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## Homework Statement

A 50.0 g ball of clay traveling east at 5.50 m/s collides and sticks together with a 40.0 g ball of clay traveling north at 6.00 m/s.

What is the speed of the resulting ball of clay?

**2. Homework Equations and 3. The Attempt at a Solution**

My work so far:

Let Ball travelling East be 1

Let ball travelling North be 2

Known

M1 = 0.05 kg

M2 = 0.04

M1 + M2 = 0.09

(Vix)1 = 5.5 m/s; (Viy)1 = 0 m/s

(Vix)2 = 0; (Viy)2 = 6 m/s

(Vfx) = 5.5 + 0 = 5.5 m/s

(Vfy) = 0 + 6 = 6 m/s

So the X components are equal to Vf* (as in, the vector V)Cos(Theta)

And the Y components are equal to Vf*Sin(Theta)

=> (M1 + M2)Vfx = (M1 + M2)VfCos(Theta) = M1(Vix)1 + M2(Vix)2 {which equals zero} = M1(Vix)1

(M1 + M2)Vfy = (M1 + M2)VfSin(Theta) = M1(Viy)1{which equals zero} + M2(Viy)2 = M2(Viy)2