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Question: In a rocket, fuel burns at the rate of \(2kg/s\). This fuel gets ejected from the rocket with a velo...

In a rocket, fuel burns at the rate of 2kg/s2kg/s. This fuel gets ejected from the rocket with a velocity of 80km/s80km/s. Force exerted on the rocket is
A)16000N B)160000N C)1600N D)16N \begin{aligned} & A)16000N \\\ & B)160000N \\\ & C)1600N \\\ & D)16N \\\ \end{aligned}

Explanation

Solution

From Newton’s second law of motion, we know that force exerted on an object is equal to the product of mass of the object and the acceleration of the object. In the case of ejection of fuel from a rocket, force exerted can also be defined as the product of rate of change of mass with respect to time and the velocity of the ejected fuel.

Formula used:
1)F=ma 2)F=dmdtv \begin{aligned} & 1)F=ma \\\ & 2)F=\dfrac{dm}{dt}v \\\ \end{aligned}

Complete step by step answer:
From Newton’s second law of motion, we know that force exerted on an object is equal to the product of mass of the object and the acceleration of the object. This is mathematically expressed as:
F=maF=ma
where
FF is the force exerted on an object
mm is the mass of the object
aa is the acceleration of the object
Let this be equation 1.
In the case of ejection of fuel from a rocket, force exerted can also be defined as the product of rate of change of mass of the fuel with respect to time and the velocity of the ejected fuel. This can be be mathematically expressed as:
F=dmdtvF=\dfrac{dm}{dt}v
where
FF is the force exerted on a rocket due to ejection of fuel
dmdt\dfrac{dm}{dt} is the rate of change of mass of fuel with respect to time
vv is the velocity of the ejected fuel
Let this be equation 2.
Coming to our question, we are given that
dmdt=2kg/s\dfrac{dm}{dt}=2kg/s
v=80×103m/sv=80\times {{10}^{3}}m/s
Substituting these values in equation 2, we haveF=dmdtv=2kgs1×80×103ms1=160000kgms2=160000NF=\dfrac{dm}{dt}v=2kg{{s}^{-1}}\times 80\times {{10}^{3}}m{{s}^{-1}}=160000kgm{{s}^{-2}}=160000N

So, the correct answer is “Option B”.

Note: The solution needs to be deduced from Newton’s second law, as already mentioned. Newtons’ second law of motion can be rearranged in different ways as given below:
F=ma=dmdtv=mdvdtF=ma=\dfrac{dm}{dt}v=m\dfrac{dv}{dt}
The correctness of the above expression can be deduced from their units, as given below:
N=kgms2=(kgs1)(ms1)=kg(ms2)N=kgm{{s}^{-2}}=(kg{{s}^{-1}})(m{{s}^{-1}})=kg(m{{s}^{-2}})