A 62-kg boy on a 1.50-kg skateboard moving at +1.2 m/s steps off and lands on the sidewalk with a velocity of +1.1 m/s. How fast is the skateboard moving?
To find the final velocity of the skateboard, we can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, as long as no external forces are acting on the system.
The momentum (p) is given by the product of mass (m) and velocity (v). So, we can calculate the momentum of the boy and the skateboard before and after the boy steps off.
Initial momentum of the boy and the skateboard:
\(p_{\text{initial}} = m_{\text{boy}} \times v_{\text{boy}} + m_{\text{skateboard}} \times v_{\text{skateboard}}\)
Final momentum of the boy and the skateboard:
\(p_{\text{final}} = m_{\text{boy}} \times v_{\text{boy}}' + m_{\text{skateboard}} \times v_{\text{skateboard}}'\)
Using the conservation of momentum principle, we can set the initial momentum equal to the final momentum:
\(m_{\text{boy}} \times v_{\text{boy}} + m_{\text{skateboard}} \times v_{\text{skateboard}} = m_{\text{boy}} \times v_{\text{boy}}' + m_{\text{skateboard}} \times v_{\text{skateboard}}'\)
Now, let's plug in the given values:
Mass of the boy, \(m_{\text{boy}} = 62 \, \text{kg}\)
Initial velocity of the boy, \(v_{\text{boy}} = +1.2 \, \text{m/s}\)
Mass of the skateboard, \(m_{\text{skateboard}} = 1.50 \, \text{kg}\)
Final velocity of the boy, \(v_{\text{boy}}' = +1.1 \, \text{m/s}\)
Let's assume the final velocity of the skateboard is \(v_{\text{skateboard}}'\). Since the boy steps off the skateboard, his final velocity is \(0 \, \text{m/s}\).
Now we can solve the equation:
\(62 \, \text{kg} \times 1.2 \, \text{m/s} + 1.50 \, \text{kg} \times v_{\text{skateboard}} = 62 \, \text{kg} \times 0 \, \text{m/s} + 1.50 \, \text{kg} \times 1.1 \, \text{m/s}\)
Simplifying the equation, we have:
\(74.4 \, \text{kg} \cdot \text{m/s} + 1.50 \, \text{kg} \times v_{\text{skateboard}} = 1.50 \, \text{kg} \times 1.65 \, \text{m/s}\)
Now, we can rearrange the equation to solve for \(v_{\text{skateboard}}\):
\(v_{\text{skateboard}} = \frac{{1.50 \, \text{kg} \times 1.65 \, \text{m/s} - 74.4 \, \text{kg} \cdot \text{m/s}}}{{1.50 \, \text{kg}}}\)
Calculating the expression:
\(v_{\text{skateboard}} \approx 0.6 \, \text{m/s}\)
Therefore, the skateboard is moving at approximately 0.6 m/s.