Is there any easier way to attack this problem?(perfectly elastic collision)

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SUMMARY

The discussion focuses on solving a perfectly elastic collision problem involving two spheres with masses of 0.1 kg and 0.05 kg, respectively. The initial height of the left sphere is calculated using the formula h = l(1 - cosθ), resulting in a height of 0.14 m. The velocities before and after the collision are derived using conservation of mechanical energy and momentum equations, yielding final velocities of 0.552 m/s for the left sphere and 2.21 m/s for the right sphere. The maximum angles after the collision are determined to be approximately 8.28 degrees for the left sphere and 42.3 degrees for the right sphere.

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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
 

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How about instead of worrying about how someone else solved the problem, you start from first principles yourself to solve the problem? In sequence:

1. What is the energy of each ball just before they're released?
2. What is the energy of each ball just before they collide?
3. What is the momentum of each ball just before they collide?
4. What is the total momentum of the system before they collide? (express as a formula)
5. What is the total momentum of the system after they collide? (express as a formula)
6. What is the total energy of the system after they collide?
7. What is the momentum and energy of each ball just after they collide?
8. What is the energy of each ball when they are at the top of the swing after the collison?

By finding expressions for everything in order and applying the conservation laws, you will determine the angles for which you are asked.
 
Okay ill try that thanks!
 

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