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(A) Let $m$ be the mass of the train,$R = 1000 \ m$ be the radius of the curve,and $	heta$ be the angle of banking.In the frame of the train,the forc

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(A) Let $m$ be the mass of the train,$R = 1000 \ m$ be the radius of the curve,and $	heta$ be the angle of banking.In the frame of the train,the forces acting along the inclined plane are the component of gravity $mg \sin 	heta$,the centrifugal force $\frac{mv^2}{R} \cos 	heta$,and the lateral force $F$ exerted by the rails.For speed $v_1 = 10 \ ms^{-1}$,the lateral force $F_1$ on the inner rail is given by:$F_1 + \frac{mv_1^2}{R} \cos 	heta = mg \sin 	heta \implies F_1 = mg \sin 	heta - \frac{mv_1^2}{R} \cos 	heta$For speed $v_2 = 20 \ ms^{-1}$,the lateral force $F_2$ on the outer rail is given by:$\frac{mv_2^2}{R} \cos 	heta = mg \sin 	heta + F_2 \implies F_2 = \frac{mv_2^2}{R} \cos 	heta - mg \sin 	heta$Given $F_1 = F_2$,we have:$mg \sin 	heta - \frac{mv_1^2}{R} \cos 	heta = \frac{mv_2^2}{R} \cos 	heta - mg \sin 	heta$$2mg \sin 	heta = \frac{m}{R} (v_1^2 + v_2^2) \cos 	heta$$2g 	an 	heta = \frac{v_1^2 + v_2^2}{R} = \frac{10^2 + 20^2}{1000} = \frac{100 + 400}{1000} = \frac{500}{1000} = 0.5$Thus,$	an 	heta = \frac{0.5}{2g} = \frac{1}{4g}$.Therefore,$4g 	an 	heta = 4g 	imes \frac{1}{4g} = 1$.

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