Question

A rocket set for vertical firing weighs $100\text{kg}$ and contains $600\text{kg}$ of fuel. It can have a maximum exhaust velocity of $4\text{km/s}$. What should be the rate of fuel consumption to give it an initial acceleration of $30{\text{m/s}}^2?$ (Take $g=10{\text{m/s}}^2)$

• A. $4\text{kg/s}$
• B. $6\text{kg/s}$
• C. $7\text{kg/s}$
• D. $9\text{kg/s}$

Answer 1

The correct option is C $7\text{kg/s}$


Given that,
mass of fuel $\left(m_f\right)=600\text{kg}$
mass of rocket $\left(m_s\right)=100\text{kg}$
Total mass of the system initially $\left(m\right)=m_g+m_s=600+100=700\text{kg}$

Required acceleration $\left(a\right)=30{\text{m/s}}^2$
Let $F_t$ be the thrust force produced by the exhaust.
$\therefore F_t-mg=ma$
$\Rightarrow a=\frac{F_t}{m}-g$
$\Rightarrow a=\frac{1}{m}(v_r\times -\frac{dm}{dt})-g$
$\Rightarrow -\frac{dm}{dt}=\frac{m(a+g)}{v_r}$

Substituting the values, we get
Rate of consumption of fuel to give an initial acceleration of $30{\text{m/s}}^2$
$-\frac{dm}{dt}=\frac{700(30+10)}{4\times {10}^3}(\because v_r=4\text{km/s})$
$-\frac{dm}{dt}=7\text{kg/s}$