The carriage travelling at a speed of v = 1m/s is to be stopped at a dead end by using a series of two springs. The carraige has a mass of 1000 kg. The stiffness of spring A is 500 N/m while that of B is 800 N/m. Determine the maximum deformation of each spring due to the collision.
1) sA = 1.104 m, sB = 0.693 m.
2) sA = 2.453 m, sB = 1.245 m.
3) sA = 1.221 m, sB = 5.213 m,

4) sA = 1.453 m, sB = 0.234 m.

Let sA and sB be the displacement of springs A and B respectively.
There is no change in the gravitational potential energy.

Conservation of energy-
T1 + V1= T2+ V2
0 + (1/2)1000(1)2= 0 + (1/2)500(sA)2 + (1/2)800(sB)2----(1)

Also, Fs = 500sA = 800sB
sA = 1.6sB ---------------------------------------(2)
sB = 0.693 m.
sA = 1.104 m.