A 65 kg skateboarder is standing on a stationary .5 kg skateboard. the skateboarder steps off the a skateboard with a horizontal veolicty of 1.5 m/s. what is the recoil velocity of the skateboard ignoring friction?
To determine the recoil velocity of the skateboard when the skateboarder steps off with a horizontal velocity of 1.5 m/s, we can apply the principle of conservation of momentum.
The momentum before the skateboarder steps off can be calculated by multiplying the mass of the skateboarder and their horizontal velocity:
Momentum before = Mass of skateboarder × Horizontal velocity of skateboarder
Momentum before = 65 kg × 1.5 m/s
The mass of the skateboard is 0.5 kg, and since the skateboarder steps off, the total mass after the skateboarder steps off includes only the skateboard:
Mass after = Mass of skateboard = 0.5 kg
According to the conservation of momentum principle, the total momentum before and after an event remains constant, provided that no external forces (like friction) act on the system.
Therefore,
Momentum before = Momentum after
(mass of skateboarder × horizontal velocity of skateboarder) = (mass after × recoil velocity of skateboard)
Plugging in the values we have:
65 kg × 1.5 m/s = 0.5 kg × recoil velocity of skateboard
Rearranging the equation to solve for the recoil velocity, we get:
Recoil velocity of skateboard = (65 kg × 1.5 m/s) / 0.5 kg
Recoil velocity of skateboard = 195 kg·m/s / 0.5 kg
Simplifying the expression, we find:
Recoil velocity of skateboard = 390 m/s
Therefore, the recoil velocity of the skateboard, disregarding friction, is 390 m/s.
To find the recoil velocity of the skateboard, we can use the principle of conservation of momentum. According to this principle, the total momentum before the skateboarder steps off the skateboard is equal to the total momentum after.
The momentum of an object is given by the product of its mass and velocity (p = m * v).
Before the skateboarder steps off, the total momentum is:
Total momentum before = (mass of skateboarder * velocity of skateboarder) + (mass of skateboard * velocity of skateboard)
Total momentum before = (65 kg * 1.5 m/s) + (0.5 kg * 0 m/s)
Total momentum before = 97.5 kg·m/s
Since there is no external force acting on the system after the skateboarder steps off (ignoring friction), the total momentum after will be the same.
Total momentum after = (mass of skateboard * velocity of skateboard)
Let's assume the recoil velocity of the skateboard is "v" m/s.
Total momentum after = (0.5 kg * v)
Equate the two equations for total momentum and solve for "v":
97.5 kg·m/s = 0.5 kg * v
Divide both sides of the equation by 0.5 kg:
v = 97.5 kg·m/s / 0.5 kg
v ≈ 195 m/s
Therefore, the recoil velocity of the skateboard, ignoring friction, is approximately 195 m/s.