A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0. At t

Q: Doesn't particle P travel a longer distance along the arc? Shouldn't it take more time?

It does travel a longer total distance! But remember, time is distance divided by speed. $P$'s speed increases so much that it more than makes up for the extra distance. To prove it rigorously, we compare their horizontal progress, where the distances are identical.

Q: Does gravity pull particle P forward horizontally?

No! Gravity acts strictly downwards. It cannot change the horizontal velocity directly. The horizontal acceleration comes entirely from the normal force exerted by the curved bowl.

Q: What happens to P's horizontal velocity after it passes the lowest point C?

Past point $C$, the normal force points inwards and backwards. This creates a negative horizontal acceleration, slowing $P$'s horizontal velocity down. However, it only slows down to its initial value $v$ exactly when it reaches $B$.

Q: Can we solve this using Conservation of Energy?

Yes! Energy conservation tells us that $P$'s total speed $V$ increases as it loses potential energy. Since $V > v$ and the angle of the bowl is relatively shallow, the horizontal component $V_x$ remains greater than $v$ throughout the journey.

Q: What if the bowl had friction?

If friction were present, it would do negative work, reducing $P$'s speed. Depending on the coefficient of friction, $P$ might lose its speed advantage, and $Q$ could potentially win the race.

Question

A particle $P$ is sliding down a frictionless hemispherical bowl. It passes the point $A$ at $t = 0$. At this instant of time, the horizontal component of its velocity is $v$. A bead $Q$ of the same mass as $P$ is ejected from $A$ at $t = 0$ along the horizontal string $AB$, with the speed $v$. Friction between the bead and the string may be neglected. Let $t_P$ and $t_Q$ be the respective times taken by $P$ and $Q$ to reach the point $B$. Then

Options:

(A) $t_P < t_Q$

(B) $t_P = t_Q$

(C) $t_P > t_Q$

(D) $\frac{t_P}{t_Q} = \frac{	ext{length of arc } ACB}{	ext{length of chord } AB}$

Sigma Percentile · Accepted Answer

### The Intuitive Trap

When you first look at this problem, your brain immediately jumps to a simple geometric fact: the curved arc $ACB$ is physically longer than the straight chord $AB$.

Since particle $P$ has to travel a longer distance, it feels obvious that it should take more time. But physics is rarely about surface-level intuition. **There is a catch here.** We must analyze the motion not by the total path length, but by breaking it down into its horizontal and vertical components.

### The Horizontal Race

Both particles start at point $A$ and end at point $B$. This means they must cover the exact same horizontal distance.

If we only look at their shadows moving along the horizontal axis, the race becomes a simple 1D kinematics problem. The time taken by either particle is purely dictated by its horizontal velocity. Let's evaluate our two racers based on this metric.

### Analyzing Particle Q: The Steady Runner

Particle $Q$ has a very straightforward journey. It moves along the horizontal string with a constant speed $v$.

Since there are no horizontal forces acting on it (friction is neglected), its horizontal velocity remains $v$ for the entire trip. The time it takes is simply the distance divided by the speed:

$$t_Q = \frac{AB}{v}$$

### Analyzing Particle P: The Gravity Assist

Particle $P$ starts at point $A$ with an initial horizontal velocity component that is also exactly $v$. At $t=0$, the race is perfectly tied.

However, as $P$ slides down the frictionless hemispherical bowl, it enters a dynamic environment. Gravity pulls it downwards, but gravity alone cannot change horizontal velocity. The real hero of this story is the **Normal Force**.

### The Secret Weapon: Normal Force

The bowl exerts a normal force $\vec{N}$ on particle $P$, pushing it perpendicular to the surface. Because the surface is a hemisphere, this force always points directly towards the center of the circle.

During the descent from $A$ to the lowest point $C$, this normal force vector points inwards and *forwards*. It has a positive horizontal component, $N_x$. According to Newton's Second Law, this creates a horizontal acceleration:

$$a_x = \frac{N_x}{m}$$

Because of this forward acceleration, $P$'s horizontal velocity $v_{Px}$ immediately becomes strictly greater than $v$. While $Q$ is jogging at a steady pace $v$, $P$ is sprinting horizontally at $v_{Px} > v$.

### The Deceleration Phase

Once $P$ passes the lowest point $C$ and begins to ascend towards $B$, the geometry flips. The normal force still points towards the center, but now that means it points inwards and *backwards*.

This creates a negative horizontal component, decelerating $P$'s horizontal motion. However, because the bowl is perfectly symmetrical, $P$'s horizontal velocity only drops back down to its initial value $v$ at the exact moment it reaches point $B$.

### The Final Verdict

Throughout the entire journey from $A$ to $B$ (except at the exact endpoints), particle $P$'s horizontal velocity was strictly greater than particle $Q$'s constant velocity $v$.

Mathematically, for all $t$ during the motion:

$$v_{Px} \ge v$$

Since $P$ is always moving faster horizontally, it will cover the horizontal distance $AB$ in less time than $Q$. Therefore, despite taking the longer physical path, the curved geometry gave $P$ the acceleration it needed to win the race.

**Final Answer:** $t_P < t_Q$

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Visualized Solution

$The Setup$

$The Horizontal Race$

$Motion of Q$

$Initial State of P$

$Forces on P (A to C)$

$The Forward Push$

$Motion from C to B$

$The Verdict$

The Sigma Insight: Motion in a Plane

The Intuitive Trap

The Horizontal Race

Analyzing Particle Q

Analyzing Particle P

The Secret Weapon

The Deceleration Phase

The Final Verdict

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Horizontal Race

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Motion of Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State of P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forces on P (A to C)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Forward Push

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Motion from C to B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Verdict