A 2.6 kg particle that is moving horizontally over a floor with velocity (-3.00 m/s)j undergoes a totally inelastic collision with a 4.00 kg particle that is moving horizontally over the floor with velocity (4.20 m/s)i. The collision occurs at xy coordinates (-0.50 m, -0.10 m). After the collision, what is the angular momentum of the stuck-together particles with respect to the origin using vector notation? I first tried using conservation of momentum, which gives me 9 as the x-momentum. This is wrong. I know that angular momentum (L) = Iw, but I'm not sure how to calculate I or w. I'm also stuck on a similar problem: At the instant the displacement of a 2.00 kg object relative to the origin is d = (2.00 m)i + (4.00 m)j - (3.00 m)k, its velocity is v = - (6.40 m/s)i + (2.90 m/s)j + (2.70 m/s)k, and it is subject to a force F = (6.40 N)i - (7.80 N)j + (4.10 N)k. Find the angular momentum of the object and the torque acting on the object about the origin using vector notation. To find momentum, I again didn't know how to find Iw, so I just tried multiplying the given mass times the velocity vector. I'm guessing this is wrong because thats for linear momentum. Any help would greatly be appreciated!