a 65 kg skateboarder is standing on a stationary .5 kg skateboard. the skateboarder steps off the skateboard with a horizontal veolicty of 1.5 m/s. what is the recoil velocity of the skateboard ignoring friction?
I need help with set up.
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, provided there are no external forces acting on the system.
In this case, the skateboarder and the skateboard can be considered as a system. Before the skateboarder steps off the skateboard, the total momentum is zero because both the skateboarder and the skateboard are stationary.
The momentum of an object is given by the product of its mass and velocity: momentum = mass × velocity.
Given:
- Mass of the skateboarder (m1) = 65 kg
- Mass of the skateboard (m2) = 0.5 kg
- Initial velocity of the skateboarder (v1) = 0 m/s (as they are stationary)
- Horizontal velocity of the skateboarder after stepping off (v2) = 1.5 m/s
To find the recoil velocity of the skateboard (v3), we can set up the conservation of momentum equation:
(mass1 × velocity1) + (mass2 × velocity2) = (mass1 × velocity3) + (mass2 × velocity3)
Plugging in the given values:
(65 kg × 0 m/s) + (0.5 kg × 1.5 m/s) = (65 kg × velocity3) + (0.5 kg × velocity3)
0 + 0.75 = 65 velocity3 + 0.5 velocity3
0.75 = 65.5 velocity3
To solve for velocity3, divide both sides of the equation by 65.5:
velocity3 = 0.75 / 65.5
Thus, the recoil velocity of the skateboard (ignoring friction) is approximately 0.0115 m/s.
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the skateboarder steps off the skateboard is equal to the total momentum after the skateboarder steps off.
Let's assume the positive direction is to the right.
The momentum before is given by the mass of the skateboarder (65 kg) multiplied by the initial horizontal velocity of the skateboarder (1.5 m/s).
Before: Momentum = (mass of skateboarder) x (initial velocity of skateboarder)
P_before = (65 kg) x (1.5 m/s)
The momentum after is given by the sum of the momentum of the skateboarder and the skateboard. The skateboard has a mass of 0.5 kg and an unknown recoil velocity (let's call it V_r).
After: Momentum = (mass of skateboarder) x (final velocity of skateboarder) + (mass of skateboard) x (recoil velocity of skateboard)
P_after = (65 kg) x (0 m/s) + (0.5 kg) x (V_r)
Since the total momentum is conserved, we have:
P_before = P_after
(65 kg) x (1.5 m/s) = (65 kg) x (0 m/s) + (0.5 kg) x (V_r)
Simplifying the equation:
97.5 kg·m/s = 0.5 kg x V_r
Now, divide both sides by 0.5 kg to solve for V_r:
V_r = 97.5 kg·m/s / 0.5 kg
V_r = 195 m/s
Therefore, the recoil velocity of the skateboard, ignoring friction, is 195 m/s.