Spring Compression (x-x0) = .00579 m

[nrip6]

Homework Statement

A spring (k = 2900 N/m ) is compressed between two blocks: block 1 of inertia 1.70 kg and block 2 of inertia 2.00 kg. The combination is held together by a string (not shown in (Figure 1) ). The combination slides without spinning across low-friction ice at 2.90 m/s when suddenly the string breaks, allowing the spring to expand and the blocks to separate. Afterward, the 2.00-kg block is observed to move at a 34.0∘ angle to its initial line of motion at a speed of 3.50 m/s, while the smaller block moves off at 3.71m/s and angle of 38.54 below it's initial line of motion. Neither block is rotating after the separation, and you can ignore the inertias of the spring and the string relative to those of the blocks. Determine the original compression of the spring, x−x0, from its relaxed length.

Homework Equations

U=1/2k(x-xo)
K=1/2mv^2
Conservation of momentum
U+K=K1+K2

The Attempt at a Solution

U+K=K1+K2
1/2k(x-xo)+1/2mv^2=1/2m1v1^2+1/2m2v2^2
k(x-xo)+mv^2=m1v1^2+m2v2^2
(2900)(x-xo)+(1.70+2.00)(2.9^2)=(1.70)(3.71^2)+(2.00)(3.50^2)
(2900)(x-xo)+31.117=23.39897+24.5
(2900)(x-xo)=16.78197
(x-xo)=.00579

 

[haruspex]

Homework Equations

U=1/2k(x-xo)

Not quite. Everything else looks ok.
It does bother me that you could introduce one more unknown, the original orientation of the system relative to its line of motion, and obtain two conservation of momentum equations. This implies the question is overspecified, perhaps inconsistently so.