Question

Your friend who has a mass of 80 kg runs with a velocity of 2

m/s and jumps onto a 2 kg stationary skateboard so that they
both move together. What is the final velocity of your friend
and the skateboard after she jumps onto it?
Assume all momentum is conserved.

Answer 1

Using the conservation of momentum, the final velocity of an 80 kg person and a 2 kg skateboard after the person jumps onto the skateboard is approximately 1.95 m/s.

Step-by-step explanation:

The subject question is based on the concept of conservation of momentum in Physics, often taught in high school. When an 80 kg person runs at 2 m/s and jumps on a stationary 2 kg skateboard, the final velocity of both moving together can be found using the principle that the total momentum before the jump equals the total momentum after the jump. For calculation, we use the formula: momentum = mass × velocity.

Before the jump, the momentum of your friend is (80 kg) × (2 m/s) = 160 kg·m/s, and the momentum of the skateboard is 0 because it is stationary. After the jump, the total mass is (80 kg + 2 kg) = 82 kg. The final velocity (v) can be determined by the equation:

Initial momentum = Final momentum
160 kg·m/s = (82 kg) × v

Solving for v, we find:

v = 160 kg·m/s / 82 kg = 1.9512 m/s (approximately).

Therefore, the final velocity of your friend and the skateboard after she jumps onto it is approximately 1.95 m/s.

Answer 2

Step-by-step explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

After she jumps on the skateboard, they have the same velocity, so v₁ = v₂ = v.

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

The skateboard is originally stationary, so u₂ = 0.

m₁ u₁ = (m₁ + m₂) v

Plugging in values and solving:

(80 kg) (2 m/s) = (80 kg + 2 kg) v

v = 1.95 m/s

Round as needed.