Recent content by clemsonguy

Just to make sure I have done this correctly X-component: 0 = (.25m)(20) + (.25m)(20)(cos 45) + (.5m)(v3)(cos θ) - (.5m)(v3)(cos θ) cos θ = (10 + 5sqrt2)/v3 Y-component: 0 = (.25m)(20)(sin 45) +(.5m)( v3)(sin θ) - (.5m)(v3)(sin θ) sin θ = 5sqrt2/v3 tan θ = (5sqrt2)/(10 +...

The 1/2m cancel so v=20cos(22.5). From that I get 18.48 cm/s.

22.5 degrees west of north? For the speed, is this correct? momentum = (m/2)V (m/2)V = sq rt[{(m/4)^2}(0.20^2) + {(m/4)^2}(0.20^2) + {2*(m/4)^2}(0.20^2)cos 45] = sq rt(2)*(m/4)*sq rt[1+cos 45] = sq rt(2)*(m/4)*sq rt[2*cos^2(45/2)] = (m/2)*cos (22.5) So V = cos(22.5) = 0.9238 m/s

Homework Statement I've been trying to solve this question for hours now, NEED HELP You are standing in a shop, holding an expensive vase of mass m. The vase accidentally slips from your hand and falls to the floor. It breaks into three pieces, two of mass 1/4m and one of mass 1/2m. The...