Physics Problems & Solutions: Rocket

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Classical Mechanics - Equation of Motion

Questions 61-62

The Equation of motion of a rocket in free space can be written where m is the rocket’s mass, v is its velocity, t is time, and u is a constant.

$m \frac{dv}{dt}+u \frac{dm}{dt}=0$

The constant u represents the speed of the

A. Rocket at t = 0
B. Rocket after its fuel is spent
C. Rocket in its instantaneous rest frame
D. Rocket’s exhaust in a stationary frame
E. Rocket’s exhaust relative to the rocket

(GR8677 #61)

Solution:

In free space → without gravity → no external force:

$\\F= \frac{dp}{dt}=\frac{dmv}{dt}=m\frac{dv}{dt}+v \frac{dm}{dt}=0 \\\\\rightarrow m\frac{dv}{dt}=-v_\text{exhaust}\frac{dm}{dt}$

The equation of motion:

$m \frac{dv}{dt}+u \frac{dm}{dt}=0 \rightarrow m \frac{dv}{dt}= -u \frac{dm}{dt}$

→ u = v_exhaust relative to the rocket.

Answer: E

Classical Mechanics - Equation of Motion

Questions 61-62

The Equation of motion of a rocket in free space can be written where m is the rocket’s mass, v is its velocity, t is time, and u is a constant.

$m \frac{dv}{dt}+u \frac{dm}{dt}=0$

The equation can be solved to give v as a function of m. If the rocket has m = m₀ and v = 0 when it starts, what is the solution?

A. u m₀/m
B. u exp(m₀/m)
C. u sin (m₀/m)
D. u tan (m₀/m)
E. None of the above

(GR8677 #62)

Solution:

$\begin{aligned} m \frac{dv}{dt}+u \frac{dm}{dt}&=0 \\m\frac{dv}{dt}&=-u \frac{dm}{dt} \\dv&=-u \frac{dm}{m} \\\int_{v_0}^{v}dv&=-u \int_{m_0}^{m}\frac{dm}{m} \\v&=-u \ln \frac{m_0}{m} \end{aligned}$

Answer: E

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