Which is correct?
`a) cos 3x = 3cosx – 4cos^3 x`
`b) cos 3x = 4cosx – 3cos^3 x`
`c) cos 3x = 3cos^3 x – 4cosx`
`d) cos 3x = 4cos^3 x – 3cosx`

Explanation

Let the velocilty of m reaching at lower end
`=v_1`
From work energy principle
`:.(1/2)xxmv_1^2-(1/2)xxm(0)^2=mgl`
`rarr v_1=sqrt(2gl)`
similarly velocity of heavy block will be ltbr. `v_2=sqt(2gl)`
`:. V_1=v_2=u(say)`
Let fiN/A,l velocity of m and 2m are `v_1 and v_2` respectively.
According to law of conservation of momenum
`rarr mxxu-2mu=mv_1+2mv_2`
`rarr v_1+2v_2=-u`
`Again v_1-v-2=[(u-v)]=-2u`
Substracting
`3v_2=u`
`rarr v_2=u/3=sqrt(2gl)/3`
substituting in ii ltbr. `v_1-v_2=-2u`
`rarr v_1=-2u+v_2`
`=-2u+(u/3)`
`=(-5)/3u=(-5)/3xxsqrt(2gl)`
`(sqrt(50gl))/3`
b. putting the work energy principle
(1/2)xx2mxx(0)^2-(1/2)xx2mxx(v_2)^2`
`=-2mxxgxxh`
`[hrarr` height reached by heavy ball]`
`rarr h=(L/9)`
similarly
`(1/2)xxmx(0)^2-(1/2)xxmv_1^2=mxxgxxh_2`
[height reached by small ball] ltbr. `(1/2)xx(50gL)/9 gxxh_2`
`h_2=(0.25L)/9`
some `h_2` is more than 2L the velocity at highest point will not be zero.