Question

At a certain instant of time mass of a rocket going up vertically is \[100\,{\text{kg}}\]. If it is ejecting \[5\,{\text{kg}}\] of gas per second at a speed of \[400\,{\text{m/s}}\], the acceleration of the rocket would be (taking \[g = 10\,{\text{m/}}{{\text{s}}^2}\])
A. \[20\,{\text{m/}}{{\text{s}}^2}\]
B. \[10\,{\text{m/}}{{\text{s}}^2}\]
C. \[2\,{\text{m/}}{{\text{s}}^2}\]
D. \[1\,{\text{m/}}{{\text{s}}^2}\]

Answer 1

Hint: Use the formula for force acting on an object in terms of the linear momentum of the object. Use the formula for linear momentum of an object. Also use the expression for Newton’s second law of motion. We have given the rate of ejection of mass of the gas. Calculate the force acting on the rocket in upward direction due to ejection of gas. Also calculate the weight of the rocket. Calculate the net force on the rocket and then calculate the acceleration of the rocket.

Formulae used:
The force \[F\] acting on an object is given by
\[F = \dfrac{{dP}}{{dt}}\] …… (1)
Here, \[dP\] is a change in momentum of the object in time \[dt\].
The momentum \[P\] of an object is
\[P = mv\] …… (2)
Here, \[m\] is the mass of the object and \[v\] is the velocity of the object.
The expression for Newton’s second law of motion is
\[{F_{net}} = ma\] …… (3)
Here, \[{F_{net}}\] is net force acting on the object, \[m\] is mass of the object and \[a\] is acceleration of the object.

Complete step by step solution:
We have given that the mass of the rocket is \[100\,{\text{kg}}\].
\[M = 100\,{\text{kg}}\]
The rate of ejection of the mass of the gas from the rocket is \[5\,{\text{kg}}\] per second.
\[\dfrac{{dm}}{{dt}} = 5\,{\text{kg/s}}\]
We have also given that the velocity of the ejection of the gas from the rocket is \[400\,{\text{m/s}}\].
\[v = 400\,{\text{m/s}}\]
The gas ejecting out of the rocket gives an upward force to the rocket for its motion in the upward direction.
This force making the rocket to move in the upward direction is given by equation (1).
\[F = \dfrac{{dP}}{{dt}}\]
Substitute \[mv\] for \[P\] in the above equation.
\[F = \dfrac{{d\left( {mv} \right)}}{{dt}}\]
\[ \Rightarrow F = v\dfrac{{dm}}{{dt}}\]
Substitute \[400\,{\text{m/s}}\] for \[v\] and \[5\,{\text{kg/s}}\] for \[\dfrac{{dm}}{{dt}}\] in the above equation.
\[ \Rightarrow F = \left( {400\,{\text{m/s}}} \right)\left( {5\,{\text{kg/s}}} \right)\]
\[ \Rightarrow F = 2000\,{\text{N}}\]
Hence, the force acting on the rocket in the upward direction is \[2000\,{\text{N}}\].
Let us now calculate the weight of the rocket in the downward direction.
\[W = Mg\]
Substitute \[100\,{\text{kg}}\] for \[M\] and \[10\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in the above equation.
\[W = \left( {100\,{\text{kg}}} \right)\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)\]
\[ \Rightarrow W = 1000\,{\text{N}}\]
Hence, the weight of the rocket acting in the downward direction is \[1000\,{\text{N}}\].
Let us now calculate the acceleration of the rocket.Hence, the net force acting on the rocket is
\[{F_{net}} = F - W\]
Substitute \[F - W\] for \[{F_{net}}\] and \[M\] for \[m\] in equation (3).
\[F - W = Ma\]
\[ \Rightarrow a = \dfrac{{F - W}}{M}\]
Substitute \[2000\,{\text{N}}\] for \[F\], \[1000\,{\text{N}}\] for \[W\] and \[100\,{\text{kg}}\] for \[M\] in the above equation.
\[ \Rightarrow a = \dfrac{{\left( {2000\,{\text{N}}} \right) - \left( {1000\,{\text{N}}} \right)}}{{100\,{\text{kg}}}}\]
\[ \therefore a = 10\,{\text{m/}}{{\text{s}}^2}\]
Therefore, the acceleration of the rocket would be \[10\,{\text{m/}}{{\text{s}}^2}\].

Hence, the correct option is B.

Note: The students should not forget to calculate the net force acting on the rocket as the force acting on the rocket is the vector addition of the upward force or thrust on the rocket and weight of the rocket in the downward direction. If this net forget is not calculated and only weight of the rocket is considered then we will end up with the incorrect value of acceleration of the rocket.