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Momentum is given by the relation -
{P=\frac{b-x^2}{a t}}  where x distance and
t is time. For variable mass what will be force from the above mention reaction -

Option: 1

\frac{a}{\left(x^2+b\right) t^2}


Option: 2

\frac{a}{\left(b-x^2\right) t}


Option: 3

\frac{a}{\left(b-x^2\right) t^2}


Option: 4

\frac{b-x^2}{a t^2}


Answers (1)

best_answer

Given,  momentum \operatorname(p)=\frac{b-x^2}{a t} 

for variable mass force is given by -
\vec{F}=\frac{d \vec{P}}{d t} \\

\vec{F}=\frac{d}{d t}(m \cdot \vec{E})
i.e.
\quad \vec{F} =\frac{d}{d t} \frac{b-x^2}{a t} \Rightarrow \frac{b-x^2}{a} \cdot \frac{d}{d t}\left(\frac{1}{t}\right) \\

F =\frac{b-x^2}{a}(-1) t^{-2} \\

F =-\frac{\left(b-x^2\right)}{a} t^{-2} \\

F =\frac{a}{\left(b-x^2\right) t^2}

Posted by

seema garhwal

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