A particle of mass m with initial kinetic ene | vedclass

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(A) According to the law of conservation of mechanical energy, the total energy remains constant: $K_i + U_i = K_f + U_f$.At $x = +\infty$, the potent

Vedclass · Accepted Answer

$\sqrt{K/m}$

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(A) According to the law of conservation of mechanical energy, the total energy remains constant: $K_i + U_i = K_f + U_f$.At $x = +\infty$, the potential energy $U_i = V(\infty) = \frac{K}{e^{\infty} + e^{-\infty}} = 0$.The initial kinetic energy is $K_i = K$.At $x = 0$, the potential energy is $U_f = V(0) = \frac{K}{e^0 + e^0} = \frac{K}{1 + 1} = \frac{K}{2}$.Let the speed at $x = 0$ be $v$. Then the final kinetic energy is $K_f = \frac{1}{2}mv^2$.Applying the conservation of energy: $K + 0 = \frac{1}{2}mv^2 + \frac{K}{2}$.Rearranging the terms: $\frac{1}{2}mv^2 = K - \frac{K}{2} = \frac{K}{2}$.Solving for $v$: $mv^2 = K \Rightarrow v^2 = \frac{K}{m} \Rightarrow v = \sqrt{\frac{K}{m}}$.Thus, the correct option is $A$.

$A$ particle of mass $m$ with initial kinetic energy $K$ approaches the origin from $x = +\infty$. Assume that a conservative force acts on it and its potential energy $V(x)$ is given by $V(x) = \frac{K}{\exp(3x/x_0) + \exp(-3x/x_0)}$, where $x_0 = 1 \ m$. The speed of the particle at $x = 0$ is

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