Question

A 37 kg person initially standing stationary on a 3.9 kg skateboard walks forward at 1.0 m/s. How fast is the skateboard ejected out behing that person? Answer to 2 decimal places.

Answer 1

The skateboard is ejected out behind the person at a speed of 0.95 m/s.

Step-by-step explanation:

To find the speed at which the skateboard is ejected behind the person, we can use the conservation of momentum. The initial momentum of the person and skateboard system is zero (since they are initially at rest), and the final momentum is also zero (due to the law of conservation of momentum). We can use the equation:

m1v1 + m2v2 = 0

Solving for v2, the speed of the skateboard, we get:

v2 = -m1v1/m2 = -(37 kg)(1.0 m/s) / 3.9 kg = -0.95 m/s

Therefore, the skateboard is ejected out behind the person at a speed of 0.95 m/s.