Answer:
12 m/s
Step-by-step explanation:
This problem involves the principle of conservation of momentum. According to this principle, the total momentum of a system of objects is conserved in the absence of external forces. We can use this principle to solve the problem as follows:
The total momentum of the system consisting of the skateboard and child before the child jumps off is given by:
p1 = m1v1 + m2v2
Where m1 and v1 are the mass and velocity of the skateboard, m2, and v2 are the mass and velocity of the child, respectively.
Substituting the given values, we have:
p1 = (10 kg)(6.0 m/s) + (30 kg)(0 m/s)
p1 = 60 kg·m/s
After the child jumps off, the momentum of the skateboard is conserved and becomes:
p2 = m1v1'
Where v1' is the velocity of the skateboard after the child jumps off.
The momentum of the child is given by:
p3 = m2v3
Where v3 is the velocity of the child after jumping off.
Using the conservation of momentum principle, we have:
p1 = p2 + p3
Substituting the given values, we have:
60 kg·m/s = (10 kg)v1' + (30 kg)(-2 m/s)
Simplifying and solving for v1', we get:
v1' = (60 kg·m/s + 60 kg·m/s) / 10 kg
v1' = 12 m/s
Therefore, the velocity of the skateboard after the child jumps off is 12 m/s.