Answer:
To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act on it.
Before the jump, the momentum of the friend is:
p1 = m1 * v1
p1 = 80 kg * 2 m/s
p1 = 160 kg m/s
The momentum of the skateboard before the jump is zero since it is stationary:
p2 = m2 * v2
p2 = 2 kg * 0 m/s
p2 = 0 kg m/s
The total momentum before the jump is:
p1 + p2 = 160 kg m/s
After the jump, the friend and the skateboard move together with a final velocity of v. The total mass of the friend and the skateboard is:
m = m1 + m2
m = 80 kg + 2 kg
m = 82 kg
Using the law of conservation of momentum, we can set the total momentum after the jump equal to the total momentum before the jump:
p1 + p2 = p
where p = m * v is the momentum of the combined system after the jump.
Solving for v, we get:
p1 + p2 = m * v
160 kg m/s = 82 kg * v
v = 160 kg m/s / 82 kg
v = 1.95 m/s (approximately)
Therefore, the final velocity of the friend and the skateboard after the jump is approximately 1.95 m/s.