A skateboarder of mass 38.0 kg is riding her 3.90-kg skateboard at a speed of 5.40 m/s. She jumps backward off her skateboard, sending the skateboard forward at a speed of 8.5 m/s. At what speed is the skateboarder moving when her feet hit the ground?
To solve this problem, we can use the principle of conservation of linear momentum.
The linear momentum of an object is given by the product of its mass and velocity. The total linear momentum before the jump is equal to the total linear momentum after the jump since there is no external force acting on the system.
Let's denote the mass of the skateboarder as M and the mass of the skateboard as m. The initial velocity of the skateboarder is v1 and the initial velocity of the skateboard is V1. After the jump, the skateboarder's velocity becomes v2, and the skateboard's velocity becomes V2.
According to the conservation of linear momentum, we can write:
(M + m) * v1 + m * V1 = M * v2 + m * V2
We are given the masses and initial velocities of the skateboarder and the skateboard. Plugging in the values:
(38.0 kg + 3.90 kg) * 5.40 m/s + 3.90 kg * 0 m/s = 38.0 kg * v2 + 3.90 kg * 8.5 m/s
Simplifying the equation:
41.9 kg * 5.40 m/s = 38.0 kg * v2 + 33.15 kg * m/s
226.26 kg·m/s = 38.0 kg * v2 + 33.15 kg * m/s
To find the velocity of the skateboarder when her feet hit the ground (v2), we need to isolate v2 on one side of the equation. First, subtract 33.15 kg*m/s from both sides:
226.26 kg·m/s - 33.15 kg·m/s = 38.0 kg * v2
193.11 kg·m/s = 38.0 kg * v2
Now, divide both sides of the equation by 38.0 kg:
(193.11 kg·m/s) / 38.0 kg = v2
v2 ≈ 5.083 m/s
Therefore, when the skateboarder's feet hit the ground, she is moving at a speed of approximately 5.083 m/s.