A rocket travelling at a speed of 200 m / s ejects its products of combustion at the speed of 1200 m / s relative to the rocket, then the speed of escaping vapours with respect to the person on the ground is:

Question

A rocket travelling at a speed of 200 m / s ejects its products of combustion at the speed of 1200 m / s relative to the rocket, then the speed of escaping vapours with respect to the person on the ground is: (A) 1000 ms -1 (B) 1200 m / s (C) 1400 m / s (D) 200 m / s

Tardigrade Admin · Accepted Answer

Relative velocity of combustion product of rocket w.r.t the motion of rocket vecVc=+1200 m / s V r =-200 m / s Velocity of vapours is V v V c = V v -( V r ) 1200= barVv-(-200) V v =1000 m / s

Q. A rocket travelling at a speed of $200 m / s$ ejects its products of combustion at the speed of $1200 m / s$ relative to the rocket, then the speed of escaping vapours with respect to the person on the ground is:

$1000 m s^{- 1}$

$1200 m / s$

$1400 m / s$

$200 m / s$

Relative velocity of combustion product of rocket
w.r.t the motion of rocket
$V_{c} = + 1200 m / s V_{r} = - 200 m / s$
Velocity of vapours is $V_{v}$
$V_{c} = V_{v} - (V_{r})$
$1200 = \overset{ˉ}{V}_{v} - (- 200)$
$V_{v} = 1000 m / s$

Q. A rocket travelling at a speed of 200m/s ejects its products of combustion at the speed of 1200m/s relative to the rocket, then the speed of escaping vapours with respect to the person on the ground is:

1000ms−1

1200m/s

1400m/s

200m/s