A block of mass m_2 = 2.8 kg is attached to a massless spring with spring consta

Question

A block of mass m_2 = 2.8 kg is attached to a massless spring with spring constant k = 1500 N/m, which is in turn attached to a fixed wall. Initially the spring is at its natural length. A second block of mass m_1 = 1.2 kg slides down from a frictionless quarter arc of radius R = 2.55 m onto a frictionless surface, and collides with m_2. What is the speed v of m_1 right before its collision with m_2? After a completely in-elastic collision, the two masses stick together and squeeze the spring. What is the velocity v' of the combined mass right after the collision? What is the maximum compression x of the spring? If the surface under the spring has a kinetic friction coefficient Mu_k = 0.57, what is the new maximum compression x' of the spring?

Explanation / Answer

part a )

v1 = sqrt(2gR)

v1 = 7.069 m/s

part b )

linear momentum remain conserve

m1v1 + m2v2 = (m1+m2)v

v2 = 0

v = m1v1/(m1+m2)

v = 2.12 m/s

part c )

1/2*kx^2 = 1/2*(m1+m2)v^2

x^2 = (2.8 + 1.2)*2.12^2/1500

x = 0.109 m

part d )

1/2kx^2 = 1/2*(m1+m2)v^2 - mu_k * (m1+m2)*g*x

750x^2= 8.9888 - 22.344x

750x^2 - 8.9888 + 22.344 x = 0

x = 0.0956 m

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