Given, m1=4 g,u1=300 m/s m2=0.8 kg=800 g,u2=0 m/s From law of conservation of momentum, m1u1+m2u2=m1v1+m2v2 Let the velocity of combined system =vm/s then, 4×300+800×0=(800+4)×v v=8041200=1.49 m/s Now, μ=0.3 (given) a=μga=0.3×10( take g=10 m/s2)=3 m/s2 then, from v2=u2+2as (1.49)2=0+2×3×s s=6(1.492);s=62.22=0.379 m