Question

In the figure here, a stationary block explodes into two pieces L and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 3.8 kg, encounters a coefficient of kinetic friction μL = 0.56 and slides to a stop in distance dL = 0.48 m. Piece R encounters a coefficient of kinetic friction μR = 0.56 and slides to a stop in distance dR = 0.31 m. What was the mass of the block?

Answer 1

consider the motion for L :

V_1i = initial speed

V_1f = final speed = 0

a = acceleration = - μ_L g = - 0.56 x 9.8 = - 5.5

d_L = stopping distance = 0.48 m

Using the equation

V_1f² = V_1i² + 2 a d_L

0² = V_1i² + 2 (- 5.5) (0.48)

V_1i = 2.3 m/s

consider the motion for R :

V_2i = initial speed

V_2f = final speed = 0

a = acceleration = - μ_R g = - 0.56 x 9.8 = - 5.5

d_R = stopping distance = 0.31 m

Using the equation

V_2f² = V_2i² + 2 a d_L

0² = V_2i² + 2 (- 5.5) (0.31)

V_2i = 1.85 m/s

m₁ = mass of L = 3.8 kg

m₂ = mass of R

using conservation of momentum

0 = m₁ V_1i - m₂ V_2i                        (since block was initially stationary)

0 = 3.8 (2.3) - m₂ (1.85)

m₂ = 4.72 kg

mass of block = m₁ + m₂ = 3.8 + 4.72 = 8.52 kg