Two identical blocks are placed on a tabletop on either side of a spring, and sq

Question

Two identical blocks are placed on a tabletop on either side of a spring, and squeezed together to compress the spring to a length of L. The blocks are then suddenly released so that the spring expands and the blocks shoot apart. The blocks each have mass m, and the spring has stiffness k and relaxed length L0.

(a) If the tabletop is frictionless, how fast are the blocks moving when they leave the spring?

(b) If the tabletop is not frictionless, and the coefficients of static and kinetic friction between the blocks and tabletop are s and k respectively, how far apart are the blocks when they come to rest?

Explanation / Answer

from momnetum conservation

initial momentum = finalmomentum

Pi = Pf

0 = m1*v1 + m2*v2

since m1 = m2

v1 = v2 = v

elastic potential energy = total kinetic energy of the blocks

(1/2)*k*L^2 = (1/2)*m1*v1^2 + (1/2)*m2*v2^2

(1/2)*k*L^2 = (1/2)*m*v^2 + (1/2)*m*v^2

(1/2)*k*L^2 = m*v^2

v = sqrt(k/2)*L

==============

(b)

work done = change in KE

uk*m*g*x1 = (1/2)*m1*v1^2

uk*m*g*x1 = (1/2)*m*L^2*k/2

x1 = L^2*k/(4*uk*g)

x2 = L^2*k/(4*uk*g)

seperation = x1 + x2 = L^2*k/(2*uk*g)

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