Question

In Figure 9-57, 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 = 0.40 and slides to a stop in distance dL = 0.15 m. Piece R encounters a coefficient of kinetic friction µR = 0.50 and slides to a stop in distance dR = 0.38 m. What was the mass of the original block?

Answer 1

let mL be the mass of the piece L

and mR be the mass of the piece R

then required mass of the original block is mL+mR

mL = 2.2 kg (Given)

according to work energy theorem

work done by the frictional force on L = change in kinetic energy of L

-fk*dL = -0.5*mL*uL^2

fk*dL = 0.5*mL*uL^2

fk = mu_L*mL*g = 0.4*2.2*9.81 = 8.64 N

8.64*0.15 = 0.5*2.2*uL^2

then initial speed of L is uL = 1.08 m/s

similarly initial speed of the R is

mu_R*dR*mR*g = 0.5*mR*uR^2

0.5*0.38*mR*9.81 = 0.5*mR*uR^2

uR = 1.93 m/s

apply law of conservation of momentum

mL*uL = -mR*uR = -mR*1.93

2.2*1.08/1.93 = -mR

mR = 1.23

totalmass of the original block is 1.23+2.2 = 3.43 kg