Question

A 63 kg skateboarder runs at a constant velocity of 10 m/s and jumps on a stationary skateboard with a mass of 12 kg. What is the velocity of the skateboard after the jump?

Answer 1

To find the velocity of the skateboard after the jump, we can use the principle of conservation of momentum. The total momentum before the jump is equal to the total momentum after the jump. Before the jump, the skateboarder has a mass of 63 kg and a velocity of 10 m/s. The skateboard has a mass of 12 kg and is stationary, so its velocity is 0 m/s. Using the equation for conservation of momentum, we find that the velocity of the skateboard after the jump is approximately 8.25 m/s.

Step-by-step explanation:

To find the velocity of the skateboard after the jump, we can use the principle of conservation of momentum. The total momentum before the jump is equal to the total momentum after the jump.

Before the jump, the skateboarder has a mass of 63 kg and a velocity of 10 m/s. The skateboard has a mass of 12 kg and is stationary, so its velocity is 0 m/s.

Using the equation for conservation of momentum, we can calculate the velocity of the skateboard after the jump:

Total momentum before the jump = Total momentum after the jump

(63 kg)(10 m/s) + (12 kg)(0 m/s) = (63 kg + 12 kg)(v)

Simplifying this equation, we find that the velocity of the skateboard after the jump is approximately 8.25 m/s.

Answer 2

Answer: We can use the law of conservation of momentum which states that the total momentum before the jump is equal to the total momentum after the jump.

The momentum before the jump is given by:

P_before = m1 * v1

where m1 is the mass of the skateboarder and v1 is their velocity before the jump.

P_before = (63 kg) * (10 m/s) = 630 kg·m/s

The momentum after the jump is given by:

P_after = (m1 + m2) * v2

where m2 is the mass of the skateboard and v2 is their velocity after the jump.

To find v2, we need to solve for it:

P_after = (63 kg + 12 kg) * v2

P_after = 75 kg * v2

v2 = P_after / 75 kg

The momentum after the jump is the same as the momentum before the jump, so:

P_before = P_after

630 kg·m/s = 75 kg * v2

v2 = 8.4 m/s

Therefore, the velocity of the skateboard after the jump is 8.4 m/s.

Step-by-step explanation: