A wooden block of mass M rests on a horizontal surface. A bullet of mass m moving in the horizontal direction strikes and gets embedded in it. The combined system covers a distance x on the surface. If the coefficient of friction between wood and the surface is μ, the speed of the bullet at the time of striking the block is (where m is mass of the bullet)

Let speed of the bullet = v Speed of the system after the collision = V By conservation of momentum mv=(m+M)V⇒V=mvM+mSo the initial K.E. acquired by the system =12(M+m) V2=12(m+M) mvM+m2=12m2v2(m+M)This kinetic energy goes against friction work done by friction = μR×x=μ(m+M)g×xBy the law of conservation of energy 12m2v2(m+M)=μ(m+M)g×x⇒v2=2μgx m+Mm2∴v=2μgx M+mm