Evaluate `sin54^@/cos36^@`.

Answer

A, B, C, D

Explanation

The mass m is given a velocity v over the larger mass M.
a. when the smaller block is travelling on the vertical part let the velocilty of the bigger block be `v_1` towards left
From law of conservation of momentum (in the horizontal direction)
` mv=(M+m)v_1`
`rarr v_1=(mv)/(M+m)`
b. When the smaller block breaks off let its result velocity is `v_2`
From law a conservation of energy
`(1/2)mv^2=(1/2)Mv_1^2+(1/2)mv_2^2+mgh`
`rarr v_2^2=v^2-M/mv_1^2-2gh`......i
`rarr v_2^2=v62[1-M/m, m^2/((M+m)^2)]-2gh`
`rarr v_2=[[(m^2+Mm+m^2)]/((M+m)^2)]-2gh]^(1/20`
c. now the vertical component of the velocity `v_2` of mass m is given by
v_y^2=v_2^I2-v_1^2` ltbr. `=((M^2+Mm+m^2)/(M+m)v^2-2gh-(m^2v^2)/((M+m)^2)`
`[:. v_1=(mv)/(M+m)]`
`rarr v_y^2=(M^2+Mm+m^2-m^2)/(M+m) v^2-gh`
`rarr v_y^2=(Mv^2)/((M+m))-2gh` .........ii
to find the maximum height from the groudn,
Let us assume the body rise to a height h over add above h
`now, (1/2)mv_y^2=mgh_1`
`rarr h_1=v_y^2/(2g)`.........iii
so total height `=h+h_1`
`=h+v_y^2/(2g)`
`=h+(Mv^2)/((M+m)2g)-h`.....iii
[from equation ii and iii ]
`rarr H=(Mv^2)/(2g(M+m))`
d. because the smaller mass has also got a horizontal componetn of velocity V at the time it breaks off from M (which has a velocity `v_1)` the block m will again land on the block M (bigger one)
Let us find out the time of flight of block m after it breaks off.
During the upward motion (BC)
`0=v_y-gt`
`rarr g_1=v/g`
`=1/g[(Mv^2)/((M+m)-2gh]`..........iv
[from equation ii]
So the time for which the smaller block was in flight is given by
`T=2t_1`
`=2/g[(Mv^2-2(M+m)gh)/(M+m)]`
so the distance trvelled by the bigger block during this time is
`S=v_1T` ltbr.gt `=(mv)/(M+m) 2/g xx[(Mv^2-(M+m)gh)]^(1/2)/((M+m)^(1/2)`
`so S=(2gv[Mv^2-2(M+m)gh)]^(1/2)/(g(M+m)^(3/2))`