- #1
Nulligan
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Here's my problem guys, I've been trying to puzzle this out for a while. Would really appreciate any help at all, thanks!
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 traveling East be 1
Let ball traveling 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
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 traveling East be 1
Let ball traveling 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