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Question

Physics Question on laws of motion

In the first second of its flight, rocket ejects 1/60 of its mass with a velocity of 2400ms12400\,\,m{{s}^{-1}} . The acceleration of the rocket is:

A

19.6ms219.6\,m{{s}^{-2}}

B

30.2ms230.2\,m{{s}^{-2}}

C

40ms240\,m{{s}^{-2}}

D

49.8ms249.8\,\,m{{s}^{-2}}

Answer

40ms240\,m{{s}^{-2}}

Explanation

Solution

Thrust force on the rocket
Ft=vr(dmdt)F_{t}=v_{r}\left(-\frac{d m}{d t}\right)
or ma=vr(dmdt)m a=v_{r}\left(-\frac{d m}{d t}\right)
where vrv_{r} is the exhaust velocity of gases and
dmdt-\frac{d m}{d t} is the mass of the ejected per unit time.
Given, vr=2400ms1,dmdt=160kg/sv_{r}=2400\,ms ^{-1},-\frac{d m}{d t}=\frac{1}{60} kg / s
m=1kgm=1\, kg
Substituting the values in above relation
1×a=2400×1601 \times a =2400 \times \frac{1}{60}
or a=40ms2a =40 \,ms ^{-2}