Question #276492

A spring is such that a 4 lb weight stretches it by 0.64 ft. The weight is pushed up 1/3 ft above

the equilibrium point and started with an initial downward velocity of 5 ft/sec. The motion takes

place in a medium which furnishes a damping force of magnitude 0.25|𝑣| at all times. Describe

the motion.

Expert's answer

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} = 6.750×10^{6 }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}

7.750×10_{6}^{ }kg•m/s = 6.750×10^{6 }kg•m/s + (250 kg)v'

(7.750×10_{6}^{ }- 6.750×10^{6 })kg•m/s= (250 kg)v'

10^{6} m/s = 250v'

v' = (10^{6}/250) m/s

v' = 4000 m/s

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