Two balls having masses m and 2m are fastened to two light strings of same length l (Figure 9-E18). The other ends of the strings are fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. (a) Find the velocities of the balls just after their collision. (b) How high will the balls rise after the collision?
Answers
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ANSWER::
(a) Let velocity of m reaching at lower end be v₁
Work Energy Principle ,
(1/2) x m x v₁² - (1/2) x m x 0² = mg l
v₁ = √(2gl)
Similarly ,
Velocity of heavy ball will be v₂ = √(2gh)
Therefore , v₁ = v₂ = u (Let)
Let final velocity of m and 2m be v₁ and v₂ respectively.
Now according to law of conservation of momentum ,
m x x₁ + 2m x v₂ = mv₁ + 2mv₂
m x u - 2 m u = mv₁ + 2mv₂
v₁ + 2v₂ = - u [Equation 1]
Now ,
v₁ - v₂ = - (v₁ - v₂)
v₁ - v₂ = - [u - u]
v₁ - v₂ = - 2u [Equation 2]
Subtracting Equation 1 and Equation 2
3v₂ = u
v₂ = u/3 = √(2gl)/3
Substituting value in Equation 2
v₁ - v₂ = - 2u
v₁ = -2u + v₂
v₁ = -2u + u/3
v₁ = -5u/3
v₁ = -5√(2gl) / 3
v₁ = - √(50gl) / 3
(b) Again using Work Energy Principle
(1/2) x 2m x 0² - (1/2) x 2m x (v₂)² = -2m x g x h [ h => height of heavy ball]
(1/2) 2g / 9 = l x h
h = l/9
Similarly ,
(1/2) x m x 0² - (1/2) x m x v₁² = m x g x h₁ [ h₁ => height of light or small ball]
h = 50l / 18
Strings are fixed at O , hence maximum height of small ball of mass m ball will be 2l.
Hope it helps!
