Let v1 and v2 are the speed of P and Q after collision. By using law of conservation of mementum, m1u1+m2u2‌‌=m1v1+m2v2 Mu+m⋅0‌‌=Mv1+mv2 ⇒‌‌‌‌‌
M(u−v1)
m
‌‌=v2 and by using law of conservation of energy, ‌‌
m1v12 Mu2+0‌‌=Mv12+mv2 ⇒‌‌M(u2−v12)‌‌=mv22 ⇒‌‌M(u−v1)(u+v1)∕m‌‌=v22 ‌ Substituting the value of ‌M‌
(u−v1)
m
‌ from Eq. (i) in ‌‌ ‌ Eq. (ii), we get ‌ v2(u+v1)‌‌=v22 u+v1‌‌=v2 ⇒‌‌M‌≫m v1‌‌=u v2‌‌=2u Eq. (ii), we get v2(u+v1)‌‌=v22 u+v1‌‌=v2 M>‌>m v1‌‌=u v2‌‌=2u Hence, option (c) is the correct.