A lighter body collides with much more massive body at rest. Prove that the direction of lighter body is reversed and massive body remains at rest.

A lighter body collides with much more massive body at rest. Prove tha

1.	A lighter body collides with much more massive body at rest. Prove that the direction of lighter body is reversed and massive body remains at rest.
Answer» Solution :Since the first BODY is very much lighter than the second body `(m_(1)ltltm_(2),(m_(1))/(m_(2))ltlt1)`, the ratio `(m_(1))/(m_(2))~~0` and also if the TARGET is at REST `(u_(2)=0)` Dividing numerator and denominator of EQUATION `v_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)+((2m_(2))/(m_(1)+m_(2)))u_(2)` by `m_(2)`, we get `v_(1)=(((m_(1))/(m_(2))-1)/((m_(1))/(m_(2))+1))u_(1)+((2)/((m_(1))/(m_(2))+1))(0)` `v_(1)=((0-1)/(0+1))u_(1)""[(m_(1))/(m_(2))=0]` `v_(1)=-u_(1)'` Similarly, dividing numerator and denominator of equation `v_(2)=((2m_(1))/(m_(1)+m_(2)))u_(1)+((m_(2)-m_(1))/(m_(1)+m_(2)))u_(2)`, by `m^(2)`, we get `v_(2)=((2(m_(1))/(m_(2)))/((m_(1))/(m_(2))+1))u_(1)+((1-(m_(1))/(m_(2)))/((m_(1))/(m_(2))+1))(0)` `v_(2)=(0)u_(1)+((1-(m_(1))/(m_(2)))/((m_(1))/(m_(2))+1))(0)` `v_(2)=0`