Question

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 2.5 kg, encounters a coefficient of kinetic friction ÎĽl = 0.43 and slides to a stop in distance dl = 0.38 m. Piece r encounters a coefficient of kinetic friction ÎĽr = 0.45 and slides to a stop in distance dr = 0.40 m. What was the mass of the block?

Answer 1

po = 0 kgm2 = pf

Therefore absolute momentum to the left = to the right

mL vL = mR vR

2 (2) = mR 0.38

mR = 2.5 kg

 

mT = 2 + 2.5 kg = 3.05 kg

Block L

                           K0   –       Ff      d    =  Kf

                       ½ 2 vo2 – μL mg (0.4)  =  0

                       vL = 2 m/s

Block R

   K0   –            Ff            d      =  Kf

½ m vo2 –       μR×mg×dx)    =  0

½ m vo2 –  (0.1+ ½ x)× mg (dx)   =  0

½ m vo2 –  mg ∫(0.1+ ½ x) ×(dx)  =  0  

½ m vo2 –  mg (0.1x + x2) ×(0 to 0.8) =  0

½ m vo2 –  0.08mg + 0.64mg      =  0

½ m vo2 =0 .72mg  

vR = 3.8 m/s