The 28 Mg freight car A and 14Mg height car B are moving tow

The 28 Mg freight car A and 14-Mg height car B are moving towards each other with the velocities shown Determine the maximum compression of the spFind theg mounted on car A. Neglect rolling resistance. Express your answer to three significant figures and include the appropriate units.

Solution

mA = 28 Mg = 28000 kg

vA = 20 km/h = 20 x 1000m / 3600s = 5.56 m/s

mB = 14000 kg

vB = - 10/3.6 = - 2.78 m/s

while maixmum compression, both car will be moving with same velocity say v.

Applying momentum conservation,

mA vA + mB vB = (mA +mB)v

28000 x 5.56 + (14000 x -2.78) = (28000 + 14000)v

v = 2.78 m/s

USing energy conservation,

initial PE + KE = final PE + KE

(28000 x 5.56^2 / 2) + (14000 x 2.78^2 / 2) + 0 = kA^2 /2 + (28000 + 14000) x 2.78^2 / 2

k = 3 M N / m = 3 x 10^6 N/m

973778 = 3 x 10^6 x A + 324593

A = 0.465 m