The vector position of a 3.00 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 5.85 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 = 2.10
(e) Find the net force exerted on the two-particle system.
____i ìN
____j ìN
r1 = 3ti + (2t^2 + 3t)j
r2 = (3-2t^2)i - 6tj
v1 = r1' = 3i + (4t+3)j
v2 = r2' = -4ti - 6j
a1 = r1'' = 4j
a2 = r2'' = -4i
F1 = m1a1 = 3*4j = 12j
F2 = m2a2 = 5.85(-4i) = -23.4i
Fnet = -23.4i + 12j
To find the net force exerted on the two-particle system, we need to calculate the total force exerted on each particle individually and then add them together.
First, let's calculate the force exerted on particle 1 using its position equation:
r1 = (3i + 3j)t + 2jt^2
To find the force, we need to find the second time derivative of the position vector, which gives us acceleration:
a1 = d^2r1/dt^2
Differentiating the position equation once with respect to time (t), we get:
v1 = dr1/dt = (3i + 3j) + (4j)t
Differentiating v1 with respect to t again, we get:
a1 = dv1/dt = 4j
Now let's calculate the force exerted on particle 2 using its position equation:
r2 = 3i - 2it^2 - 6jt
Similar to particle 1, we differentiate the position equation twice to find the acceleration:
v2 = dr2/dt = -4it - 6j
a2 = dv2/dt = -4i
Now that we have the accelerations for both particles, we can calculate the forces using Newton's second law:
F1 = m1 * a1
F2 = m2 * a2
Substituting the given masses, where m1 = 3.00 g = 0.00300 kg and m2 = 5.85 g = 0.00585 kg:
F1 = 0.00300 kg * 4j = 0.0120j N
F2 = 0.00585 kg * -4i = -0.0234i N
Finally, to find the net force exerted on the two-particle system, we add together the forces:
Net Force = F1 + F2
Net Force = 0.0120j N + (-0.0234i N)
Therefore, the net force exerted on the two-particle system is:
Net Force = -0.0234i N + 0.0120j N.