A massless spring with spring constant k is compressed by an amount xo. A block

Question

A massless spring with spring constant k is compressed by an amount xo. A block of mass m is placed on the compressed spring and released from rest on a frictionless surface. This block then strikes another block of mass 2m, which is hung on a massless cord of length L as shown in the figure to the right. The masses stick together after the collision and they both swing up to a height h.

Calculate max, the maximum angle from the vertical position that is reached by the masses, if you are given the following values: k = 1500. N/m, xo = 0.100 m, m = 2.00 kg, L = 1.00 m andg = 9.80 m/s2. Give your answer in degrees.

Explanation / Answer

initially spring energy = 1/.k.x0.^2

let V0 be initial velocity of mass 'm' after released from the spring,

then by energy conservation

1/.k.x0.^2 = 1/2.m.V0^2    V0 = k.X0^2 / m = 1500 x 0.1^2 / 2 = 2.74 m/s

now by momentum conservation after striking

m.Vo = (2m + m). Vf {Vf = combined velocity of mass m and 2m}

Vf = V0/3 =2.74/3 = 0.913 m/s

now at max thevelocitu of the mass (2m+m) will become 0

then applying kinematic equation : 0 = Vf2 - 2.g.h

here h = l(1-cos)

0.9132 = 2 x 9.81 x 1 (1 - cos ) cos = 0.957

max = 16.760

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