First slide

Third law

Question

If force on a rocket having exhaust velocity of 300 m/sec is 210 N, then rate of combustion of the fuel is

Easy

A

0.7 kg/s
B

1.4 kg/s
C

0.07 kg/s
D

10.7 kg/s

Solution

$F = u (\frac{dm}{dt}) \Rightarrow \frac{dm}{dt} = \frac{F}{u} = \frac{210}{300} = 0 .7 kg / s$

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