Figure (9-E22) 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 vertical. The smaller mass is pushed on the longer one at a speed v and the system is left to itself. Assume that all the surfaces are frictionless. (a) Find the 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 mass at height h. (c) Find the maximum height (from the ground) that the smaller mass ascends (d) Show that the smaller mass will again land on the bigger one. Find the distance traversed by the bigger block during the time when the smaller block was in its fight under gravity.

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 velocity of the bigger block be $v_1$ towards left.

From law of conservation of momentum (in the horizontal direction)

$mv=(m+m)v_1$

$\Rightarrow v_1=\frac{mv}{M+m}$

(b) When the smaller block breaks off, let its resultant velocity is $v_2.$

From law a of conservation of energy.

$\left(\frac{1}{2}\right)m{v}^2=\left(\frac{1}{2}\right)M{v}^2+\left(\frac{1}{2}\right)m{v}_2^2+mgh$

$\Rightarrow v_2^2=v^2-\frac{M}{m}{v}_1^2-2gh...\left(i\right)$

$\Rightarrow v_2^2=v^2[1-\frac{M}{m}.\frac{m^2}{{(M+M)}^2}]-2gh$

$\Rightarrow v_2=[\frac{\left[(m^2+Mm+m^2)\right]}{{(M+m)}^2}-2gh]$

(c) Now, the vertical component of the velocity $v_2$ of mass 'm' is given by,

$v_y^2=v_2^2-v_1^2$

$=\frac{(M^2+Mm+m^2)}{(M+m)}{v}^2-2gh-\frac{m^2{v}^2}{{(M+m)}^2}$

$[\therefore v_1=\frac{mv}{M+m}]$

$\Rightarrow v_y^2=\frac{M^2+Mm+m^2-m^2}{(M+m)}{v}^2-gh$

$\Rightarrow v_y^2=\frac{M{v}^2}{(M+m)}-2gh...\left(ii\right)$

To find the maximum height (from the ground),

Let us assume the body rise to a height 'h' over and above 'h',

Now, $\left(\frac{1}{2}\right)m{v}_y^2=mg{h}_1$

$\Rightarrow h_1=\frac{v_y^2}{2g}...\left(ii\right)$

So, Total height $=h+h_1$

$=h+\frac{v_y^2}{2g}$

$=h+\frac{M{v}^2}{(M+m)2g}-h...\left(iii\right)$

[from equation (ii) and (iii)]

$\Rightarrow H=\frac{M{v}^2}{2g(M+m)}$

(d) Because the smaller mass has also got a horrizontal component 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),

$0=v_y-gt$

$\Rightarrow t_1=\frac{v}{g}$

$=\frac{1}{g}[\frac{M{v}^2}{(M+m)}-2gh]...\left(iv\right)$

[From equation (ii],

So, the time for which the smaller block was in flight is given by,

$T=2t_1$

$=\frac{2}{g}\left[\frac{M{v}^2-2(M+m)gh}{(M+m)}\right]$

So, the distance travelled by the bigger block during this time is,

$S=v_{1}T$

$=\frac{mv}{M+m}\frac{2}{g}\times \frac{{[M{v}^2-2(M+m)gh]}^{1/2}}{{(M+m)}^{1/2}}$

or $S=\frac{2gv{[M{v}^2-2(M+m)gh]}^{1/2}}{g{(M+m)}^{3/2}}$