Two blocks of mass ` m_(1) and m_(2)` lie on smooth horizontal table in contact with each other as shown in figure

If a force F is applied to the mass `m_(1)` then the contact force between the block will be

Answer

` f =(m_(2) xxF)/(m_(1) + m_(2))` .

Explanation

Let a be the common acceleration of the combination of blocks
`:. A = (F)/(m_(1) m_(2))`
If f is force exerted by `m_(1)` on `m_(2)`
then force exerted by `m_(2)` on `m_(1)` is (-) f , as shown in
As block of mass `m_(2)` has accelera- tio a
`:. f = m_(2) a = (m_(2) xx F)/(m _(1) + m_(2)`
.