A 65 kg person runs with a horizontal velocity of 3.5 m/s and jumps onto an 0.9 kg skateboard which is initially at rest.
Assuming negligible friction, what will be the final velocity of the skateboard and the person?
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the jump should be equal to the total momentum after the jump.
The momentum of an object is given by the product of its mass and velocity. Therefore, we can calculate the initial momentum of the person and the final momentum of the person and the skateboard together.
Initial momentum of the person = mass of the person * velocity of the person
Final momentum of the person and the skateboard = (mass of the person + mass of the skateboard) * final velocity of both
Since momentum is conserved, we can set up the equation:
mass of the person * velocity of the person = (mass of the person + mass of the skateboard) * final velocity of both
Plugging in the given values:
(65 kg) * (3.5 m/s) = (65 kg + 0.9 kg) * final velocity of both
Now, we can solve for the final velocity of both:
Final velocity of both = (65 kg * 3.5 m/s) / (65 kg + 0.9 kg)
Final velocity of both ≈ 3.43 m/s
Therefore, the final velocity of both the person and the skateboard, assuming negligible friction, will be approximately 3.43 m/s.