Question

An experiment is done on a horizontal surface. A block of mass m_1 is initially held in place compressing a spring with spring constant k by a distance x_0. The other end of the spring is firmly attached to the wall and does not move. What is the final speed of m_1 after it is released and loses contact with the spring? After it reaches its final speed, m_1 collides and sticks to a second block of mass m_2. What kind of collision is this? What is the final speed of the two masses? (What does "final" mean here?) After the collision the 2-block system is slowed down by friction between the blocks and the surface on which they are moving. If they travel a distance d from the point of the collision to the point when they come to a stop, what is the coefficient of kinetic friction between the blocks and the surface?

Answer 1

a)

using conservation of energy

kinetic energy = spring potential energy

(0.5) m v² = (0.5) k x_o²

m₁ v² = k x_o²

v = x_o sqrt(k/m₁)

b)

since the blocks sticks together after the collision , hence the collision is inelastic collision

using conservation of momentum

m₁ v_1i + m₂ v_2i = (m₁ + m₂) V

m₁ x_o sqrt(k/m₁) + m₂ (0) = (m₁ + m₂) V

V = x_o sqrt(km₁) /(m₁ + m₂)

c)

d = stopping distance

a = acceleration = - $\mu$ _k g

V_f = final velocity = 0 m/s

V_i = initial velocity = x_o sqrt(km₁) /(m₁ + m₂)

using the equation

V_f² = V_i² + 2 a d

0² = x²_o (km₁)² /(m₁ + m₂)² + 2 (- $\mu$ _k g) d

$\mu$ _k = x²_o (km₁)² /(m₁ + m₂)² /(2gd)