Problem 3. The cylindrical block A of mass mA-1.5 kg is released from rest at B

Question

Problem 3. The cylindrical block A of mass mA-1.5 kg is released from rest at B and slides down the smooth circular guide, then strikes the block C with mc = 2.25 kg and becomes embedded in it. Both blocks slide the horizontal path. If the coefficient of kinetic friction between the block and the horizontal surface is k = 0.25 and radius of the circular guide is r 1.35 m, MA (a) Estimate the energy loss during impact. (b What is the distance s travel by the embedded blocks when they reach half their velocity after impact?

Explanation / Answer

a)Inital energy of block A=mA=1.5*9.8*1.35=19.845 Joules

velocity of the block just before hitting block C=sqrt(2*9.8*1.35)=5.14 m/s

mass of block C=2.25 kg

From momentumconservation, initial momentum and final momentum will be conserved

momentum of system just before block A hits block C=1.5*5.14=7.71 kg-m/s

let velocity of system just after collision takes place be v

momentum after collision = (1.5+2.25)*v=3.75v

3.75v=7.71

v=2.056 m/s

energy of system just after impact=0.5*3.75*2.0562=7.926 Joules

Energy lost due to impact=19.845-7.926=11.92 Joules

Answer is 11.92 joules

b)deceleration of the blocks due to friction=0.25*9.8=2.45 m/s2

velocity just after impact=2.056 m/s

half of the velocity=2.056/2=1.028 m/s

let distnace travelled be s

1.0282=2.0562+(2*-2.45*s)

s=0.647m

Answer is 0.647 m

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