A particle (m=1kg) slides down a frictionless track (AOC) starting from rest at a point A (height 2

Q: Why don't we need to calculate the velocity at point C where it leaves the track?

That's the beauty of the Conservation of Energy! Energy is a state function, meaning it only depends on the initial and final states, not the path taken. We can completely skip point C and directly compare point A to point P.

Q: Isn't the velocity at the highest point of a projectile zero? Why does it have kinetic energy?

A common trap! Only the *vertical* component of velocity is zero at the highest point. The particle still has a *horizontal* component of velocity, which means it still has kinetic energy.

Q: What if the mass of the particle was $5 \text{ kg}$ instead of $1 \text{ kg}$?

The total energy would be higher, and the final kinetic energy would be $50 \text{ J}$. However, if you were asked for the *velocity* at point P, the mass would cancel out of the equation, and the velocity would remain exactly the same!

Q: Why can't we just use the kinematic equations ($v^2 = u^2 + 2as$) to solve this?

The track AOC is curved, which means the acceleration is constantly changing direction and magnitude. Kinematic equations only work for *constant* acceleration. Energy conservation is the only practical way to solve this.

Q: Does the shape of the track matter at all?

As long as the track is completely frictionless, its shape does not matter one bit! The final speed and kinetic energy depend solely on the vertical height difference between the starting and ending points.

Question

A particle ($m=1	ext{kg}$) slides down a frictionless track ($AOC$) starting from rest at a point $A$ (height 2 m). After reaching $C$, the particle continues to move freely in air as a projectile. When it reaching its highest point $P$ (height 1 m), the kinetic energy of the particle (in J) is : (Figure drawn is schematic and not to scale; take $g=10	ext{ms}^{-2}$) ......... .

Sigma Percentile · Accepted Answer

### The Setup: A Frictionless Rollercoaster

Imagine you are standing at the top of a massive, perfectly smooth rollercoaster track. You hold a small particle of mass $m = 1 	ext{ kg}$ at point $A$, which is exactly $2 	ext{ m}$ above the ground. You let it go from rest.

The particle swoops down the frictionless track, feeling the rush of gravity. It reaches the bottom at point $O$, speeds up the other side, and launches off the edge at point $C$. Now, it is no longer constrained by the track; it becomes a free-flying projectile soaring through the air.

Our mission is to analyze the particle at the absolute peak of its flight, point $P$, which is $1 	ext{ m}$ above the ground. Specifically, we need to find its kinetic energy at this exact moment.

### The Master Key: Energy Conservation

In physics, whenever you hear the word "frictionless" and see a particle moving under the influence of gravity alone, your brain should immediately scream: **Conservation of Mechanical Energy!**

Because there are no non-conservative forces (like friction or air resistance) doing work on our particle, the total mechanical energy of the system remains perfectly constant. The energy it has at the very beginning is the exact same amount of energy it has at the very end, and at every single point in between.

Mathematically, we can write this as:
$$E_A = E_P$$
$$U_A + K_A = U_P + K_P$$

### Crunching the Numbers

Let's break down the energy at our two points of interest.

At the starting point $A$, the particle is released from rest. This means its initial velocity is zero, and therefore, its initial kinetic energy ($K_A$) is zero. All of its energy is stored as gravitational potential energy ($U_A$), which depends on its height $h_A = 2 	ext{ m}$.

$$U_A = mgh_A = (1 	ext{ kg})(10 	ext{ m/s}^2)(2 	ext{ m}) = 20 	ext{ J}$$

Now, let's look at the peak of the flight, point $P$. The particle is at a height $h_P = 1 	ext{ m}$. It has some potential energy, and because it is moving horizontally at the peak of its projectile motion, it also has the kinetic energy ($K_P$) we are trying to find.

$$U_P = mgh_P = (1 	ext{ kg})(10 	ext{ m/s}^2)(1 	ext{ m}) = 10 	ext{ J}$$

### The Final Reveal

Now, we simply plug these values back into our conservation of energy equation:

$$20 	ext{ J} + 0 = 10 	ext{ J} + K_P$$

To isolate the kinetic energy, we subtract $10 	ext{ J}$ from both sides:

$$K_P = 20 	ext{ J} - 10 	ext{ J}$$
$$K_P = 10 	ext{ J}$$

And there we have it! The kinetic energy of the particle at its highest point is exactly **$10 	ext{ J}$**. The elegance of energy conservation allowed us to completely bypass the complex kinematics of the curved track and the projectile motion. We just looked at the start, looked at the end, and let the universe's bookkeeping do the rest.

Enter Numerical Value:

Visualized Solution

$Visualizing the Journey$

$Conservation of Energy$

$Setting up the Equation$

$Substituting Values$

$Simplifying$

$Final Kinetic Energy$

$Food for Thought$

The Sigma Insight: Conservation of Mechanical Energy

The Setup

The Master Key

Crunching the Numbers

The Final Reveal

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