A 4g bullet is fired horizontally with a speed of 300 m/s into 0.8kg block of wood at rest on a table and gets embedded in the block. If the coefficient of friction between the block and the table is 0.3, how far will the block slide approximately ?

first, apply momentum conservation on the bullet and block

m_1​v_1​ = m₂​v₂​

m₁​ = mass of bullet

v₁​=velocity of bullet 

m₂​=mass of block

v_2​=velocity of block

momentum conservation.

(4/1000​)×300=0.8×v₂

​v₂ ​= 1.5m/s

now there is friction that resists the motion of the block

friction retardation

a = −μg = −0.3×9.8

a = −2.94m/s²

by using newtons law of motion

v²=u²−2aS

(v=0 final velocity of block)

0=(1.5)²−2×(2.94)×S

S=0.379m