Here is the question. On July 25, 1956, the liner Andrea Doria, of mass 4.1Times

Question

Here is the question.

On July 25, 1956, the liner Andrea Doria, of mass 4.1Times107 kg, was heading west at 40 km/h. It collided off Nantucket Island with the Stockholm, of mass 1.7 times 107 kg, sailing at 30 km/h in the direction 20degree east of north. The bow of the Stockholm temporarily lodged in the side of the Andrea Doria; that is, the collision was completely inelastic. (a) Find their common velocity just after the collision. (b) What was the loss in kinetic energy due to the collision? (The Andrea Doria subsequently sank.)

Explanation / Answer

m1 = 4.1*10^7 kg

v1 = 40kmph west = 11.11 -i m/s

m2 = 1.7*10^7 kg

v2 = 30 kmph 20 east of north = 2.85i + 7.83j m/s

m1v1 + m2v2 = (m1+m2)v

v = 7.02i + 2.3j m/s = 26.6kmph 18.5 north of west

loss in kinetic energy = 0.5*10^7*(4.1*11.11^2 + 1.7*8.33^2 - 5.8*7.38^2)

= 1.54*10^9 Kgm^2/s^2

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