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A block of mass "m" is released from the top of an incline of mass "M" as shown in the figure, where all the surfaces in contact are frictionless. The displacement of the incline, when the block has reached bottom is :
$$\begin{array}{l}\mathrm{(a)}\phantom{\rule{4pt}{0ex}}\frac{mL}{m+M}\phantom{\rule{4pt}{0ex}}\mathrm{(b)}\phantom{\rule{4pt}{0ex}}\frac{mML}{m+M}\phantom{\rule{4pt}{0ex}}\mathrm{(c)}\phantom{\rule{4pt}{0ex}}\frac{mL}{M-m}\phantom{\rule{4pt}{0ex}}\mathrm{(d)}\phantom{\rule{4pt}{0ex}}\frac{mML}{M-m}\end{array}$$
This question makes use of the concept of COM to a great effect. Answer to this question involving force analysis would have been difficult. Here, we note that the block and incline is at rest in the beginning. This means that COM of the system is at rest before the block is released. When the block is released, the system is subjected to external force due to gravity. This means that COM should move along the direction of applied external force. As the direction of external force is vertically downward, the COM will also be accelerated downward. Importantly, COM will have no displacement in horizontal direction. We shall make use of this fact in solving this problem.
We can estimate here that the block slides down towards left. In order that the COM has no horizontal displacement, we can conclude that incline moves towards right. Let us consider that the incline moves to right by a distance "x".
The horizintal component of displacement of block on reaching bottom is given as "L" with respect to incline. This component of displacement with respect to ground,"d", is (note that in the meantime incline has also moved in opposite direction and the block is carried along with the incline) :
$$\begin{array}{l}\Rightarrow d=L-x\end{array}$$
As COM of the "block and incline" system has no displacement in horizontal direction,
$$\begin{array}{l}{x}_{\mathrm{COM}}=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}}=0\end{array}$$
$$\begin{array}{l}{m}_{1}{x}_{1}+{m}_{2}{x}_{2}=0\\ \Rightarrow m(L-x)-Mx=0\end{array}$$
$$\begin{array}{l}\Rightarrow x=\frac{mL}{m+M}\end{array}$$
Hence, option (a) is correct.
Two particles are projected with speeds ${v}_{1}$ and ${v}_{2}$ making angles ${\theta}_{1}$ and ${\theta}_{2}$ with horizontal at the same instant as shown in the figure. Neglecting air resistance, the trajectory of the COM of the two particles :
(a) can be a vertical straight line
(b) can be a horizontal straight line
(c) is a straight line
(d) is a parabola
To answer this question, we need to analyze the speeds of COM of two particles in horizontal (x) and vertical (y) directions. Here,
$$\begin{array}{l}{v}_{\mathrm{COMx}}=\frac{{m}_{1}{v}_{\mathrm{1x}}+{m}_{2}{v}_{\mathrm{2x}}}{{m}_{1}+{m}_{2}}=\Rightarrow {v}_{\mathrm{COMx}}=\frac{{v}_{\mathrm{1x}}+{v}_{\mathrm{2x}}}{2}\end{array}$$
$$\begin{array}{l}\Rightarrow {v}_{\mathrm{COMx}}=\frac{{v}_{1}\mathrm{cos}{\theta}_{1}-{v}_{2}\mathrm{cos}{\theta}_{2}}{2}=\text{a constant}\end{array}$$
and
$$\begin{array}{l}{v}_{\mathrm{COMy}}=\frac{{m}_{1}{v}_{\mathrm{1y}}+{m}_{2}{v}_{\mathrm{2y}}}{{m}_{1}+{m}_{2}}=\Rightarrow {v}_{\mathrm{COMy}}=\frac{{v}_{\mathrm{1y}}+{v}_{\mathrm{2y}}}{2}\end{array}$$
$$\begin{array}{l}\Rightarrow {v}_{\mathrm{COMy}}=\frac{{v}_{1}\mathrm{cos}{\theta}_{1}-gt+{v}_{2}\mathrm{cos}{\theta}_{2}-gt}{2}\end{array}$$
$$\begin{array}{l}\Rightarrow {v}_{\mathrm{COMy}}=\frac{{v}_{1}\mathrm{cos}{\theta}_{1}+{v}_{2}\mathrm{cos}{\theta}_{2}-2gt}{2}\end{array}$$
On inspection of the speed of COM in x-direction, we see that it can be zero if :
$$\begin{array}{l}{v}_{1}\mathrm{cos}{\theta}_{1}={v}_{2}\mathrm{cos}{\theta}_{2}\end{array}$$
For this condition, COM has only speed in vertical direction and varies with time. For all other values of attributes of motion, which do not satisfy the equality, the COM has both components of velocity and, therefore, moves in parabolic trajectory.
Hence, option (a) is correct.
The wedge "C" is placed on a smooth horizontal plane. Two blocks "A" and "B" connected with "mass-less" string is released over a double sided wedge "C". The masses of blocks "A","B" and "C" are m, 2m and 3m respectively. After the release of the blocks, the block "B" moves on the incline by 10 cm, before it is stopped. Neglecting friction between all surfaces, the displacement of wedge "C", in the meantime, is (in cm):
$$\begin{array}{l}\mathrm{(a)}\phantom{\rule{4pt}{0ex}}5\phantom{\rule{4pt}{0ex}}\mathrm{(b)}\phantom{\rule{4pt}{0ex}}\frac{5}{\surd 2}\phantom{\rule{4pt}{0ex}}\mathrm{(c)}\phantom{\rule{4pt}{0ex}}\frac{5}{2\surd 2}\phantom{\rule{4pt}{0ex}}\mathrm{(d)}\phantom{\rule{4pt}{0ex}}10\end{array}$$
This question makes use of the concept of COM to a great effect. Answer to this question involving force analysis would have been difficult. Here, we note that The system of blocks and wedge is at rest in the beginning. This means that COM of the system is at rest before the block is released. When the blocks are released, the system is subjected to external force due to gravity. This means that COM should move along the direction of applied external force. As the direction of external force is vertically downward, the COM will also be accelerated downward. Importantly, COM will have no displacement in horizontal direction. We shall make use of this fact in solving this problem.
We can estimate here that the heavier block "B" slides down, whereas lighter block "A" slides up the incline. It means that block "B" moves towards right. In order that the COM has no horizontal displacement, we can conclude that wedge "C" moves towards left. Let us consider that the wedge "C" moves to left by a distance "x".
Now, there is yet another twist in the question. The displacement of block "B" is given with respect to wedge "C". The horizontal component of this displacement with respect to wedge ,d, is :
$$\begin{array}{l}d=\mathrm{cos}{45}^{0}=5\surd 2\end{array}$$
The horizontal component of displacement of the block "B" with respect to ground, d', is :
$$\begin{array}{l}\mathrm{d\text{'}}=d-x=5\surd 2-x\end{array}$$
As COM has no displacement in horizontal direction,
$$\begin{array}{l}{x}_{\mathrm{COM}}=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}}=0\end{array}$$
$$\begin{array}{l}{m}_{1}{x}_{1}+{m}_{2}{x}_{2}=0\\ \Rightarrow m(5\surd 2-x)-3mx=0\end{array}$$
$$\begin{array}{l}\Rightarrow x=\frac{5}{2\surd 2}\end{array}$$
Hence, option (c) is correct.
Two blocks of equal mass are released over a fixed double incline as shown in the figure. Neglecting friction between incline and blocks, the acceleration of the COM of the blocks and double incline system is :
$$\begin{array}{l}\mathrm{(a)}\phantom{\rule{4pt}{0ex}}\frac{\surd 2-1}{2\surd 3}g\phantom{\rule{4pt}{0ex}}\mathrm{(b)}\phantom{\rule{4pt}{0ex}}\frac{\surd 2+1}{4\surd 3}g\phantom{\rule{4pt}{0ex}}\mathrm{(c)}\phantom{\rule{4pt}{0ex}}\frac{\surd 3-1}{4\surd 3}g\phantom{\rule{4pt}{0ex}}\mathrm{(d)}\phantom{\rule{4pt}{0ex}}\frac{\surd 3+1}{2\surd 3}g\end{array}$$
Each of the block behaves as particle. Thus, the acceleration of the COM of two blocks connected with "mass-less" string is given by :
$$\begin{array}{l}{\mathbf{a}}_{\mathrm{COM}}=\frac{{m}_{1}{\mathbf{a}}_{1}+{m}_{2}{\mathbf{a}}_{2}}{{m}_{1}+{m}_{2}}\end{array}$$
The two blocks have same mass. Thus,
$$\begin{array}{l}{\mathbf{a}}_{\mathrm{COM}}=\frac{({\mathbf{a}}_{1}+{\mathbf{a}}_{2})}{2}\end{array}$$
The two blocks are connected by a taut string having acceleration of same magnitude, say, "a". Now, as individual accelerations are at right angles, the magnitude of resultant acceleration is :
$$\begin{array}{l}|{\mathbf{a}}_{1}+{\mathbf{a}}_{2}|=\surd ({a}^{2}+{a}^{2}=a\surd 2\end{array}$$
Thus, magnitude of acceleration of the COM is :
$$\begin{array}{l}{a}_{\mathrm{COM}}=\frac{a\surd 2}{2}=\frac{a}{\surd 2}\end{array}$$
We need to find the common acceleration of the individual blocks by force analysis :
For first block :
$$\begin{array}{l}\Rightarrow \sum {F}_{x}=mg\mathrm{sin}{60}^{0}-T=ma\end{array}$$
For second block :
$$\begin{array}{l}\Rightarrow \sum {F}_{x}=T-mg\mathrm{sin}{30}^{0}=ma\end{array}$$
Solving two equations, we have :
$$\begin{array}{l}{a}_{\mathrm{COM}}=\frac{g(\mathrm{sin}{60}^{0}-\mathrm{sin}{30}^{0})}{2}\end{array}$$
$$\begin{array}{l}a=\frac{\surd 3-1}{4}\end{array}$$
Substituting this value in the equation of acceleration of COM, we have :
$$\begin{array}{l}a=\frac{\surd 3-1}{4\surd 3}g\end{array}$$
Hence, option (c) is correct.
1. (d) 2. (b) 3. (d) 4. (a) 5. (a)
6. (a) 7. (c) 8. (c)
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