Question

A rocket of mass $40\text{kg}$ has $160\text{kg}$ fuel. The exhaust velocity of the fuel is $2\text{km/s}$. The rate of consumption of fuel is $4\text{kg/s}$. Calculate the ultimate vertical speed gained by the rocket. Assume that value of $g$ does not vary.
$(g=10{\text{m/s}}^2\text{and}\ln 5=1.609)$

• A. $3218\text{m/s}$
• B. $2818\text{m/s}$
• C. $2418\text{m/s}$
• D. $2018\text{m/s}$

Answer 1

The correct option is B $2818\text{m/s}$

Total mass of rocket initially,

$M_0=40+160=200\text{k}g$

Mass of empty rocket, $m=40\text{kg}$

The rate of consumption of fuel, $\mu =4\text{kg/s}$

The exhaust velocity of fuel, $v_{rel}=2\text{km/s}=2000\text{m/s}$

Time taken by rocket to burn all the $160\text{kg}$ fuel.

$t=\frac{160}{\mu }=\frac{160}{4}=40\text{s}$

Ultimate speed,

$V=(u-gt)+v_{rel}\ln \left(\frac{M_0}{m}\right)$

$=[0-(10\times 40\left)\right]+2000\ln \left(\frac{200}{40}\right)$

$=-400+(2000\times 1.609)=2818\text{m/s}$

Hence, $\left(\text{B}\right)$ is the correct answer.