Question

please solve 2

QUESTION 2 a) (5 p) Interpret the rocket equation dv(t)M(1)=-udMO [EQ.1) within the framework of the law of momentan conservation, written in a closed system here M(t) is the rocket mass, at time t, whereas dMt) is by definition, dM(t)-M(t+dt)-M(t): -SM(t)-M(!), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is still by definition, dy(t){t+dt)-v(t), i.e. the increase in the velocity of the rocker through the period of time dt; u is the relative velocity of the exhaust gas, with respect to the rocket. b) (10 p) with EQ.1) show that aM(O-R(EQ.2). Here RisdM(ty/dt, i.e. the mass of the gas thrown out by the rocket through the unit period of time of concern; interpret this equation, and justify it, quickly. (10p) Suppose that, R is a constant. Then calculate to start to lift a payload of 100 tones against Earth's dcceleration gu -2.5 km/s, g"10 m's? Nore, in order to produce force one has to spend energy, and using it, throw out momentum. Attach File Browse My Computer Browse Content Collection

Answer 1

1 AP M = mass of the rocket at M wp-W t tudt dm (a) Suppose Initially, Mo = initial mass of rocket + fuel Võ = initial velocity of socket any instant t during its flight. V = Velocity acquised by socket at that instant wort ground ejected!! dm = mass of the fuel ejected with form of gases gases vg by socket in time at Vg = Velocity of exhaust gases with respect to ground. J = increase in the velocity of rocket in time dt. Now the linear momentum of rocket at time t is given P = M(t) After time (ttdt), the momentum of rocket and ejected gases P2 = (m-dm(t) (3 + dv) tamct) (Vg) 08, P2 = M(t) + MCE) -TMCE) - DMCU) dū - VgdMCL) ox, P2 = MCHỳ + MCt) dvd MCE) (7+Vg) - dMCE) dū - © According to law of conservation of momentum P = P2 Equating ear 1 and ② we get MCE) Ū = MC+ + MCt) -dact) Cv+) - JMC+) since dm (t) du very small. So, we can ignore these quantities. Finally we get Mt) düet) = d MCL)(V+ ) are

suppose ut ug: - put this in eqr & we get M (t) du(t)= - U AM (E). Hence proved (6) Now From eqn@ du(t) = -0 DM(t) MCE) At t=0, MCE)Mo and V = Vo At tot, MCH) - MCU) and V=VCt). Integrating eqn ③ we get Y av =-us dmct) M vo Mo MCE) -- vo Mo MCE) [v] [enm] m or, V(t)- Vo [ en M(t)- en mo 08, VCE) -Vo = v[ en Mo - In MCt)] 08, V(t) – Vo = enamo of V(t) = Votuen when t=0, Vo=0 So, V(t)= ven(mo Upthrust of rocket: Now dividing eqn by dt we get мо MCE) V MCE) dv - UdMct) dt dt or, M (t) a = - dM (t) dt where R = 1-dM dt dt 1 - DMCA So, M(t)a : UR

Thus we conclude that the thrust F on the rocket at any instant is the product of velocity of ex haust gases and the rate of Combustion of the fuel at that instant. Negative sign indicates that the thoust and Velocity of gases are in opposite directions. CC) M = 100 tonnes = 100 x 103 kg = 105 kg U = 2.5 km/se - 2500 m/s g=10m/s² 2 Now F=UR or, Mg = UR ox, R = Mg (105kg) (10m/s2) 400 kg (Ans) 2500 m/s