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.
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To determine the vector position of the center of mass at t = 2.80, we need to find the position vectors of both particles at that time and then calculate the center of mass using the formula:
Center of mass = (m1 * r1 + m2 * r2) / (m1 + m2)
where m1 and m2 are the masses of the particles, and r1 and r2 are the position vectors of the particles.
Let's first find the position vector for the first particle at t = 2.80:
1 = 3 + 3t + 2t^2, where t = 2.80
1 = 3 + 3(2.80) + 2(2.80)^2
1 = 3 + 8.4 + 2(7.84)
1 = 3 + 8.4 + 15.68
1 = 26.08 cm
So, the position vector for the first particle at t = 2.80 is 1 = 26.08 cm.
Now, let's find the position vector for the second particle at t = 2.80:
2 = 3 - 2t^2 - 6t, where t = 2.80
2 = 3 - 2(2.80)^2 - 6(2.80)
2 = 3 - 2(7.84) - 6(2.80)
2 = 3 - 15.68 - 16.8
2 = -29.48 cm
So, the position vector for the second particle at t = 2.80 is 2 = -29.48 cm.
Now, let's calculate the center of mass:
Center of mass = (m1 * r1 + m2 * r2) / (m1 + m2)
Given:
m1 = 3.45 g = 0.00345 kg
m2 = 5.00 g = 0.00500 kg
r1 = 26.08 cm = 0.2608 m
r2 = -29.48 cm = -0.2948 m
Center of mass = (0.00345 * 0.2608 + 0.00500 * -0.2948) / (0.00345 + 0.00500)
Center of mass = (0.00089916 - 0.0014730) / 0.00845
Center of mass = -0.00057384 / 0.00845
Center of mass = -0.0679 m
Therefore, the vector position of the center of mass at t = 2.80 is approximately -0.0679 m.