Physics: Linear Momentum Not homework?

Question

Particles A and B, of masses 0.15kg and 0.2kg respectively, are free to move on a horizontal surface. Air resistance may be ignored. At a particular instant A is moving with speed 2ms-1 towards B, which is stationary at a point 4m from A. Particle A collides directly with particle B.

It is given instead that the coefficient of friction between A and the surface is 0.05. A is again brought to rest by the collision.

Find the speed of B immediately after the collision?

Answer 1

first apply work enrgy theoram to get the velocity of a just before the collision
ie.
1/2*0.15(v^2-4)=0.05*0.15*9.8*4

after finding v
apply conservation of momentum and u will get velocity of b after collision.

*u may also remember that though friction is an impulsive force,u may apply conservation of momentum for a short time during collision

Answer 2

according to the law of reserve in momentum:
mA*vA+mB*vB=mA*v'A+mB*v'B:
0.15*2+0=0+0.2*v'B( B was stationary at first then it was A turn)
=> v'B=(0.15*2)/.2=1.5 (m/s).