JEE Advanced Model Paper 6 with solutions for online practice

R =

2υxυy

g

, T =

2υy

g

After first collision
υx′ = eυx
υy′ = υy

Distance covered before 2nd collision
d1 = υx′

T

2

= eυx

T

2

=

eυxυy

g

After second collision
υx′′ = υx′
υy′′ = eυy′
d2 =

2υx′′υy′′

g

=

2eυxeυy

g

d1+d2 =

eυxυy

g

(1 + 2e)
But C2 = C2E = 4µF × 3V = 12µC
e (1 + 2e) = 1 ; 2e2 + e - 1 = 0
e =

−1±√1+8

4

=

−1±3

4

Rejecting -ve value e = 1/2