Question

A diwali rocket is ejecting 0.05 kg of gases per second at a velocity 200m/s .The accelerating force on the rocket will be?​

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Correct \: Question}}} \mid}}$

A diwali rocket is ejecting 0.05 kg/second of gases per second at a velocity 200m/s .The accelerating force on the rocket will be?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Velocity of Rocket (v) = 200 m/s.
• Rocket is ejecting Feul at a rate of 0.05 Kg/sec.

$\huge{\bold{\underline{Explanation:-}}}$

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From Newton's Second law,

$\large{\boxed{\tt F = \dfrac{\Delta P}{t}}}$

$\large{\tt \leadsto F = \dfrac{\Delta P}{t}}$

As we know ΔP = mv.

$\large{\tt \leadsto F = \dfrac{mv}{t}}$

In the case of rockets, the Velocity is constant and mass keeps on varying.

Therefore,

$\large{\tt \leadsto F = v \times \dfrac{m}{t}}$

$\large{\tt \leadsto F = v .\dfrac{m}{t} \: ------(1)}$

$\large{\boxed{\tt F = v .\dfrac{m}{t}}}$

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Now, Here

Rate of Feul ejection = m/t

I.e. m/t = 0.05 kg/sec.

Substituting the values in equation (1).

$\large{\tt \leadsto F = v \times \dfrac{m}{t}}$

$\large{\tt \leadsto F = 200 \times 0.05}$

$\large{\tt \leadsto F = 200 \times \dfrac{5}{100}}$

$\large{\tt \leadsto F = \cancel{200} \times \dfrac{5}{\cancel{100}}}$

$\large{\tt \leadsto F = 2 \times 5}$

$\huge{\boxed{\boxed{\tt F = 10 \: N}}}$

So, the Accelerating Force on the rocket is 10N.

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Answer 2

Answer:

The accelerating force on the rocket will be 10 N.

Explanation:

Given;-

   Rate at which fuel is burning, dm/dt = 0.05 kg/s

     Velocity of gases, v = 200 m/s

Let the accelerating force on the rocket be 'F'

We know by formula that;-

              F = v dm/dt

Putting the given values in the formula, we get;-

               F = 200 × 0.05

               F = 10 N

Hence, the accelerating force on the rocket will be 10 N.

Hope it helps! ;-))