Lab Report Physics Essay

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Mass of Puck 1 = m1 = 0.80 kg Mass of Puck 2 = m2 = 0.40 kg Charge of Puck 1 = q1 = 3.0 × 10-4 C Charge of Puck 2 = q2 = 3.0 × 10-4 C Velocity of Puck 1 = vi1 = 12.0 ms-1 Velocity of Puck 2 = vi2 = 8.0 ms-1 Initial Distance = ri = 20.0m To Find: Minimum separation between 2 pucks = r =? Solution: We know that, PTO = PTF This can be written as m1v1 + m2v2 = (m1 + m2) v Rearranging the equation to find v as followed v = m1v1 + m2¬v2 / m1 + m2 Putting the given values in the equation as followed v = (0.80 kg × 12.0 ms-1) + (0.40 kg × -8.0 ms-1) / (0.80 kg + 0.40kg) v = 9.6 kgms-1 – 3.2 kgms-1 / 1.2kg v = 6.4 kgms-1 / 1.2 kg v = 5.3 ms-1 To find the minimum separation the equation for conversion of energy is used. Ek1 + Ek2 + EE = E’E + E’k This

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