XwakeriderX
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Homework Statement
(see attachment)
The work the did gives the correct answer but i do not understand where a lot of that stuff came from! My first question about the work is how they derived 1/2[(m_1)((v_i)_1)^2] =m_1gh
Homework Equations
Below is the solution
The Attempt at a Solution
length of the string l =1.5 m
angle =25 degrees
height h_1 =h_2 =h =l(1-cosθ) =0.14 m
mass of the left side sphere m_1 =100 gm =0.1 kg
mass of the right side sphere m_2 =50 gm =0.05 kg
when left side ball is at height h (mass m_1) ,apply conservation of mechanical energy ,we get
1/2[(m_1)((v_i)_1)^2] =m_1gh
velocity of the left side ball (v_i)_1,just before the collision (v_i)_1=sq.root (2gh) =1.657 m/s
after the collision:
b)
(v_f)_1 =[m_1-m_2/m_1+m_2](v_i)_1
substitute the given data in above equation we get,
velocity of the left side ball after the collision =(v_f)_1 =0.552 m/s
(v_f)_2 =[(2m_1)/m_1+m_2](v_i)_1
substitute the given data in above equation we get,
velocity of the right side ball after the collision =(v_f)_2 =2.21 m/s
c)
after the collision ,the height of m_1 is
( h_f)_1 =l(1-cosθ) =[[(v_f)_1]^2]/2g
l(1-cosθ) =[[(v_f)_1]^2]/2g
(1-cosθ) =[[(v_f)_1]^2]/2gl
cosθ =1-[[(v_f)_1]^2]/2gl
substitute the given data in above equation we get,
maximum angle for left side ball ,θ =8.26 deg≈ 8.28 deg
after the collision ,the height of m_2 is
( h_f)_2 =[[(v_f)_2]^2]/2g
cosθ =1-[[(v_f)_2]^2]/2gl
substitute the given data in above equation we get,
maximum angle for right side ball ,θ =42.25 deg≈ 42.3 deg