Question

A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown in figure (9-E21). Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.

Answer 1

The block 'm' will slide down the inclined plane of mass M with acceleration $a_1=gsin\alpha$ (relative) to the inclined plane.

The horizontal component of $a_1$ will be $a_x=gsin\alpha cos\alpha ,$ for which the block M will accelerate towards left, left the acceleration be $a_2.$

According to the concept of centre of mass, (in the horizontal direction external force is zero).

$m{a}_x=(M+m)a_2$

$\Rightarrow a_2=\frac{m{a}_x}{M+m}$

$=\frac{mgsin\alpha cos\alpha }{M+m}...\left(i\right)$

So, the absolute (Resultant) acceleration of 'm' on the block 'M' along the direction of the incline will be,

$a=gsin\alpha -a_{2}cos\alpha$

$=gsin\alpha -\frac{mgsin\alpha co{s}^2\alpha }{M+m}$

$=gsin\alpha [1-\frac{mco{s}^2\alpha }{M+m}]$

$=gsin\alpha \left[\frac{M+m-mco{s}^2\alpha }{M+m}\right]$

So, $a=gsin\alpha \left[\frac{M+msi{n}^2\alpha }{M+m}\right]...\left(ii\right)$

Let, the time taken by the block 'm' to reach the bottom end be 't'

Now, $S=ut+\left(\frac{1}{2}\right)a{t}^2$

$\Rightarrow \frac{h}{sin\alpha }=\left(\frac{1}{2}\right)a{t}^2$

$\Rightarrow t=\sqrt{\frac{2h}{asin\alpha }}$

So, the velocity of the bigger block after time 't' will be,

$v_m=u=a_{2}t$

$=\frac{mgsin\alpha cos\alpha }{M+m}\sqrt{\frac{2h}{asin\alpha }}$

$={\left[\frac{2m^2{g}^{2}hsi{n}^2\alpha c{o}^2\alpha }{(M+m{)}^{2}asin\alpha }\right]}^{1/2}$

Now substituting the value of 'a' from equation (ii), we get,

$V_M=[\frac{2m^2{g}^{2}hsi{n}^2\alpha }{(M+m{)}^{2}sin\alpha }\times \frac{co{s}^2\alpha }{gsina}\frac{(M+m)}{(M+msi{n}^2\alpha )}]$

or $V_M={\left[\frac{2m^{2}ghco{s}^2\alpha }{(M+m)(M+msi{n}^2\alpha )}\right]}^{1/2}$