The vector position of a 3.80 g particle moving in the xy plane varies in time according to the following equation.
r1=(3i+3j)t+2jt^2
At the same time, the vector position of a 5.35 g particle varies according to the following equation.
r2=3i-2it^2-6jt
For each equation, t is in s and r is in cm. Solve the following when t = 2.60
Find the vector position of the center of mass.(i+j)
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To find the vector position of the center of mass, we need to calculate the weighted average of the positions of the two particles, where the weights are the masses of the particles.
1. Start by calculating the position of each particle at t = 2.60.
For particle 1:
r1 = (3i + 3j)t + 2jt^2
r1 = (3i + 3j)(2.60) + 2j(2.60)^2
r1 = (7.8i + 7.8j) + (2(2.60)^2)j
r1 = (7.8i + 7.8j) + (13.52j)
r1 = (7.8i + 21.32j)
For particle 2:
r2 = 3i - 2it^2 - 6jt
r2 = 3i - 2i(2.60)^2 - 6j(2.60)
r2 = 3i - 2i(6.76) - 15.6j
r2 = 3i - 13.52i - 15.6j
r2 = (-10.52i - 15.6j)
2. Calculate the center of mass position using the formula:
r_cm = (m1 * r1 + m2 * r2) / (m1 + m2)
where m1 and m2 are the masses of particle 1 and particle 2, respectively.
Given that particle 1 has a mass of 3.80 g and particle 2 has a mass of 5.35 g, convert these masses to kilograms:
m1 = 3.80 g = 0.00380 kg
m2 = 5.35 g = 0.00535 kg
3. Substitute the values into the equation:
r_cm = (0.00380 kg * (7.8i + 21.32j) + 0.00535 kg * (-10.52i - 15.6j)) / (0.00380 kg + 0.00535 kg)
r_cm = (0.02964i + 0.08102j - 0.05607i - 0.08346j) / 0.00915 kg
r_cm = (0.02557i - 0.00244j) / 0.00915 kg
r_cm = 2.8009i - 0.267 j
Therefore, at t = 2.60, the vector position of the center of mass is approximately 2.8009i - 0.267j cm.