A bob of mass $0.1\ \text{kg}$ hung from the ceiling of a room by a string $2\ \text{m}$ long is set into oscillation. The speed of the bob at its mean position is $1\ \text{m s}^{-1}$. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.

A bob of mass 0.1kg hung from the ceiling of a room by a string 2m long is set into oscillation. The speed of

Exercises · 4.12

0.1\ \text{kg}

(a) at one of its extreme positions,

(b) at its mean position.

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Concept understanding — Projectile Motion

Projectile Motion — From Intuition to Precision

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it rises, slows down, then curves downward and falls. That curved path is a projectile's trajectory. The ball is a projectile: any object that is launched into the air and then moves only under the influence of gravity (and air resistance, which we ignore for now).

The key intuition: once the ball leaves your hand, the only force acting on it is gravity pulling it straight down. There is no forward force after release. The ball keeps moving forward because of inertia — it wants to keep going in a straight line at constant speed. But gravity keeps pulling it down, so the forward motion and downward acceleration combine to produce a curved path.

The Precise Statement

g \approx 9.8 \, \text{m/s}^2

We break the motion into two independent components:

Horizontal motion: No acceleration ( $a_x = 0$ ). So horizontal velocity $v_x$ is constant.
Vertical motion: Constant downward acceleration ( $a_y = -g$ ). So vertical velocity $v_y$ changes linearly with time.

The independence of these components is the central idea — what happens vertically does not affect what happens horizontally, and vice versa.

The Equations (for a projectile launched with initial speed $u$ at angle $\theta$ above horizontal)

First, resolve the initial velocity:

u_x = u \cos\theta, \quad u_y = u \sin\theta

Horizontal motion (constant velocity):

x = u_x t = (u \cos\theta) t

Vertical motion (constant acceleration $-g$ ):

v_y = u_y - gt = u \sin\theta - gt

y = u_y t - \frac{1}{2} g t^2 = (u \sin\theta) t - \frac{1}{2} g t^2

Key Results You Must Know

Important

Time of flight $T$ : total time the projectile stays in the air (until $y=0$ again).

$T = \frac{2u \sin\theta}{g}$

Important

Maximum height $H$ : the highest vertical position reached (when $v_y = 0$ ).

$H = \frac{u^2 \sin^2\theta}{2g}$

Important

Range $R$ : the horizontal distance covered when it returns to launch height.

$R = \frac{u^2 \sin 2\theta}{g}$

Note

The range is maximum when $\sin 2\theta = 1$ , i.e., $\theta = 45^\circ$ . For a given speed, $45^\circ$ gives the farthest throw.

The Trajectory Equation (Path Shape)

Eliminate $t$ from the $x$ and $y$ equations to get $y$ as a function of $x$ :

y = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}

This is a parabola — the signature shape of projectile motion.

Common Mistake to Avoid …

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When the string is cut, the bob becomes a projectile with whatever velocity it had at that instant. At an extreme position the bob is momentarily at rest, so it falls vertically; at the mean position it has horizontal velocity $1\ \text{m s}^{-1}$ , so it follows a parabolic path.

Why the trajectory changes: understanding projectile motion after release

A pendulum bob swings back and forth because the string tension and gravity together provide the restoring force. The moment you cut the string, tension vanishes. From that instant onward, only gravity acts on the bob—it becomes a projectile.

The trajectory of any projectile is determined entirely by its velocity at the moment of release. If the bob has zero velocity, it simply falls straight down under gravity. If it has a horizontal component of velocity, it traces a parabola (the classic projectile path). The key is to identify the bob's velocity at the instant the string is cut.

(a) String cut at an extreme position

At an extreme position of a pendulum's swing, the bob momentarily comes to rest before reversing direction. This is the turning point of the oscillation.

Velocity at the extreme: The speed is zero, $v = 0$ .
Initial conditions when the string is cut:
- Horizontal velocity: $v_x = 0$
- Vertical velocity: $v_y = 0$
Motion after release: With no initial velocity in any direction, the bob is subject only to gravitational acceleration $g$ downward. It falls vertically along a straight line.

The trajectory is a vertical straight line (free fall).

(b) String cut at the mean position

At the mean (equilibrium) position, the bob is moving with its maximum speed. For a pendulum, this velocity is horizontal—tangent to the circular arc, which at the lowest point is perpendicular to the vertical string.

Velocity at the mean position: The bob has speed $v = 1\ \text{m s}^{-1}$ , directed horizontally (let's say along the positive $x$ -direction).
Initial conditions when the string is cut:
- Horizontal velocity: $v_x = 1\ \text{m s}^{-1}$
- Vertical velocity: $v_y = 0$ (the bob is moving horizontally at the lowest point) …

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Q.A bob of mass $0.1\ \text{kg}$ hung from the ceiling of a room by a string $2\ \text{m}$ long is set into oscillation. The speed of the bob at its mean position is $1\ \text{m s}^{-1}$ . What is the trajectory of the bob if the string is cut when the bob is

Concept understanding — Projectile Motion

Projectile Motion — From Intuition to Precision

The Precise Statement

The Equations (for a projectile launched with initial speed $u$ at angle $\theta$ above horizontal)

Key Results You Must Know

The Trajectory Equation (Path Shape)

Common Mistake to Avoid …

Why the trajectory changes: understanding projectile motion after release

(a) String cut at an extreme position

(b) String cut at the mean position

Concept: Projectile Motion Once a Constraint (String Tension) Is Removed

Q.A bob of mass 0.1 kg0.1\ \text{kg}0.1 kg hung from the ceiling of a room by a string 2 m2\ \text{m}2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s−11\ \text{m s}^{-1}1 m s−1. What is the trajectory of the bob if the string is cut when the bob is

Why the trajectory changes: understanding projectile motion after release

(a) String cut at an extreme position

(b) String cut at the mean position

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Q.A bob of mass $0.1\ \text{kg}$ hung from the ceiling of a room by a string $2\ \text{m}$ long is set into oscillation. The speed of the bob at its mean position is $1\ \text{m s}^{-1}$ . What is the trajectory of the bob if the string is cut when the bob is

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