Answer to Question #276493 in Civil and Environmental Engineering for Fernando

Question

A spring is such that a 2 lb weight stretches it by 6 in. There is a damping force present, with

the magnitude equal to 0.75|𝑣|. An impressed force of F(t) = 2 sin 8t is acting on the spring. If

the 2-lb weight is released from a point 3 in below the equilibrium point, describe the motion

Answer 1

Momentum after release of steam P.F._f 

P_f = mv + m'v'       (1)

Where mass of empty launch steam m=7500 kg

m' is mass of satellite , m'=250 kg

v is final velocity of launch steam after releasing satellite, v=900 m/s

v' is final velocity of the satellite after release

Putting all the values in equation (1), we get

P_f =(7500 kg)(900 m/s) + (250 kg)v'

P_f = 6.750×10⁶ kg•m/s + (250 kg)v'

Now applying conservation of momentum

Momentum before release of satellite = momentum after release of satellite

P_i = P_f

7.750×10₆ kg•m/s = 6.750×10⁶ kg•m/s + (250 kg)v'

(7.750×10₆ - 6.750×10⁶ )kg•m/s= (250 kg)v'

10⁶ m/s = 250v'

v' = (10⁶/250) m/s

v' = 4000 m/s