The vector position of a 3.75 g particle moving in the xy plane varies in time according to the following equation.

Question

The vector position of a 3.75 g particle moving in the xy plane varies in time according to the following equation.

At the same time, the vector position of a 5.20 g particle varies according to the following equation.

For each equation, t is in s and r is in cm. Solve the following when t = 2.00
(a) Find the vector position of the center of mass.
cm
cm

(b) Find the linear momentum of the system.
g-cm/s
g-cm/s

(c) Find the velocity of the center of mass.
cm/s
cm/s

(d) Find the acceleration of the center of mass.
cm/s2
cm/s2

(e) Find the net force exerted on the two-particle system.
μN
μN

Answer 1

Where is the "following Eq"?

Answer 2

To solve this problem, let's start by finding the vector position of each particle at t = 2.00.

Particle 1:
Given equation: r₁ = 3.75t î + (2.00t² + 1.00) ĵ
Substituting t = 2.00:
r₁ = 3.75(2.00) î + (2.00(2.00)² + 1.00) ĵ
r₁ = 7.50 î + (4.00 + 1.00) ĵ
r₁ = 7.50 î + 5.00 ĵ

Particle 2:
Given equation: r₂ = (3.00t² - 4.50) î + 2.50 ĵ
Substituting t = 2.00:
r₂ = (3.00(2.00)² - 4.50) î + 2.50 ĵ
r₂ = (12.00 - 4.50) î + 2.50 ĵ
r₂ = 7.50 î + 2.50 ĵ

(a) To find the vector position of the center of mass, we need to find the average position of the two particles:
R = (1/M)(m₁r₁ + m₂r₂), where M = m₁ + m₂

Given m₁ = 3.75 g and m₂ = 5.20 g, we need to convert these masses to kg:
m₁ = 3.75 g = 0.00375 kg
m₂ = 5.20 g = 0.00520 kg

Substituting the values, we get:
R = (1/(0.00375 + 0.00520))((0.00375)(7.50 î + 5.00 ĵ) + (0.00520)(7.50 î + 2.50 ĵ))

Now, calculate R to find the vector position of the center of mass.

(b) The linear momentum of the system is given by the equation:
p = m₁v₁ + m₂v₂

To find the linear momentum, we need to find the velocities of the particles at t = 2.00.

Given v₁ = 3.75 î + (4.00t) ĵ,
Substituting t = 2.00:
v₁ = 3.75 î + (4.00(2.00)) ĵ
v₁ = 3.75 î + 8.00 ĵ

Given v₂ = (6.00t) î + (2.50) ĵ,
Substituting t = 2.00:
v₂ = (6.00(2.00)) î + 2.50 ĵ
v₂ = 12.00 î + 2.50 ĵ

Substituting the values into the equation, we get:
p = (0.00375)(3.75 î + 8.00 ĵ) + (0.00520)(12.00 î + 2.50 ĵ)

Now, calculate p to find the linear momentum of the system.

(c) The velocity of the center of mass can be found by differentiating R with respect to time:
V_cm = (dR/dt)

(d) The acceleration of the center of mass can be found by differentiating V_cm with respect to time:
A_cm = (d²R/dt²)

(e) To find the net force exerted on the two-particle system, we can use Newton's Second Law:
F_net = d(p)/dt

Now that we have the setup, you can proceed with the calculations to find the answers to parts (a) to (e) of the problem.