Let v1 and v2 are the speed of P and Q after collision. By using law of conservation of mementum, m1u1+m2u2=m1v1+m2v2Mu+m⋅0=Mv1+mv2⇒mM(u−v1)=v2 and by using law of conservation of energy, 21m1u12+21m2u22=21m1v12+21m2v22⇒Mu2+0=Mv12+mv22⇒M(u2−v12)=mv22⇒mM(u−v1)(u+v1)=v2221m1u12+21m2u22=21m1v12Mu2+0=Mv12+mv2⇒M(u2−v12)=mv22⇒mM(u−v1)(u+v1)=v22Substituting the value of Mmu−v1 from Eq. (i) in Eq. (ii), we getv2(u+v1)=v22u+v1=v2⇒M≫mv1=uv2=2u Eq. (ii), we get v2(u+v1)=v22u+v1=v2M>>mv1=uv2=2u Hence, option (c) is the correct.