In the figure heavy mass m moves down the smooth surface of a wedge making an angle `alpha` with the horizontal. The wedge at rest `t =0` is on a smooth surface. The mass of the wedge is `M` the direaction of motion of the mass m makes an angle `beta` with the horizontal then, `"Tan" beta` is
.

Options

A. `m/M tan alpha`
B. `M/m tan alpha`
C. `(1+m/M)tan alpha`
D. `(1+M/m)tan alpha`

Answer

C

Explanation

From free body diagrams
`Nsin alpha=Ma`…(i)
`N+mA sin alpha=mgcos alpha` ….(ii)
`mg sin alpha+mA cos alpha=ma` ….(iii)
On solving equations (i), (ii) and (iii)
`A=(mg cos alpha sin alpha)/(M+m sin^(2)alpha), a=((M+m)g sin alpha)/(M+msin^(2)alpha)`

Now, `vec(a)_("block")=vec(a)_("block/wedge")+vec(a)_("wedge")`
`:. vec(a)_("block")=a cosalphahat(i)-a sin alphahat(j)-Ahat(i)`
`(a cos alpha-A)hat(i)-a sin alphahat(j)`
`:. tanbeta=(a sin alpha)/(a cos alpha-A)=(1+m/M)tanalpha`