Arun
Last Activity: 6 Years ago
Rocket without fuel: Mass M = 0.1 kg, Fuel mass m = 0.02 kg
fuel burning time = T = 3 sec,
Rate of fuel burning = dm/dt = r = 0.02/3 kg/s
Mass of total rocket at time t = M + m - r t
Velocity of rocket = v
Acceleration of rocket forward = a = dv/dt
Relative velocity of burnt and ejected gases from fuel relative to rocket = u
Reduction in mass of rocket = dm = r * dt
Applying the law of conservation of linear momentum, we get:
u dm = (M + m - r t) dv
u * dm/dt = (M + m - r t) dv/dt
a(t) = dv/dt = u *dm/dt /(M + m- r t)
Thrust = (M+m- rt) * a = u dm/dt = u * r
velocity of rocket = v(T) = integral a(t) dt from t = 0 to T
v(T) = integral of { u r /(M+m- r t) } dt from t = 0 to T
v(T)= u * Ln [(M+m) / (M+m- r T)]
When t = T = 3 sec
Hence, 20 m/s = u * Ln [ (0.1+0.02) /(0.1 + 0.02 - 002/3 * 3)]
u = 20 / Ln(0.12/0.1) = 109.69 m/s
Thrust forward: u * r = 109.69 * 0.02/3 = 0.731 N
