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On a frictionless horizontal air table, puck A (with mass 0.250 kg ) is moving toward puck B (with mass 0.374 kg), which is i

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Answer #1

here,

mass of puck A , mA = 0.25 kg

mass of puck B, mB = 0.374 kg

final velocity of A , vA = 0.125 m/s

final velocity of B , vB = 0.649 m/s

a)

let the speed of the particle before the collison be uA

using conservation of momentum

mA * uA = mA * vA + mB * vB

0.25 * uA = 0.25 * 0.125 + 0.374 * 0.649

uA = 0.846 m/s

b)

the change in total kinetic energy of the system , KE = KEf - KEi

KE = (0.5 * mA * vA^2 + 0.5 * mB * vB^2) - (0.5 * mA * uA^2)

KE = 0.5 * ( 0.25 * 0.125^2 + 0.374 * 0.649^2 - 0.25 * 0.846^2) J

KE = - 8.75 * 10^-3 J

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