Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

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, Let, the acceleration be $a_2$.

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

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

$\Rightarrow a_2=\frac{m{a}_x}{M+m}=\frac{mgsin\alpha cos\alpha }{M+m}$ .......(1)

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

a = g sin $\alpha -a_{2}cos\alpha$

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

= g sin $\alpha$ $\left[\frac{M+m-co{s}^2\alpha }{M+m}\right]$

So, a = g sin $\alpha$ $\left[\frac{M+msi{n}^4\alpha }{M+m}\right]$ ........ (2)

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

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

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

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

$V_m=u+a_{2}t=\frac{mgsin\alpha cos\alpha }{M+m}\sqrt{\frac{2}{asin\alpha }}=\sqrt{\frac{2m^2{g}^{2}hsi{n}^2\alpha co{s}^2\alpha }{(M+m{)}^{2}asin\alpha }}$

Now, subtracting the value of a from equation (2), we get,

$V_m={[\frac{2m^2{g}^{2}hsi{n}^2\alpha co{s}^2\alpha }{(M+m{)}^{2}sin\alpha }\times \frac{(M+m)}{gsin\alpha (M=msi{n}^2\alpha }]}^{\frac{1}{2}}$

or $V_m={\left[\frac{2m^2{g}^{2}hco{s}^2\alpha }{(M+m)(M+msi{n}^2\alpha )}\right]}^{\frac{1}{2}}$