Question

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

Answer 1

Correct Answer - C
From free body diagrams
`N sin alpha =Ma----(1)`
`N + mA sin alpha = mg cos alpha ----(2)`
`mg sin alpha + mA cos alpha =ma----(3)`
on solving (1) (2) (3)
`A=(mg cos alphasinalpha)/(M+msin^(2)alpha),a=((M+m)gsinalpha)/(M+msin^(2)alpha)`
Now `overset(rarr)(block)=overset(rarr)(block//"wedge")+overset(rarr)(a_(wegde)`
`:.vecablock=(acosalphahati-asinalpha)hatj-Ahati`
`(acosalpha-A)hati-asinalphahatj`
`:.tanbeta=(asinalpha)/(acosalpha-A)=(1+(m)/(A))tanalpha`

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