Answer to Question #156462 in Classical Mechanics for Afroja Akhi

Question

A rocket has an initial mass of 2 x 104 kg, a mass ratio of 3, a burning rate of 100 kg/sec, and an exhaust velocity of 980 m/sec. The rocket is fired vertically from the surface of the earth. How long after ignition of the engines will the rocket leave the ground? (

Answer 1

The impulse force exerted on the rocket by the burning fuel can be written as follows:

$$F=\dfrac{d}{dt}(mv)=v\dfrac{dm}{dt},$$

here, $\dfrac{dm}{dt}=100\ \dfrac{kg}{s}$ is the burning rate of the rocket, $v=980\ \dfrac{m}{s}$ is the exhaust velocity.

The weight of the remaining mass of the rocket can be written as follows:

$$W=mg=(m_0-\dfrac{dm}{dt}t)g.$$

The rocket will leave the ground when the impulse force exerted on the rocket by the burning fuel exceeds the weight of the remaining mass of the rocket. Therefore, we can equate both expressions:

$$v\dfrac{dm}{dt}=(m_0-\dfrac{dm}{dt}t)g.$$

From this equation we can find time that the rocket takes to leave the ground after the ignition of the engines:

$$t=\dfrac{m_0g-V\dfrac{dm}{dt}}{\dfrac{dm}{dt}g},$$ $$t=\dfrac{2\cdot10^4\ kg\cdot9.8\ \dfrac{m}{s^2}-980\ \dfrac{m}{s}\cdot100\ \dfrac{kg}{s}}{100\ \dfrac{kg}{s}\cdot9.8\ \dfrac{m}{s^2}}=100\ s.$$

Answer:

$t=100\ s.$