Question

If the thrust on a rocket moving with a velocity of $300\text{m/s}$ is $210\text{N}$, the rate of combustion of the fuel is

• A. $0.007\text{kg/s}$
• B. $1.4\text{kg/s}$
• C. $0.7\text{kg/s}$
• D. $10.7\text{kg/s}$

Answer 1

The correct option is C $0.7\text{kg/s}$


From newton's $2^{\text{nd}}$ law:
$F=\frac{dP}{dt}=\frac{d}{dt}\left(mv\right)...\left(i\right)$
let the rocket rate of combustion is $x\text{kg/s}$, hence time rate of change of mass:
$\Rightarrow \frac{dm}{dt}=x\text{kg/s}$
Given: $F=210\text{N},v=300\text{m/s}$
since velocity is constant hence from Eq. $\left(i\right)$
$\Rightarrow F=v\frac{dm}{dt}$
$210=300\times x$
$\therefore x=\frac{210}{300}=0.7\text{kg/s}$