Posted by **engmath** on Tuesday, February 16, 2010 at 9:04pm.

The vector position of a 3.65 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 5.60 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 = 2.90

(a) Find the vector position of the center of mass.

i-hat__________ cm

j-hat___________cm

(b) Find the linear momentum of the system.

i-hat___________ g-cm/s

j-hat____________g-cm/s

(c) Find the velocity of the center of mass.

i-hat_________cm/s

j-hat__________ cm/s

(d) Find the acceleration of the center of mass.

i-hat____________cm/s2

j-hat_____________cm/s2

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

i-hat___________ μN

j-hat___________μN

----------------------

Maybe someone could just explain to me how to solve for each part because I'm thoroughly confused. THANK YOU.

## Answer this Question

## Related Questions

physics - The vector position of a 3.80 g particle moving in the xy plane varies...

Physics - The vector position of a 3.80 g particle moving in the xy plane varies...

Physics - The vector position of a 3.00 g particle moving in the xy plane varies...

Physics - The vector position of a 3.45 g particle moving in the xy plane ...

Physics - The vector position of a 3.45 g particle moving in the xy plane ...

Physics - The vector position of a 3.45 g particle moving in the xy plane ...

Physics - The vector position of a 3.45 g particle moving in the xy plane ...

Physics - The vector position of a 3.40 g particle moving in the xy plane ...

physics - The vector position of a 3.75 g particle moving in the xy plane ...

physics - The acceleration of a particle moving only on a horizontal plane is ...