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

Question

A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 ms−1 . To give an initial upward acceleration of 20 ms−2 , the amount of gas ejected per second to supply the needed thrust will be (g = 10 ms−2)  (A)  127.5 kg s−1  (B)  187.5 kg s−1  (C)  185.5 kg s−1  (D)  137.5 kg s−1

Tardigrade Admin · Accepted Answer

Thrust force on the rocket Ft=vr(-(d m/d t)) (upwards) Weight of the rocket w=m g (downwards) Net force on the rocket F text net =Ft-w ⇒ m a =vr((-d m/d t))-m g ⇒ ((-d m/d t)) =(m(g+a)/vr) ∴ Rate of gas ejected per second =(5000(10+20)/800)=(5000 × 30/800) =187.5 kg s -1

Q. 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 $(g = 10 m s^{- 2})$

$127.5 k g s^{- 1}$

$187.5 k g s^{- 1}$

$185.5 k g s^{- 1}$

$137.5 k g s^{- 1}$

Q. 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)

127.5kgs−1

187.5kgs−1

185.5kgs−1

137.5kgs−1

Thrust force on the rocket Ft​=vr​(−dtdm​)(upwards) Weight of the rocket w=mg(downwards) Net force on the rocket Fnet ​=Ft​−w ⇒ma=vr​(dt−dm​)−mg ⇒(dt−dm​)=vr​m(g+a)​ ∴ Rate of gas ejected per second =8005000(10+20)​=8005000×30​ =187.5kgs−1

Q. 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 $(g = 10 m s^{- 2})$

$127.5 k g s^{- 1}$

$187.5 k g s^{- 1}$

$185.5 k g s^{- 1}$

$137.5 k g s^{- 1}$