Question

Toy rocket of mass 0.1 kg has a small fuel of mass 0.02 kg which it burns in 3s. Starting from rest on the horizontal smooth track it gets a speed of 20m\s after the fuel is burned out. What is the thrust of the rocket?

Answer 1

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

Answer 2

Answer:

Explanation:

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