AL PHY SHM

A trolley of mass 1kg is attacted to two fixed supports by identical stretched springs each of stiffness (restoring force per unit extension ) equal to 2.5Nm^-1


圖片參考：http://i33.photobucket.com/albums/d51/lop0102/IMG_1738.jpg


a) The trolley is displaced from its equ. postion by 0.1m (both springs remain in tension) and released from rest.
Calculated the time for it to complete one quarter of an oscillation (neglect friction)

b) The trolley is again displaced by 0.1m from its equilibrium postion and released from rest . However , this time a 1kg load is dropped from just above the trolley and lands on the trolley exactly when it passed through the equilibrium position . Assuming that the load sticks to the trolley , calculate the new amplitude of oscillation.

i have done the part a of this question.
but i have some question on part b
the answer of part b is 0.071m

But in my calculation , I tried this method to find the answer.

Since the trolley is displaced by 0.1m.
so the elastic P.E : (0.5)(k)(A)^2
(0.5)(k)(0.1)^2 = 0.025 J

by energy conservation, the elastic PE will change into KE
PE lost = KE gained.
0.025 = (0.5)(m)(v)^2
0.025 = (0.5)(0.1)(v)^2
v = 0.707 ms^-1

since there is a load dropped and stick to the trolley
by conservation of momentum,
m1v1 = m2v2
1*0.707 = (1+1)(v2)
v2 = 0.353 ms^-1

since now the new KE is (0.5)(1+1)(0.353)^2 = 0.124 J
and the KE will change into elastic PE

0.124 = 0.5 * 5 * A^2
so A = 0.224m

but my answer is different from the correct question,
where is my mistakes?

please point out my mistakes

if necessary , you can explain it in chinese.