- #1
sophzilla
- 20
- 0
A big olive (m = 0.11 kg) lies at the origin of an xy coordinate system, and a big Brazil nut (M = 0.82 kg) lies at the point (0.99, 2.1) m. At t = 0, a force Fo = (4i + 4j) N begins to act on the olive, and a force Fn = (-4i -3j) N begins to act on the nut. What is the (a)x and (b)y displacement of the center of mass of the olive-nut system at t = 4.6 s, with respect to its position at t = 0?
I first started approaching the problem by doing E(sigma)mixi/Emi, and the same for the y-direction. So, for x-direction, it would be:
(.99molive + 0mnut)/(.82kg + .11kg)
for the y-direction, it would be:
(2.1molive + 0mnut)/(.82kg + .11kg)
I don't even know if I did those correctly.
For the rest, they give you the force in both directions and the duration time (4.6 sec). I have to find the displaceent, which means I first have to find the center of mass for 0 seconds and then for 4.6 seconds.
Can someone help me with how to approach this problem? Thank you.
I first started approaching the problem by doing E(sigma)mixi/Emi, and the same for the y-direction. So, for x-direction, it would be:
(.99molive + 0mnut)/(.82kg + .11kg)
for the y-direction, it would be:
(2.1molive + 0mnut)/(.82kg + .11kg)
I don't even know if I did those correctly.
For the rest, they give you the force in both directions and the duration time (4.6 sec). I have to find the displaceent, which means I first have to find the center of mass for 0 seconds and then for 4.6 seconds.
Can someone help me with how to approach this problem? Thank you.