The vector position of a 3.45 g particle moving in the xy plane varies in time according to
1 = 3 + 3t + 2t2
where t is in seconds and is in centimeters. At the same time, the vector position of a 5.00 g particle varies as 2 = 3 − 2t2 − 6t.
Determine the vector position of the center of mass at t = 2.80.
To determine the vector position of the center of mass, we need to find the weighted average of the positions of the two particles. The weight for each particle is given by its mass.
Given that the positions of the two particles at time t are:
Particle 1: r₁ = 3 + 3t + 2t²
Particle 2: r₂ = 3 − 2t² − 6t
We can 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 the particles.
Given that m₁ = 3.45 g and m₂ = 5.00 g, we can substitute these values and calculate the center of mass position at t = 2.80.
First, let's calculate the position of each particle at t = 2.80:
Particle 1:
r₁ = 3 + 3(2.80) + 2(2.80)² = 3 + 8.4 + 15.68 ≈ 27.08 cm
Particle 2:
r₂ = 3 − 2(2.80)² − 6(2.80) = 3 − 15.68 − 16.8 ≈ -29.48 cm
Now, let's calculate the center of mass position:
r(cm) = (m₁ * r₁ + m₂ * r₂) / (m₁ + m₂)
= (3.45 * 27.08 + 5.00 * (-29.48)) / (3.45 + 5.00)
= (93.426 + (-147.4)) / 8.45
= -53.974 / 8.45
≈ -6.39 cm
Therefore, the vector position of the center of mass at t = 2.80 is approximately -6.39 cm.