Question

A rocket of mass $m\circ $ has attained a speed equal to its exhaust speed and at that time the mass of the rocket is $m\circ $ the ratio $\dfrac{m\circ }{m}$ is (neglects gravity).
a. 2.718
b. 7.8
c. 3.14
d. 4

Answer 1

Hint: The changing mass of an object can be seen in special relativity cases, when an object is moving at a certain seed with respect to the stationary or moving frame of reference. The ratio between the mass of the object at different times can be obtained by the speed of the object at any time t.
As per the given data,
Initial speed of rocket $(u)$= exhaust speed$(V_{ex})$.
Acceleration due to gravity $(g)= 0$
Formula used:
Speed of rocket at any time t$(V_{ex})$:
\[{{V}_{r}}=u-gt+{{V}_{ex}}\ln \text{ }\dfrac{m\circ }{m}\]

Complete answer:
Here we have a case where the mass of an object is changing. We analyze the motion of a rocket, which changes its velocity by ejecting burned fuel gases, thus causing it to accelerate in the opposite direction of the velocity of the ejected fuel.
Speed of rocket at any time t$(V_{ex})$:
\[{{V}_{r}}=u-gt+{{V}_{ex}}\ln \dfrac{m\circ }{m}\]
  Where,
$u=$ Initial speed of the rocket
 $g=$Acceleration due to gravity
\[V_{ex}=\] Exhausted speed of the fuel
 $m\circ =$Mass at initial stage
  $m=$Mass at time t
By taking consideration of the given data. The equation of speed at any time t will be:
    ${{V}_{r}}=0-0t+{{V}_{ex}}\ln \dfrac{m\circ }{m}$
\[{{V}_{r}}={{V}_{ex}}\ln \dfrac{m\circ }{m}\]
According to the question rocket attains speed equal to that of exhaust speed. Therefore;
\[{{V}_{r}}={{V}_{ex}}\].
So we come to a point where;
\[1=\ln \dfrac{m\circ }{m}\]
 To balance the LHS and RHS sides:
$\dfrac{m\circ }{m}=e$
=2.718

So, the correct answer is “Option A”.

Note:
Read the question carefully. Everything is there only in the question only. Don’t forget to take consideration of the situation applied (g=0). As the given phenomenon is a kind of closed loop system we can apply the theory of conservation of momentum.