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Tutor profile: Karthik M.

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Karthik M.
Adjunct Physics Instructor
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Questions

Subject: Physics (Newtonian Mechanics)

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Question:

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

Inactive
Karthik M.
Answer:

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.
Answer:

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.
Answer:

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

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