Problem 46 - Master Resource Book in JEE Main Physics

Question

A particle is projected with velocity \(v_{0}\) along \(x\)-axis. The deceleration on the particle is proportional to the square of the distance from the origin, ie., \(a=\alpha x^{2}\), the distance at which the particle stop is (a) \(\sqrt{\frac{3 v_{0}}{2 \alpha}}\) (b) \(\left(\frac{3 v_{0}}{2 \alpha}\right)^{1 / 3}\) (c) \(\sqrt{\frac{2 v_{0}^{2}}{3 \alpha}}\) (d) \(\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{1 / 3}\)

Step-by-Step Solution

Answer

Option (d) \( \left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{1 / 3} \) is correct.

Step 1: Understanding the problem

The problem states that a particle is projected along the x-axis with an initial velocity \( v_0 \), and it experiences a deceleration proportional to the square of the distance from the origin, given by \( a = \alpha x^2 \). We need to find the distance from the origin at which the particle stops, \( x_{f} \).

Step 2: Express acceleration in terms of velocity and position

We know the relationship between acceleration and velocity: \( a = \frac{dv}{dt} = v \frac{dv}{dx} \). Using \( a = \alpha x^2 \) and \( a = v \frac{dv}{dx} \), we can write \( v \frac{dv}{dx} = -\alpha x^2 \).

Step 3: Solve the differential equation for velocity

Separate variables and integrate: \[ v dv = -\alpha x^2 dx \]. Integrating both sides yields: \[ \frac{v^2}{2} = -\frac{\alpha x^3}{3} + C \].

Step 4: Apply initial conditions to find the constant

At \( x = 0 \), the velocity \( v = v_{0} \). Substituting in the equation, \( \frac{v_0^2}{2} = C \). Therefore, the expression for velocity becomes \( \frac{v^2}{2} = -\frac{\alpha x^3}{3} + \frac{v_0^2}{2} \).

Step 5: Determine the stopping condition

The particle stops when \( v = 0 \). Substituting this into the velocity equation: \( -\frac{\alpha x_{f}^3}{3} + \frac{v_0^2}{2} = 0 \).

Step 6: Solve for the stopping distance

Rearrange to solve for \( x_{f} \, : \, \frac{\alpha x_{f}^3}{3} = \frac{v_0^2}{2} \). Solving for \( x_{f} \), we find: \[ x_{f} = \left( \frac{3 v_0^2}{2 \alpha} \right)^{1/3} \].

Step 7: Choose the correct option

The correct option corresponding to the derived expression for \( x_{f} \) is (d), \( \left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{1 / 3} \).

Key Concepts

Projectile motion

Projectile motion is an essential concept in kinematics that describes the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. In simple terms, it's the kind of motion you observe when you see a ball thrown or a firework launched. Although our exercise deals with motion along a single axis (the x-axis), the basic principles and equations of projectile motion still apply here. When studying projectile motion:

• The motion has two components: horizontal and vertical.
• Gravity is the only force acting once a projectile is in motion (assuming air resistance is negligible).
• The horizontal motion is uniform, meaning a constant velocity, since no horizontal forces act (again, assuming no air resistance).
• Vertical motion is uniformly accelerated, with gravity as the accelerating force.

In our problem, the focus is not on the typical vertical motion against gravity, but rather on a horizontal deceleration proportional to the square of the distance from the origin, a unique twist compared to usual projectile motion problems.

Differential equations

Differential equations are equations that involve an unknown function and its derivatives. They are crucial in modeling various real-world phenomena, including the problem we are exploring here. In kinematics, they help link acceleration, velocity, and position - all fundamental aspects of understanding motion. For this problem, we use the relationship between acceleration and velocity, given by the differential equation:\[a = \frac{dv}{dt} = v \frac{dv}{dx}\]This equation connects the rate of change of velocity (i.e., acceleration) to velocity and distance. By rearranging and integrating this equation with the given condition that acceleration, \(a\), is proportional to \(x^2\), we obtain:\[v \frac{dv}{dx} = -\alpha x^2\]This is a separable differential equation that can be solved using standard techniques: separate the variables, integrate both sides, and apply initial conditions to find specific constants. In our solution, this led us to determine how the velocity changes over time and, ultimately, find the stopping distance.

Deceleration

Deceleration is simply negative acceleration; it signifies an object's reduction in speed. In our exercise, the particle experiences deceleration proportional to the square of its distance from the origin.Examining this relationship:

• Deceleration is defined by \(a = \alpha x^2\).
• The negative sign in \(v \frac{dv}{dx} = -\alpha x^2\) indicates the velocity is decreasing over time as the particle moves away from the origin.
• This quadratic dependency on distance means that as the particle moves further away from the origin, the deceleration increases.

Ultimately, you calculate when the velocity becomes zero, the particle stops moving. By solving this unique quadratic deceleration, we find the stopping distance using standard integration techniques for differential equations in motion.