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 2.2 kg, encounters a coefficient of kinetic friction μL = 0.52 and slides to a stop in distance dL = 0.41 m. Piece R encounters a coefficient of kinetic friction μR = 0.33 and slides to a stop in distance dR = 0.44 m. What was the mass of the block?
Let the initial velocity of L and R be uL and uR respectively, then by conservation of momentum,
mLuL = mRuR
For L:
acceleration, a= -Lg; final velocity, vL=0; sL=0.41 m
using, vL2 - uL2 = 2as
uL= 2.0452 m/s
For R:
acceleration, a= -Rg; final velocity, vR=0; sR=0.44 m
using, vR2 - uR2 = 2as
uR= 1.687 m/s
putting the values of uL and uR and mLin the first equation,
2.2*2.0452 - mR*1.687 = 0 {negative sign indicates that the velocities are in the opposite direction}
mR= 2.6671 kg
mass of the block, m= 2.2+2.66
m= 4.86 kg
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