Figure shows a small body of mass m placed over a larger mass M whose surface is horizontal near the smaller mass and gradually curves to become verticle. The smaller mass is pushed on the longer on at a speed v and the system is left to itself. Assume thast all the surfaces are frictionless. a. find teh speed of the larger block when the smaller block is sliding on the vertical part. b. find the speed of the smaller mass when it breaks off the larger mas at height h. c. find the maximum height (from the ground) that the smaller mass seconds. d show that the smaller mass will again land on the bigger one. Find the distane traversed by the bigger block during the time when the smaller block was in its flight under gravity.

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))`