In Figure, a stationary block explodes into two pieces and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.6 kg, encounters a coefficient of kinetic friction μL=0.40 and slides to a stop in distance deal = 0.15 m. Piece R encounters a coefficient of kinetic friction μR=0.50 and slides to a stop in distanced R=0.30 m. What was the mass of the original block?

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toroztatG · Accepted Answer

Original stationary block mass - M&amp;nbsp;mL, left going mass an its velocity be&amp;nbsp;vL,&amp;nbsp;mR&amp;nbsp;is right going mass and its velocity be&amp;nbsp;vR&amp;nbsp;Momentum before impact = momentum after impact&amp;nbsp;M×0=mLvL+mRvR&amp;nbsp;vLvR=−mRmL=−M−2.62.6&amp;nbsp;KE of left mass is equal to the work done against frictional force&amp;nbsp;12mLvL2=frdL&amp;nbsp;=μLmLgdL&amp;nbsp;vL2=2×0.4×9.8×0.15&amp;nbsp;vL=1.0844ms&amp;nbsp;The kinetic energy of right mass is utilised inovercoming friction&amp;nbsp;12(M−2.6)vR2=frdR=μR(M−2.6)g×dR&amp;nbsp;vR2=2×.5×9.8×.3&amp;nbsp;vR=1.7146ms&amp;nbsp;M−2.6=vLvR2.6&amp;nbsp;M=4.244kg

In Figure, a stationary block explodes into two pieces and R that slide across a frictionless floor and then into regions with friction

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