- #1
naianator
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Homework Statement
A block of mass m is attached to a spring with a force constant k, as in the above diagram. Initially, the spring is compressed a distance x from the equilibrium and the block is held at rest. Another block, of mass 2m, is placed a distance x/2 from the equilibrium as shown. After the spring is released, the blocks collide inelastically and slide together. How far (Δx) would the blocks slide beyond the collision point? Neglect friction between the blocks and the horizontal surface.
Homework Equations
U_spring = 1/2*k*x^2
K = 1/2*m*v^2
F = m*a
v_f^2 = v_0^2+2*a*x
The Attempt at a Solution
E_i = 1/2*k*x^2 = E_f = 1/2*m*v^2 + 1/2*k*x_f^2
at x/2 where m collides with 2m:
1/2*k*x^2 = 1/2*3m*v^2 + 1/8*k*x^2
k*x^2 = 3m*v^2 + 1/4*k*x^2
3/4*k*x^2 = 3*m*v^2
yielding
v = sqrt(k*x^2/(4*m))
by Newtons second law a = -k*x/(3*m)
and using kinematics (v_f^2 = v_0^2+2*a*x)
0 = k*x^2/(4*m) - 2*(k*x/(3*m))*delta(x)
x/4 = 2/3*delta(x)
delta(x) = 3/8*x
Am I way off base with this approach?