1. The problem statement, all variables and given/known data An open container of mass 50.0 kg is rolling to the left at speed of 5.0 m/s on a frictionless surface. A 15.0 kg box slides down a 37° inclined plane and leaves the end of the plane with speed of 3.0 m/s. The end of the plane is a vertical distance of 4.0 m above the bottom of the container (see figure). The box lands in the container and they roll off together. a- What are the velocity components of the box just before it lands in the container? b- What is the magnitude and direction of the final velocity of the container? 2. Relevant equations U1+K1=U2+K2 Pi=Pf 3. The attempt at a solution a)U1+K1=U2+K2 (U2=0) mgh=.5mVo^2=.5mv^2 then i get V=9.35 m/s b)Vx=(15*7.47)-50*5)/65=-2.12 m/s Vy=(15*-5.627)/65 = -1.3 m/s so in (b) i get V=2.12 m/s to the left (removed the Vy because the floor is horizontal) and in (a) is it right to have U2=0 ?? because the question wasn't specific if the container in the ground or not ,, anyway the picture is in the attachment,,are my answers right ?