
In the first second of its flight, rocket ejects \[\dfrac{1}{{60}}\] of its mass with a velocity of \[2400{\text{ }}m/s\]. The acceleration of this rocket is:
A. \[19.6m/{s^2}\]
B. \[30.2m/{s^2}\]
C. \[40m/{s^2}\]
D. \[49.8m/{s^2}\]
Answer
583.8k+ views
Hint: The rate of change of mass will provide the necessary exhaust speed to the rocket.
Complete step by step solution:
Let \[m\] be the mass of the rocket and \[a\] be the acceleration. Let $\dfrac{{dm}}{{dt}}$ be the rate at which mass is ejecting. This rate of change of mass will provide the necessary exhaust speed to the rocket. If $v$ is the rate of change of mass and be the exhaust speed, then
Force acting on the rocket =$v\left( { - \dfrac{{dm}}{{dt}}} \right)$ , the negative sign shows that the mass is reducing.
\[ \Rightarrow ma = v\left( { - \dfrac{{dm}}{{dt}}} \right) \Rightarrow a = \dfrac{v}{m}\left( { - \dfrac{{dm}}{{dt}}} \right)\]
It is given that $\left( { - \dfrac{{dm}}{{dt}}} \right) = \dfrac{1}{{60}}$ , \[m = 1\,kg\] and
Putting these values in the above equation, we get
\[a = \dfrac{{2400}}{1}.\dfrac{1}{{60}} \Rightarrow a = 40\,m{s^{ - 2}}\]
Hence the acceleration of the rocket is \[40m/{s^2}\]
The correct option is (C).
Note: In a rocket engine stored fuel and stored oxidizer are ignited in a combustion chamber. This combustion produces great amounts of exhaust gas at high temperature and pressure. The hot exhaust is passed through a nozzle which accelerates the flow. Thrust is produced according to Newton's third law of motion. The amount of thrust produced by the rocket depends on the mass flow rate through the engine, the exit velocity of the exhaust and the pressure. In the motion of a rocket, the burned fuel gases change its velocity and hence its momentum, thus causing it to accelerate in the opposite direction of the velocity of the ejected fuel. As the rocket engines operate, it continuously ejects burned fuel gases, which have both mass and velocity and therefore some momentum. By conservation of momentum, the rocket’s momentum changes by this same amount.
Complete step by step solution:
Let \[m\] be the mass of the rocket and \[a\] be the acceleration. Let $\dfrac{{dm}}{{dt}}$ be the rate at which mass is ejecting. This rate of change of mass will provide the necessary exhaust speed to the rocket. If $v$ is the rate of change of mass and be the exhaust speed, then
Force acting on the rocket =$v\left( { - \dfrac{{dm}}{{dt}}} \right)$ , the negative sign shows that the mass is reducing.
\[ \Rightarrow ma = v\left( { - \dfrac{{dm}}{{dt}}} \right) \Rightarrow a = \dfrac{v}{m}\left( { - \dfrac{{dm}}{{dt}}} \right)\]
It is given that $\left( { - \dfrac{{dm}}{{dt}}} \right) = \dfrac{1}{{60}}$ , \[m = 1\,kg\] and
Putting these values in the above equation, we get
\[a = \dfrac{{2400}}{1}.\dfrac{1}{{60}} \Rightarrow a = 40\,m{s^{ - 2}}\]
Hence the acceleration of the rocket is \[40m/{s^2}\]
The correct option is (C).
Note: In a rocket engine stored fuel and stored oxidizer are ignited in a combustion chamber. This combustion produces great amounts of exhaust gas at high temperature and pressure. The hot exhaust is passed through a nozzle which accelerates the flow. Thrust is produced according to Newton's third law of motion. The amount of thrust produced by the rocket depends on the mass flow rate through the engine, the exit velocity of the exhaust and the pressure. In the motion of a rocket, the burned fuel gases change its velocity and hence its momentum, thus causing it to accelerate in the opposite direction of the velocity of the ejected fuel. As the rocket engines operate, it continuously ejects burned fuel gases, which have both mass and velocity and therefore some momentum. By conservation of momentum, the rocket’s momentum changes by this same amount.
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