QJEEAdv20Ph5037: JEE Advanced 2020 Physics - Class 11 JEEAdv20Ph5037 Solution: Starting at time t = 0 from the origin with speed 1 ms -1 , a…

Given:

At time $t = 0$ , the particle starts from the origin $(0, 0)$ with a speed of $v = 1\, \text{m/s}$ .

The particle moves along the path defined by $y = \frac{x^2}{2}$ .

Understanding the Particle's Motion

Since the particle follows the path $y = \frac{x^2}{2}$ , we can find relationships between its velocity and acceleration components.

1. Velocity Components

Let $x(t)$ and $y(t)$ be the particle's position as functions of time. Then:

$y = \frac{x^2}{2}$

Differentiate both sides with respect to time $t$ :

$\frac{dy}{dt} = x \frac{dx}{dt}$

The velocity components are:

$v_x = \frac{dx}{dt}$

$v_y = \frac{dy}{dt} = x v_x$

2. Acceleration Components

The acceleration components are:

$a_x = \frac{dv_x}{dt}$

$a_y = \frac{dv_y}{dt} = \frac{d}{dt}(x v_x) = v_x^2 + x a_x$

Analyzing Each Option

Option A

Statement: $a_x = 1\, \text{m/s}^2$ implies that when the particle is at the origin, $a_y = 1\, \text{m/s}^2$ .

Verification:

At the origin $x = 0$ , and from initial conditions, $v_x = 1\, \text{m/s}$ .

Using the expression for $a_y$ :

$a_y = v_x^2 + x a_x = (1\, \text{m/s})^2 + 0 \times a_x = 1\, \text{m/s}^2$

Conclusion: Option A is true.

Option B

Statement: $a_x = 0$ implies $a_y = 1\, \text{m/s}^2$ at all times.

Verification:

If $a_x = 0$ , then $v_x = \text{constant} = 1\, \text{m/s}$ .

Position along $x$ : $x = v_x t = t$ .

Then $v_y = x v_x = t \times 1\, \text{m/s} = t\, \text{m/s}$ .

Therefore, $a_y = \frac{dv_y}{dt} = 1\, \text{m/s}^2$ .

Conclusion: Option B is true.

Option C

Statement: At $t = 0$ , the particle's velocity points in the $x$ -direction.

Verification:

The slope of the path is $\frac{dy}{dx} = x$ .

At $x = 0$ , $\frac{dy}{dx} = 0$ , meaning the path is horizontal.

Thus, $v_y = 0$ at $t = 0$ , and $v_x = 1\, \text{m/s}$ .

Conclusion: Option C is true.

Option D

Statement: $a_x = 0$ implies that at $t = 1\, \text{s}$ , the angle between the particle's velocity and the $x$ -axis is $45^\circ$ .

Verification:

With $a_x = 0$ , $v_x = 1\, \text{m/s}$ .

At $t = 1\, \text{s}$ , $x = 1\, \text{m}$ .

Then $v_y = x v_x = 1\, \text{m} \times 1\, \text{m/s} = 1\, \text{m/s}$ .

The angle $\theta$ between $\vec{v}$ and the $x$ -axis is:

$\tan \theta = \frac{v_y}{v_x} = \frac{1\, \text{m/s}}{1\, \text{m/s}} = 1 \implies \theta = 45^\circ$

Conclusion: Option D is true.

Final Answer

All options A, B, C, and D are true.