Question

A 3000 kg rocket is fired. If the exhaust speed is
500 ms-, how much gas must be ejected per
second to supply the thrust needed to give the rocket an
initial acceleration of 19.6 ms-2?​

Answer 1

Answer:

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Answer 2

Answer:

• Rate of ejection of gas = 176.4 kg/s

Explanation:

Given

• Mass of rocket, M = 3000 kg
• Exhaust velocity/speed, v = 500 m/s
• Initial acceleration needed by the rocket, a = 19.6 m/s²

To find

• Rate of ejection of gas = d(m) / d(t) =?

Knowledge required:

• Force is given by mass times acceleration.

        Force = mass · acceleration

• Force is known as the rate of change of momentum.

        Force = d(p) / d(t)

• Formula to calculate momentum

        momentum = mass · velocity

[ p is momentum, t is the time ]

Solution:

In the given condition

The thrust force (T), (upward) produced by the ejection of gas will be,

→ T = d(p) / d(t)

[ since, p = m v ]

→ T = d(mv) / d(t)

[ Since, velocity (v) is constant ]

→ T = v · d(m) / d(t)

Also, the weight of the rocket (W) would be acting downwards.

→ W = M g

We need to produce an initial acceleration 'a' = 19.6 m/s² (upward) in the rocket.

So, the force required (F) to move the rocket with the given acceleration would be,

→ F = M a

( Thrust force 'T' and force required 'F' to produce the given acceleration are acting upward. while Weight 'W' of Rocket is acting downward. )

Also the force required 'F' to produce the acceleration 'a' would be,

→ F = T - W

→ M a = v · d(m) / d(t) - [ M g ]

→ 3000 · 19.6 = 500 · d(m) / d(t) - [ 3000 · 9.8 ]

→ 500 · d(m) / d(t) = 3000 ( 19.6 + 9.8 )

→ 500 · d(m) / d(t) = 3000 · 29.4

→ d(m) / d(t) = 3000 · 29.4 / 500

→ d(m) / d(t) = 176.4  kg/s

Therefore,

• 176.4 kg gas must be ejected per second to supply the thrust needed for providing the rocket an acceleration of 19.6 m/s².