A 2.0 kg mass, shown in the figure above, is attached to a horizontal Hooke\'s L

Question

A 2.0 kg mass, shown in the figure above, is attached to a horizontal Hooke's Law spring with a spring constant of 600 N/m and displaced 1.0 m to the right from the equilibrium point. The 2.0 kg mass is released and, at a position, 0.50 m to the left of the equilibrium point, it collides with a 4.0 kg mass and sticks. Both masses rest on a frictionless surface.

What is the amplitude of the simple harmonic motion after the collision. Hint: You will need to calculate the velocity of the 2.0 kg mass the instant before the collison and the velocity of the pair after the collision. Again you can assume that both masses have physical dimensions which can be neglected.

0.71 m

0.65 m

0.59 m

0.78 m

1.00 m

0.71 m

0.65 m

0.59 m

0.78 m

1.00 m

Explanation / Answer

change in spring potential energy = kinetic energy

1/2 * kx^2 = 1/2 * mv^2

v = sqrt (k/m) * x

v = 17.32 m/s

after collison momentm conserved

m1v1 + m2v2 = (m1 + m2) v'

v2 = 0

v' = 2 x 17.32 / 6 = 5.77 m/s

total energy = 1/2 * k * A^2

1/2 * k * A^2 = 1/2 * k x^2 + 1/2 (m1+m2)v'^2

A = 1.15 m

2nd block at 0.5 m left from equlibrium

A =1.15 - 0.5 = 0.65 m

B option

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