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 2.3 kg, encounters a coefficient of kinetic friction ?L = 0.55 and slides to a stop in distance dL = 0.35 m. Piece R encounters a coefficient of kinetic friction ?R = 0.30 and slides to a stop in distance dR = 0.43 m. What was the mass of the block?

Answer 1

Given that
         mJ = 2.3 kg
         dL = 0.35 m
         ?L = 0.55
         ?R = 0.30
         dR = 0.43 m


According to work - energy theorem we have
          0.5 mL vL² = ?L mL g dL
==>     v_L = ? [ 2 ?_L g d_L ]
                = ? [ 2 * 0.55 * 9.8 * 0.35 ]
                = - 1.942 m/s ( negative direction is because it is moving left side)
Similarly for R
       v_R = ? [ 2 ?_R g d_R ]
                  = ?[ 2 * 0.30 * 9.8 * 0.43]
                  = 1.5900 m/s
According to conservation of energy we have
                0 = mL vL + mR vR
==>       mL / mR = - vR / vL
==>       mR = - m L vL / vR
                     = - 2.3 * -1.942 / 1.5900
                     = 2.809 kg
Total mass M = mL + mR = 2.3 kg + 2.809 kg = 5.109kg