A model rocket with an initial mass of 2.0 kg is launched horizontally by burning and expelling 980 g of fuel with a velocity of 140 m/s. What is the velocity of the rocket after the fuel is expelled? (Assume the propellant is expelled instantaneously.)
To find the velocity of the rocket after the fuel is expelled, we can use the principle of conservation of momentum.
The initial momentum of the rocket and the fuel is given by the equation:
\(m_{\text{initial}} \cdot v_{\text{initial}} = m_{\text{final}} \cdot v_{\text{final}}\)
Where:
\(m_{\text{initial}}\) = Initial mass of the rocket + mass of the fuel = 2.0 kg + 0.980 kg = 2.980 kg
\(v_{\text{initial}}\) = Initial velocity of the rocket + velocity of the expelled fuel = 0 + 140 m/s = 140 m/s
\(m_{\text{final}}\) = Final mass of the rocket after the fuel is expelled = 2.0 kg
\(v_{\text{final}}\) = Final velocity of the rocket after the fuel is expelled (unknown)
Plugging in the values, we get:
\(2.980 \cdot 140 = 2.0 \cdot v_{\text{final}}\)
Simplifying the equation:
\(v_{\text{final}} = \frac{2.980 \cdot 140}{2.0} = \frac{417.2}{2.0} \approx \boxed{208.6 \, \text{m/s}}\)