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# Tutor profile: Karthik M.

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Karthik M.
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

$$\textit{Imagine a cubical body of mass m placed on a wedge of mass M which makes angle} \,\alpha\,\textit{with the base.}\textit{ All surfaces are frictionless and the system is left to evolve.}$$ $$\textit{(m<<M i.e m is not enough to produce any torque on the wedge).}$$ $$\textit{With what acceleration(both magnitude and direction) does the wedge move?}$$

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Karthik M.

We need to draw the free body diagrams for both the bodies.(which I am not able to do here because of restrictions) From FBD of m: We get $$N_m=mg\cos\alpha(\textit{where}\,$$$$N_m$$$$\textit{ is the Normal reaction on the body of mass m})$$ $$\textit{There is an unbalanced force mg}$$$$\sin\alpha$$$$\,\textit{parallel to the inclined surface of the wedge}$$ From FBD of M: We get $$N_M=Mg+N_m\cos\alpha(\textit{where}\,$$$$N_M$$$$\textit{ is the Normal reaction on the wedge of M})$$ $$\textit{There is an unbalanced force}\,$$$$N_m\sin\alpha$$$$\,\textit{acting towards the -x direction on the wedge}$$ $$\textit{This force is responsible for the movement of the wedge and the body}$$ $$N_m\sin\alpha=(M+m)a$$ $$\displaystyle{a=\frac{N_msin\alpha}{M+m}}$$ $$\textit{But}$$ $$N_m=mg\cos\alpha$$ $$\textit{So}$$$$\,a=\displaystyle{\frac{mg\cos\alpha\sin\alpha}{M+m}}$$

### Subject:Calculus

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Question:

$$\int\limits_{-1}^1\left([x]-|x|\right) \,dx$$

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Karthik M.

For this question, we first need to know the definitions of step function and modulus function. $$|x|=\displaystyle{{\displaystyle{\,\,\,\,\,\,x\,\,:\,\, x\ge0\\ -x\,\,:\,\,x<0}}}$$ $$[x]= \displaystyle{\,\,\,\,\,n\,\,:\,\,x\ge n, where \, n \, is \,the\,closest\,integer\,possible}$$ $$For Example \\1)\,\,x=3.4,\,\,then \,\,[x]=3 \\2)\,\,x=-3.4,\,\, then\,\,[x]=-4$$ $$Now\, to \, compute\,this\,integral \, we\,need\,to\,split \,the\, integral\, into\, two\, parts\, using\, the \,following \,rule\,of \,integration\\ \int\limits_{a}^{b}f(x)\,dx=\int\limits_{a}^{c}f(x)\,dx+\int\limits_{c}^{b}f(x)\,dx\,\,\,\,where\,\,\,a,b,c \,\, \in \mathbb{R} \,\, and \,\,a<c<b$$ $$\int\limits_{-1}^1\left([x]-|x|\right) \,dx = \int\limits_{-1}^0\left([x]-|x|\right) \,dx \, + \int\limits_{0}^1\left([x]-|x|\right) \,dx$$ $$=\int\limits_{-1}^0\left(-1-(-x)\right) \,dx\,+\, \int\limits_{0}^1\left(0-(x)\right) \,dx$$ $$=\int\limits_{-1}^0\left(-1+x\right) \,dx\,+\, \int\limits_{0}^1\left(-x\right) \,dx$$ $$=\left[-x+\frac{x^2}{2}\right]_{-1}^{0}\,\,+\, \left[-\frac{x^2}{2}\right]_{0}^{1}$$ $$=\{0-[-(-1)+\frac{1}{2}]\}+\{[-\frac{1}{2}]-0\}$$ $$=-\frac{3}{2}-\frac{1}{2}$$ $$=-2$$ Now,this is what one would expect to be the answer, but this is $$\textbf{NOT THE ANSWER}$$. For a function to be integrable within certain limits, it should be continuous and differentiable through out those limits. $$[x]$$ is not continuous through out the limits of integration. It is discontinuous at 0. $$|x|$$ is not differentiable at $$x=0$$ So, the above solution is wrong and the correct answer is $$\textbf{This function is not integrable within the specified limits}$$

### Subject:Physics

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Question:

Imagine a meteorite burning in the atmosphere before reaching the earth surface. What can you say about its momentum?

Inactive
Karthik M.

It loses its momentum to the air molecules in the atmosphere.

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