A balloon having mass ' m ' is filled with gas and is held in hands of a boy. Then suddenly it get released and
gas starts coming out of it with a constant rate. The velocities of the ejected gases is also constant 2 m/s
with respect to the balloon. Find out the velocity of the balloon when the mass of gas is reduced to half.
(neglect Buoyancy and Gravity)
Answers
Initial momentum of gas + balloon together = 0.
Initial mass of gas inside = M0 -- this data should be given..
Mass of balloon = m
There is no external force on them. Due to pressure difference gas comes out at dM/dt = k kg/sec. Using Newtons’ third law, the force exerted by outgoing gas on balloon = force exerted by balloon on gas.
Relative velocity of gas wrt balloon = v_rel m/s towards right.
Let absolute velocity of balloon towards left = v m/s
Absolute velocity of gas = v – v_rel in direction towards left. Or, v_rel – v in the direction towards right. We neglect buoyancy, drag, air resistance and gravity effects here.
Mass of gas left in the balloon at time t = M (t) = M = M0 – k t.
Gas inside balloon moves at velocity v towards left, changes direction and velocity to v_rel. Total change in momentum in dt time of gas: dp = (k dt) (v+v_rel).
So Force on gas leaking out = dp/dt = k (v+v_rel)
Impulse imparted to balloon towards left = (m + M0 – k t – k dt) dv. We neglect dt dv term as it is too small compared to others.
Force on balloon towards left = (m+M0- kt) dv/dt
k (v+ v_rel) = (m+M0 –kt ) dv/dt
dv/(v+v_rel) = k dt /(M0+m- kt)
Integrating from t = 0 to t = t and v = 0 to v = v,
Ln [ (v + v_rel)/v_rel ] = Ln [(M0+m) / (M0+m – kt) ]
1 + v/v_rel = (M+m0) / (M0+m- k t)
v/v_rel = k t /(M0 +m – k t)
To find v when mass of gas inside balloon becomes half. So
M0 – k t = M0/2 => t = M0/(2k)
v = M0 v_rel / (2 m+M0) =2 /[ 1 + 2 (m/M0) ] m/sec
v < v_rel or 2 m/sec
