Question
A 50 g ball of clay traveling east at 3.5 m/s collides and sticks together with a 50 g ball of clay traveling north at 2.0 m/s. What is the speed of the resulting ball of clay?
Answers
((m1)(v1))i+((m2)(v2))i =(m1+m2)Vf
[(0.05kg)(0)+(0.05kg)(2.0m/s)]
/[(0.05kg+0.05kg)]=1.0 m/s .........(east)
[(0.05kg)(3.5m/s)+(0.05kg)(0m/s)]
/[0.05kg+0.05kg)]= 17 m/s............(north)
(1)^2+(17)^2 = Vf^2
Vf=Sqrt[(1^2)+(17^2)]
Vf= 17.6 m/s
[(0.05kg)(0)+(0.05kg)(2.0m/s)]
/[(0.05kg+0.05kg)]=1.0 m/s .........(east)
[(0.05kg)(3.5m/s)+(0.05kg)(0m/s)]
/[0.05kg+0.05kg)]= 17 m/s............(north)
(1)^2+(17)^2 = Vf^2
Vf=Sqrt[(1^2)+(17^2)]
Vf= 17.6 m/s
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