PHYSICS HELP PLEASE!! grade 11, thank you?

Question

PHYSICS HELP PLEASE!! grade 11, thank you?
on a level ice surface( coefficient of friction = 0.1) two boys stan facing each other 0.15m apart. one has a mass of 45kg and the other 65kg. the smaller boy pushes the bigger boy with a force of 55N for o.4s. At the en of the push how far apart will the boys be? THANK YOU

Answer 1

Force acting on big boy = 55 N.
Friction acting on him = μ mg = 0.98 x 65 = 63.7 > 55N.
Friction cannot exceed the applied force.
It is limited to 55N and hence the big boy is at rest,

Force acting on smaller boy is 55 N {of course in the opposite direction}
Friction acting on him is = μ mg = 0.98 x 45 = 44.1N
Net force acting on him is 55 - 44.1 = 10.9 N
Acceleration = 10.9 / 45 = 4.128 m/s^2.
Distance moved = 0.5 x 4.128 x (0.4)^2 = 0.33024 m.
Since the bigger one is at rest,
distance of separation is 0.33 m

Answer 2

s = (1/2)(55 - 0.1*45*9.80665)(0.4^2)/45 + (1/2)(55 - 0.1*65*9.80665)(0.4^2)/65
s = (1/2)(55/45 - 0.1*9.80665)(0.4^2) + (1/2)(55/65 - 0.1*9.80665)(0.4^2)
s = (1/2)(0.4^2)(55/45 + 55/65 - 0.2*9.80665)
s = (1/2)(0.4^2)(55*110/(45*65) - 0.2*9.80665)
s ≈ 0.9441463 m ≈ 0.94 m