Let final velocity of m and 2m be v₁ and v₂ respectively.
Now according to law of conservation of momentum ,
m×X1+2m×V2=mV1+2mV2
m×u−2mu=mV1+2mV2
V1=2V2=−u................(1)
now,
V1−V2=−(V1−V2)
V1−V2=−(u−u)
V1−V2=−2u........................(2)
Subtracting Equation 1 and Equation 2
3V2=U
V2=U3=√2gl3
Substituting value in Equation 2
V1−V2=−2U
V1=−2U+V2
V1=−2U+U3
V1=−5√2gl3
V1=−√50gl3
(b) Again using Work Energy Principle
12×2mg2−12×V22=−2mghh= height of heavy ball.
12×(2g9)=l×h
h=l9
Similarly ,
12×mg2−12mV21=mgh1 h_1 = height of light or small ball
h=50l18
Strings are fixed at O , hence maximum height of small ball of mass m ball will be 2l.