Lab Report Physics Essay

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