Question

A particle of mass m is given a velocity $v_0$ on a rough horizontal surface. The coefficient of friction between the particle and surface is $\mu$. There is also a variable external force acts on the particle given as $|F|=k|v|$ where k is a positive constant and v is instantaneous velocity. The directions of force at any instant is perpendicular to velocity.

• A. the time taken by the particle to stop is $\frac{v_0}{\mu g}$
• B. the time taken to change the angle between the acceleration and the velocity from $120^\circ$ to $150^\circ$ is $\frac{2m}{\sqrt{3}k}$
• C. the total distance covered by the particle is $\frac{v_0^2}{2\mu g}$
• D. The radius of curvature of the path at any time t is $\frac{m}{k}(v_0-\frac{\mu g t}{2})$

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The problem describes the motion of a particle under two forces: kinetic friction and a variable external force.

1. Kinetic Friction ($f_k$): This force acts opposite to the direction of motion (tangential) and its magnitude is constant, $f_k = \mu N = \mu mg$, where $N$ is the normal force, equal to $mg$ for a horizontal surface. This force causes deceleration.

2. Variable External Force ($F$): Its magnitude is given as $F = kv$, where $v$ is the instantaneous speed. Its direction is always perpendicular to the velocity. This means the force acts as a centripetal force, changing the direction of motion but not the speed.

Let's analyze the acceleration components:

• Tangential acceleration ($a_t$): This component is responsible for changing the speed of the particle. It is due to the friction force. $m a_t = -f_k \implies m \frac{dv}{dt} = -\mu mg$ $a_t = \frac{dv}{dt} = -\mu g$ (constant deceleration)

• Normal (Centripetal) acceleration ($a_n$): This component is responsible for changing the direction of the particle's velocity, causing it to move in a curved path. It is due to the external force $F$. $m a_n = F \implies m a_n = kv$ $a_n = \frac{kv}{m}$

Now let's evaluate each option:

(A) The time taken by the particle to stop is $\frac{v_0}{\mu g}$

From the tangential acceleration equation: $\frac{dv}{dt} = -\mu g$ Integrating with initial velocity $v_0$ at $t=0$: $\int_{v_0}^{v} dv = \int_{0}^{t} -\mu g dt$ $v - v_0 = -\mu g t$ $v(t) = v_0 - \mu g t$ The particle stops when $v(t) = 0$. Let $t_{stop}$ be this time. $0 = v_0 - \mu g t_{stop}$ $t_{stop} = \frac{v_0}{\mu g}$ This statement is correct.

(C) The total distance covered by the particle is $\frac{v_0^2}{2\mu g}$

The distance covered is $s = \int_{0}^{t_{stop}} v(t) dt$. $s = \int_{0}^{v_0/\mu g} (v_0 - \mu g t) dt$ $s = \left[ v_0 t - \frac{1}{2}\mu g t^2 \right]_{0}^{v_0/\mu g}$ $s = v_0 \left(\frac{v_0}{\mu g}\right) - \frac{1}{2}\mu g \left(\frac{v_0}{\mu g}\right)^2$ $s = \frac{v_0^2}{\mu g} - \frac{1}{2}\mu g \frac{v_0^2}{\mu^2 g^2}$ $s = \frac{v_0^2}{\mu g} - \frac{v_0^2}{2\mu g}$ $s = \frac{v_0^2}{2\mu g}$ This statement is correct.

(B) The time taken to change the angle between the acceleration and the velocity from $120^\circ$ to $150^\circ$ is $\frac{2m}{\sqrt{3}k}$

Let $\vec{v}$ be the velocity vector. The tangential acceleration $\vec{a_t}$ is opposite to $\vec{v}$ (due to friction), so the angle between $\vec{a_t}$ and $\vec{v}$ is $180^\circ$. The normal acceleration $\vec{a_n}$ is perpendicular to $\vec{v}$. The total acceleration $\vec{a} = \vec{a_t} + \vec{a_n}$. Let $\theta$ be the angle between $\vec{a}$ and $\vec{v}$. Consider a right triangle formed by the magnitudes $a_t = \mu g$ and $a_n = \frac{kv}{m}$. The angle $\alpha$ between $\vec{a}$ and $\vec{a_t}$ (which is in the direction of $-\vec{v}$) is given by: $\tan \alpha = \frac{a_n}{a_t} = \frac{kv/m}{\mu g} = \frac{kv}{\mu mg}$ Since $\vec{a_t}$ is in the direction opposite to $\vec{v}$, the angle $\theta$ between $\vec{a}$ and $\vec{v}$ is $\theta = 180^\circ - \alpha$. So, $\tan(180^\circ - \theta) = \tan \alpha = \frac{kv}{\mu mg}$. Since $\tan(180^\circ - \theta) = -\tan \theta$, we have: $\tan \theta = -\frac{kv}{\mu mg}$

For $\theta_1 = 120^\circ$: $\tan 120^\circ = -\sqrt{3}$ So, $-\sqrt{3} = -\frac{kv_1}{\mu mg} \implies v_1 = \frac{\sqrt{3} \mu mg}{k}$

For $\theta_2 = 150^\circ$: $\tan 150^\circ = -\frac{1}{\sqrt{3}}$ So, $-\frac{1}{\sqrt{3}} = -\frac{kv_2}{\mu mg} \implies v_2 = \frac{\mu mg}{\sqrt{3}k}$

The time taken to change the speed from $v_1$ to $v_2$ is $\Delta t$. Using $v_2 = v_1 - \mu g \Delta t$: $\Delta t = \frac{v_1 - v_2}{\mu g}$ $\Delta t = \frac{1}{\mu g} \left( \frac{\sqrt{3} \mu mg}{k} - \frac{\mu mg}{\sqrt{3}k} \right)$ $\Delta t = \frac{m}{k} \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right)$ $\Delta t = \frac{m}{k} \left( \frac{3 - 1}{\sqrt{3}} \right)$ $\Delta t = \frac{2m}{\sqrt{3}k}$ This statement is correct.

(D) The radius of curvature of the path at any time t is $\frac{m}{k}(v_0-\frac{\mu g t}{2})$

The radius of curvature $R$ is related to the normal acceleration $a_n$ and speed $v$ by the formula: $a_n = \frac{v^2}{R}$ So, $R = \frac{v^2}{a_n}$. We know $a_n = \frac{kv}{m}$. Substituting $a_n$: $R = \frac{v^2}{kv/m} = \frac{mv^2}{kv} = \frac{mv}{k}$ Now, substitute the expression for $v(t)$: $R(t) = \frac{m}{k} (v_0 - \mu g t)$ Comparing this with the given option: $\frac{m}{k}(v_0-\frac{\mu g t}{2})$, we see they are different. This statement is incorrect.

Therefore, options (A), (B), and (C) are correct.