Contraction of spring colliding with balls

[devanlevin]

on a vertical spring (k=100n/m) there is a box with a mass of 5kg, a ball of 1kg is dropped into it from a height of 1m what is the maximum contraction of the spring if the collosion is plastic.

i hope that my terminology is correct as i am not studying this in english and am translating

what i did was find the velcocity of the colision, ie the velocity of the ball after falling 1m, came to 4.27[m/s]

then using momentum and collision laws

m1v1+m2v2=(m1+m2)U

M1-the ball
M2-the box

m2v2=0(stationary object)

U=m1v1/(m1+m2)
U=(1*4.27)/6=0.7378[m/s]

then what i did was, using energy i said the energy is conserved from after the impact till the springs maximum contraction so:

E=0.5(m1+m2)U^2=0.5Kx^2
3*0.7378^2=50x^2

x=0.18m

then onto that i added another0.49m which is the original contraction from the weight of the box, and i get
x=0.67m

but the answer in the book is 0.79m and i just can't see how they got it

 

[Winzer]

Hello devinlevin!
I would agree with your attack until your energy equation:E=0.5(m1+m2)U^2=0.5Kx^2
3*0.7378^2=50x^2- specifically the left side.

First try setting up a general equation, not plugging in numbers and tell me what you get.

 

[baba89]

I would say that your approach and calculations seem reasonable and accurate. However, it is important to note that there may be some minor differences in the calculations due to rounding or slight variations in the values used. It is also possible that the answer in the book is incorrect, as mistakes can happen. I would suggest double-checking your calculations and if they are correct, then your answer of 0.67m is most likely the accurate maximum contraction of the spring.