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. Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.
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The block ‘m’ will slide down the inclined plane of mass M with acceleration $\text{a}_1\text{g}\sin\alpha$ (relative) to the inclined plane. The horizontal component of $a_1$ will be, $\text{a}_{\text{x}}=\text{g}\sin\alpha\cos\alpha,$ 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).
$ ma_x = (M + m)a_2$_ $\Rightarrow\text{a}_2=\frac{\text{ma}_{\text{x}}}{\text{M}+\text{m}}=\frac{\text{mg}\sin\alpha\cos\alpha}{\text{M}+\text{m}}\ \dots(1)$ So, the absolute (Resultant) acceleration of ‘m’ on the block ‘M’ along the direction of the incline will be, $\text{a}=\text{g}\sin\alpha-\text{a}_2\cos\alpha$
$=\text{g}\sin\alpha-\frac{\text{mg}\sin\alpha\cos^2\alpha}{\text{M}+\text{m}}$
$=\text{g}\sin\alpha\Big[1-\frac{\text{m}\cos^2\alpha}{\text{M}+\text{m}}\Big]$
$=\text{g}\sin\alpha\Big[\frac{\text{M}-\text{m}-\text{m}\cos^2\alpha}{\text{M}+\text{m}}\Big]$ So, $\text{a}=\text{g}\sin\alpha\Big[\frac{\text{M}+\text{m}\sin^2\alpha}{\text{M}+\text{m}}\Big]\ \dots(2)$ Let, the time taken by the block ‘m’ to reach the bottom end be ‘t’.
Now, $\text{S}=\text{ut}+\Big(\frac{1}{2}\Big)\text{at}^2$
$\Rightarrow\frac{\text{h}}{\sin\alpha}=\Big(\frac{1}2{}\Big)\text{at}^2$
$\Rightarrow\text{t}=\sqrt{\frac{2}{\text{a}\sin\alpha}}$ So, the velocity of the bigger block after time ‘t’ will be. $\text{V}_{\text{m}}=\text{u}+\text{a}_2\text{t}$
$=\frac{\text{mg}\sin\alpha\cos\alpha}{\text{M}+\text{m}}\sqrt{\frac{2}{\text{a}\sin\alpha}}$
$=\sqrt{\frac{\text{2m}^2\text{g}^2\text{h}\sin^2\alpha\cos^2\alpha}{(\text{M}+\text{m})^2\text{a}\sin\alpha}}$ Now, subtracting the value of a from equation (2) we get, $\text{V}_\text{m}=\bigg[\frac{\text{2m}^2\text{g}^2\text{h}\sin^2\alpha\cos^2\alpha}{(\text{M}+\text{m})^2\text{a}\sin\alpha}\times\frac{(\text{M}+\text{m})}{\text{g}\sin\alpha(\text{M}+\text{m}\sin^2 \alpha)}\bigg]^{\frac{1}{2}}$ or $\text{V}_{\text{m}}=\bigg[\frac{2\text{m}^2\text{g}^2\text{h}\cos^2\alpha}{(\text{M}+\text{m})(\text{M}+\text{m}\sin^2\alpha)}\bigg]^{\frac{1}{2}}$
$ ma_x = (M + m)a_2$_ $\Rightarrow\text{a}_2=\frac{\text{ma}_{\text{x}}}{\text{M}+\text{m}}=\frac{\text{mg}\sin\alpha\cos\alpha}{\text{M}+\text{m}}\ \dots(1)$ So, the absolute (Resultant) acceleration of ‘m’ on the block ‘M’ along the direction of the incline will be, $\text{a}=\text{g}\sin\alpha-\text{a}_2\cos\alpha$
$=\text{g}\sin\alpha-\frac{\text{mg}\sin\alpha\cos^2\alpha}{\text{M}+\text{m}}$
$=\text{g}\sin\alpha\Big[1-\frac{\text{m}\cos^2\alpha}{\text{M}+\text{m}}\Big]$
$=\text{g}\sin\alpha\Big[\frac{\text{M}-\text{m}-\text{m}\cos^2\alpha}{\text{M}+\text{m}}\Big]$ So, $\text{a}=\text{g}\sin\alpha\Big[\frac{\text{M}+\text{m}\sin^2\alpha}{\text{M}+\text{m}}\Big]\ \dots(2)$ Let, the time taken by the block ‘m’ to reach the bottom end be ‘t’.
Now, $\text{S}=\text{ut}+\Big(\frac{1}{2}\Big)\text{at}^2$
$\Rightarrow\frac{\text{h}}{\sin\alpha}=\Big(\frac{1}2{}\Big)\text{at}^2$
$\Rightarrow\text{t}=\sqrt{\frac{2}{\text{a}\sin\alpha}}$ So, the velocity of the bigger block after time ‘t’ will be. $\text{V}_{\text{m}}=\text{u}+\text{a}_2\text{t}$
$=\frac{\text{mg}\sin\alpha\cos\alpha}{\text{M}+\text{m}}\sqrt{\frac{2}{\text{a}\sin\alpha}}$
$=\sqrt{\frac{\text{2m}^2\text{g}^2\text{h}\sin^2\alpha\cos^2\alpha}{(\text{M}+\text{m})^2\text{a}\sin\alpha}}$ Now, subtracting the value of a from equation (2) we get, $\text{V}_\text{m}=\bigg[\frac{\text{2m}^2\text{g}^2\text{h}\sin^2\alpha\cos^2\alpha}{(\text{M}+\text{m})^2\text{a}\sin\alpha}\times\frac{(\text{M}+\text{m})}{\text{g}\sin\alpha(\text{M}+\text{m}\sin^2 \alpha)}\bigg]^{\frac{1}{2}}$ or $\text{V}_{\text{m}}=\bigg[\frac{2\text{m}^2\text{g}^2\text{h}\cos^2\alpha}{(\text{M}+\text{m})(\text{M}+\text{m}\sin^2\alpha)}\bigg]^{\frac{1}{2}}$
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