Comprehension # 1
If net force on a system in a particular direction is zero (say in horizontal direction) we can apply:
`Sigmam_(R )x_(R )= Sigmam_(L)x_(L), Sigmam_(R )v_(R )= Sigmam_(L)v_(L)` and `Sigmam_(R )a_(R ) = Sigmam_(L)a_(L)`
Here R stands for the masses which are moving towards right and L for the masses towards left, x is displacement, v is veloctiy and a the acceleration (all with respect to ground). A small block of mass `m = 1 kg` is placed over a wedge of mass `M = 4 kg` as shown in figure. Mass m is released from rest. All surface are smooth. Origin O is as shown.

Normal reaction between the two blocks at an instant when absolute acceleration of `m` is `5 sqrt3 m//s^(2)` at `60^(@)` with horizontal is ......... N. Normal reaction at this instant is making `30^(@)` with horizontal :-

Options

A. 6
B. 10
C. 4
D. 5

Answer

D

Explanation

`a_(cm) = (m_(1)a_(1) + m_(2)a_(2))/(m_(1) + m_(2))`
(1) `a cos 60 = 4a`
`a_(M) = (5dsqrt3)/(8) m//s^(2)`
`N cos 30 = 4 ms_(M)`
`N(sqrt3)/(2) = 4(1) (5sqrt3)/(8)`
`rArr N = 5` Newton