Question

A particle of mass $10^{-2}$ kg is moving along the positive X-axis under the influence of a force $F(x) = -k/2x^2$ where $k = 10^{-2} Nm^2$. At time $t = 0$, it is at $x = 1.0$ m and its velocity $v = 0$

(a) Find its velocity when it reaches $x = 0.5$ m.

(b) Find the time at which it reaches $x = 0.25$ m.(IIT-JEE 1998)

Answer 1

(a) -1 m/s, (b) $\pi/3 + \sqrt{3}/4$ s

Answer 2

Solution: The problem describes a particle of mass $m = 10^{-2}$ kg moving along the positive X-axis under the influence of a force $F(x) = -k/(2x^2)$, where $k = 10^{-2} Nm^2$. At $t=0$, the particle is at $x_0 = 1.0$ m with velocity $v_0 = 0$.

Part (a): Find its velocity when it reaches $x = 0.5$ m. The force is conservative, so we can use the work-energy theorem or conservation of mechanical energy. The work done by the force $F(x)$ as the particle moves from $x_1$ to $x_2$ is $W = \int_{x_1}^{x_2} F(x) dx$. According to the work-energy theorem, $W = \Delta KE = KE_f - KE_i$. The initial state is at $x_0 = 1.0$ m with $v_0 = 0$. The initial kinetic energy is $KE_i = (1/2) m v_0^2 = 0$. The final state is at $x_f = 0.5$ m with velocity $v_f$. The final kinetic energy is $KE_f = (1/2) m v_f^2$. The work done is $W = \int_{1.0}^{0.5} \frac{-k}{2x^2} dx = -\frac{k}{2} \int_{1.0}^{0.5} x^{-2} dx = -\frac{k}{2} [-x^{-1}]_{1.0}^{0.5} = -\frac{k}{2} [-\frac{1}{0.5} - (-\frac{1}{1.0})] = -\frac{k}{2} [-2 + 1] = -\frac{k}{2} [-1] = \frac{k}{2}$. Using the work-energy theorem: $W = KE_f - KE_i$. $\frac{k}{2} = \frac{1}{2} m v_f^2 - 0$. $v_f^2 = \frac{k}{m}$. Given $k = 10^{-2} Nm^2$ and $m = 10^{-2}$ kg, $v_f^2 = \frac{10^{-2}}{10^{-2}} = 1$. So, $v_f = \pm 1$ m/s. The force $F(x) = -k/(2x^2)$ is always negative for $x > 0$, meaning it acts towards the origin. Since the particle starts at $x=1.0$ m and moves towards $x=0.5$ m, its position is decreasing, which means its velocity must be negative. Thus, the velocity when it reaches $x = 0.5$ m is $v_f = -1$ m/s.

Part (b): Find the time at which it reaches $x = 0.25$ m. We need to find the velocity as a function of position first. Using conservation of mechanical energy is convenient. The potential energy $U(x)$ is given by $F(x) = -dU/dx$. $U(x) = -\int F(x) dx = -\int \frac{-k}{2x^2} dx = \frac{k}{2} \int x^{-2} dx = \frac{k}{2} (-\frac{1}{x}) + C = -\frac{k}{2x} + C$. Let's set the potential energy to zero at infinity, so $C=0$. $U(x) = -k/(2x)$. The total mechanical energy $E = KE + PE = \frac{1}{2} m v^2 + U(x)$. The initial energy at $t=0$, $x_0 = 1.0$ m, $v_0 = 0$: $E_0 = \frac{1}{2} m (0)^2 + U(1.0) = 0 - \frac{k}{2(1.0)} = -\frac{k}{2}$. By conservation of energy, the total energy $E$ is constant and equal to $E_0$. $\frac{1}{2} m v^2 + U(x) = -\frac{k}{2}$. $\frac{1}{2} m v^2 - \frac{k}{2x} = -\frac{k}{2}$. $\frac{1}{2} m v^2 = \frac{k}{2x} - \frac{k}{2} = \frac{k}{2} (\frac{1}{x} - 1)$. $v^2 = \frac{k}{m} (\frac{1}{x} - 1)$. Since the particle is moving towards the origin from $x=1.0$, the velocity is negative: $v = \frac{dx}{dt} = -\sqrt{\frac{k}{m} (\frac{1}{x} - 1)}$. We need to find the time taken to go from $x_0 = 1.0$ m to $x_f = 0.25$ m. $dt = \frac{dx}{v} = \frac{dx}{-\sqrt{\frac{k}{m} (\frac{1}{x} - 1)}} = -\sqrt{\frac{m}{k}} \frac{dx}{\sqrt{\frac{1-x}{x}}} = -\sqrt{\frac{m}{k}} \frac{\sqrt{x}}{\sqrt{1-x}} dx$. Integrate from $t=0$ to $t$ and from $x=1.0$ to $x=0.25$: $\int_0^t dt = \int_{1.0}^{0.25} -\sqrt{\frac{m}{k}} \frac{\sqrt{x}}{\sqrt{1-x}} dx$. $t = -\sqrt{\frac{m}{k}} \int_{1.0}^{0.25} \frac{\sqrt{x}}{\sqrt{1-x}} dx$. Substitute $m = 10^{-2}$ and $k = 10^{-2}$, so $\sqrt{m/k} = \sqrt{1} = 1$. $t = - \int_{1.0}^{0.25} \frac{\sqrt{x}}{\sqrt{1-x}} dx$. Let's evaluate the integral $\int \frac{\sqrt{x}}{\sqrt{1-x}} dx$. Substitute $x = \sin^2 \theta$. Then $dx = 2 \sin \theta \cos \theta d\theta$. $\sqrt{x} = \sin \theta$ and $\sqrt{1-x} = \sqrt{1-\sin^2 \theta} = \cos \theta$ (for $x \in [0, 1)$). $\int \frac{\sin \theta}{\cos \theta} (2 \sin \theta \cos \theta) d\theta = \int 2 \sin^2 \theta d\theta = \int (1 - \cos 2\theta) d\theta = \theta - \frac{1}{2} \sin 2\theta + C = \theta - \sin \theta \cos \theta + C$. Substitute back $x = \sin^2 \theta$, so $\theta = \arcsin(\sqrt{x})$. The integral is $\arcsin(\sqrt{x}) - \sqrt{x}\sqrt{1-x} + C$. Now evaluate the definite integral from $x=1.0$ to $x=0.25$. When $x=1.0$, $\sqrt{x}=1$, $\arcsin(1) = \pi/2$. $\sqrt{x}\sqrt{1-x} = 1 \times 0 = 0$. Value is $\pi/2$. When $x=0.25$, $\sqrt{x}=0.5$, $\arcsin(0.5) = \pi/6$. $\sqrt{x}\sqrt{1-x} = 0.5 \times \sqrt{1-0.25} = 0.5 \times \sqrt{0.75} = 0.5 \times \sqrt{3}/2 = \sqrt{3}/4$. Value is $\pi/6 - \sqrt{3}/4$. The definite integral $\int_{1.0}^{0.25} \frac{\sqrt{x}}{\sqrt{1-x}} dx = [\arcsin(\sqrt{x}) - \sqrt{x}\sqrt{1-x}]_{1.0}^{0.25}$ $= (\arcsin(\sqrt{0.25}) - \sqrt{0.25}\sqrt{1-0.25}) - (\arcsin(\sqrt{1.0}) - \sqrt{1.0}\sqrt{1-1.0})$ $= (\arcsin(0.5) - 0.5 \times \sqrt{0.75}) - (\arcsin(1) - 1 \times 0)$ $= (\pi/6 - 0.5 \times \sqrt{3}/2) - (\pi/2 - 0)$ $= \pi/6 - \sqrt{3}/4 - \pi/2 = (\pi - 3\pi)/6 - \sqrt{3}/4 = -2\pi/6 - \sqrt{3}/4 = -\pi/3 - \sqrt{3}/4$. The time $t = - \times (-\pi/3 - \sqrt{3}/4) = \pi/3 + \sqrt{3}/4$.

The final answer is $\pi/3 + \sqrt{3}/4$ seconds.

Explanation of the solution: (a) Use the work-energy theorem $\int_{x_i}^{x_f} F(x) dx = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$. Calculate the integral of the force from $x=1.0$ m to $x=0.5$ m. Set $v_i=0$ at $x_i=1.0$ m and solve for $v_f$ at $x_f=0.5$ m. Determine the sign of the velocity based on the direction of motion. (b) Use the definition of velocity $v = dx/dt$. From energy conservation, find the expression for $v(x)$. Separate variables to get $dt = dx/v(x)$. Integrate $dt$ from $t=0$ to $t$ and $dx$ from $x=1.0$ m to $x=0.25$ m. Evaluate the definite integral using an appropriate substitution (like $x=\sin^2 \theta$).