Question

A 15,000 kg rocket traveling at +230 m/s turns on its engines. Over a 6.0 s period it burns 1,000 kg of fuel. An observer on the ground measures the velocity of the expelled gases to be −1,200 m/s.

Answer 1

Answer:

Sample question;

Calculate the final velocity

The answer to the question is;

The final velocity is 312.79 m/s.

Explanation:

To solve the question, we note that Tsiolkovsky rocket equation is given by

Δv = v$_e$ ×㏑$(\frac{m_0}{m_f} )$

Where:

Δv = The maximum velocity change of the rocket = v - v₀

m₀ = The total mass of the rocket at the initial stage = 15,000 kg

m$_f$ = The final sum of the mass of the rocket after exhausting the propellant

v$_e$ = The exhaust velocity = -1200 m/s

v₀ = Initial velocity of the rocket = 230 m/s

Therefore we have

m$_f$  = m₀ - mass of propellant = 15,000 kg - 1,000 kg = 14,000 kg

and Δv = 1200 ×㏑$(\frac{15000}{14000} )$ = 82.79 m/s

Therefore v - v₀ =  82.79 m/s ⇒ v = 82.79 m/s + 230 m/s = 312.79 m/s.

Answer 2

Answer:

a) $v = 312.791\,\frac{m}{s}$, b) $a = 13.333\,\frac{m}{s^{2}}$

Explanation:

The problem is asking the rocket velocity and acceleration at t = 6 s.

a) The general equation of the rocket is:

$v=v_{o} -v_{ex}\cdot \ln \frac{m}{m_{o}}$

$v = 230\,\frac{m}{s}-(1200\,\frac{m}{s} )\cdot \ln \frac{14000\,kg}{15000\,kg}$

$v = 312.791\,\frac{m}{s}$

b) The acceleration experimented by the rocket is:

$a = \frac{v_{ex}}{m_{o}}\cdot \frac{dm}{dt}$

$a = \frac{1200\,\frac{m}{s} }{15000\,kg}\cdot \frac{1000\,kg}{6\,s}$

$a = 13.333\,\frac{m}{s^{2}}$