A block `B` of mass `0.6kg` slides down the smooth face `PR` of a wedge `A` of mass `1.7kg` which can move freely on a smooth horizontal surface.
The inclination of the face `PR` to the horizontal is `45^(@)` . Then `:`

Options

A. the acceleration of `A` is `3g//20`
B. the vertical component of the acceleration of `B` is `23g//40`
C. the horizontal component of the acceleration of `B` is `17g//40`
D. none of these

Answer

A, B, C

Explanation

`F.B.D.` of block `B w.r.t.` wedge

for block `A` ltbr. `N cos 45^(@)=1.7a ….(1)`
for block `B` ,br. `0.6g sin 45^(@)=0.6a cos 45^(@)=0.6b …..(b)`
`N+0.6a cos 45^(@)=0.6g cos 45^(@) ….(ii)`
by solving `(i), (ii) & (iii)`
`a=(3g)/(20) ` and `b=(23g)/(20sqrt(2))`
Now vertical component acceleration of
`B=b cos 45^(@)=(23g)/(40)`
and horizontal component of acceleration of
`B=b sin 45^(@)-a=(17g)/(40)`