SATHEE: Laws of Motion - Result Question 11

Laws of Motion - Result Question 11

11. A $5000 k g$ rocket is set for vertical firing. The exhaust speed is $800 m s^{- 1}$ . To give an initial upward acceleration of $20 m s^{- 2}$ , the amount of gas ejected per second to supply the needed thrust will be $(g = 10 m s^{- 2})$

(a) $127.5 k g s^{- 1}$

(b) $187.5 k g s^{- 1}$

(d) $137.5 k g s^{- 1}$

[1998]

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Answer:

Correct Answer: 11. (b)

Solution:

(b) Given : Mass of rocket (m) $= 5000 k g$

Exhaust speed $(v) = 800 m / s$

Acceleration of rocket $(a) = 20 m / s^{2}$

Gravitational acceleration $(g) = 10 m / s^{2}$

Thrust, $\Rightarrow \frac{v d m}{d t}$

We know that upward force,

$F = m (g + a) = 5000 (10 + 20)$

$= 5000 \times 30 = 150000 N$

This thrust gives upward force, $F = \frac{v d m}{d t}$

We also know that amount of gas ejected

$\Rightarrow (\frac{d m}{d t}) = \frac{F}{v} = \frac{150000}{800} = 187.5 k g / s$

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Laws of Motion - Result Question 11

11. A 5000kg rocket is set for vertical firing. The exhaust speed is 800ms−1. To give an initial upward acceleration of 20ms−2, the amount of gas ejected per second to supply the needed thrust will be (g=10ms−2)

Answer:

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11. A $5000 k g$ rocket is set for vertical firing. The exhaust speed is $800 m s^{- 1}$ . To give an initial upward acceleration of $20 m s^{- 2}$ , the amount of gas ejected per second to supply the needed thrust will be $(g = 10 m s^{- 2})$