Question

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

Answer 1

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