First slide

Third law

Question

A rocket set for vertical firing weighs 50 kg and contains 450 kg of fuel. The product of combustion can have a maximum velocity of 2 km/s. What should be the minimum rate of fuel consumption to give it an acceleration of $20 m / s^{2}$ ? Take $g = 9.8 m / s^{2} .$

Moderate

A

4.74 kg/s
B

6.25 kg/s
C

3.45 kg/s
D

7.45 kg/s

Solution

$ma = - mg + V_{r} (- \frac{dm}{dt}) ∴ - \frac{dm}{dt} = \frac{m (a + g)}{V_{r}} = \frac{(450 + 50) (20 + 9 .8)}{2 \times 10^{3}} kg / s \Rightarrow - \frac{dm}{dt} = 7 .45 kg / s$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App