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 2.2 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?

Answer #1

Let the initial velocity of L and R be u_{L} and
u_{R} respectively, then by conservation of momentum,

m_{L}u_{L} = m_{R}u_{R}

For L:

acceleration, a= -_{L}g; final
velocity, v_{L}=0; s_{L}=0.41 m

using, v_{L}^{2} - u_{L}^{2} =
2as

u_{L}= 2.0452 m/s

For R:

acceleration, a= -_{R}g; final
velocity, v_{R}=0; s_{R}=0.44 m

using, v_{R}^{2} - u_{R}^{2} =
2as

u_{R}= 1.687 m/s

putting the values of u_{L} and u_{R} and
m_{L}in the first equation,

2.2*2.0452 - m_{R}*1.687 = 0 {negative sign indicates
that the velocities are in the opposite direction}

m_{R}= 2.6671 kg

mass of the block, m= 2.2+2.66

m= 4.86 kg

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