to be frictionless, so we can assume no external forces act on the two skaters.
The momentum (p) of an object is given by the product of its mass (m) and its velocity (v), i.e., p = mv.
Using this formula, we can calculate the momentum of the first skater:
p1 = m1 * v1
= 80 kg * 2.5 m/s
= 200 kg*m/s
Since the total momentum of the system is conserved, the total momentum before the skaters push each other apart should be equal to the total momentum after they push apart. Therefore, the total momentum of the system after the skaters push should also be 200 kg*m/s.
Let the unknown velocity of the second skater be v2. The momentum of the second skater is then given by:
p2 = m2 * v2
= 58 kg * v2
According to the conservation of momentum, the total momentum of the system after the skaters push is given by:
Total momentum = p1 + p2
= 200 kg*m/s + 58 kg * v2
Since the total momentum is conserved, this expression should be equal to 200 kg*m/s:
200 kg*m/s + 58 kg * v2 = 200 kg*m/s
Simplifying the equation:
58 kg * v2 = 0 kg*m/s
Dividing both sides by 58 kg:
v2 = 0 kg*m/s
Therefore, the velocity of the second skater after being pushed is 0 m/s.
To find the velocity of the second skater, we need to use the principle of conservation of momentum. This principle states that the total momentum before an event is equal to the total momentum after the event, as long as there are no external forces acting on the system.
In this case, the total initial momentum is zero since both skaters are motionless. After they push each other apart, we need to find the velocity of the second skater.
We can express the momentum of an object as the product of its mass and velocity (p = mv). Let's assign a positive direction to the right and a negative direction to the left. The initial momentum of the first skater is:
p1_initial = (mass1)(velocity1) = (80 kg)(2.5 m/s) = 200 kg·m/s (to the right)
Let's call the velocity of the second skater v2. The initial momentum of the second skater is:
p2_initial = (mass2)(velocity2) = (58 kg)(v2)
According to the conservation of momentum:
p1_initial + p2_initial = p1_final + p2_final
Since the first skater moves to the right and the second skater moves to the left, their velocities have opposite signs. The final momentum of the first skater is:
p1_final = (mass1)(velocity1) = (80 kg)(2.5 m/s) = 200 kg·m/s (to the right)
The final momentum of the second skater is:
p2_final = (mass2)(velocity2) = (58 kg)(-v2) = -58v2 kg·m/s (to the left)
Substituting the values into the conservation of momentum equation, we have:
200 kg·m/s + (58 kg)(v2) = 200 kg·m/s - 58v2 kg·m/s
Simplifying the equation, we get:
58v2 = -58v2
Dividing both sides by 58, we have:
v2 = -v2
Multiplying both sides by -1, we get:
-v2 = v2
This equation tells us that the velocity of the second skater is equal in magnitude but opposite in direction to the velocity of the first skater. Therefore, the velocity of the second skater is 2.5 m/s to the left.