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. The block will strike the x-axis at `x =` ............ `m` :-

Options

A. `4.2`
B. `7.6`
C. `5.6`
D. `6.8`

Answer

D

Explanation

When 'm' leavers the wedge 'M' then wedge moves distance 'x' in left side
`:. M (4 - x) = Mx rArr = 0.8 m`
`:.` Co-ordinate where block will be leave wedge
`x = 4 - 0.8 = 3.2`
Time for m will strike the ground is `= sqrt((2 xx 2)/(10))`
`:. x_(f) = 3.2 + 4 sqrt2 xx (2)/(sqrt10) = 6.8 m`