4 Most SAQ’s of Motion in a Plane Chapter in Inter 1st Year Physics (TS/AP)

4 Marks

SAQ-1 : Show that the trajectory of an object thrown at certain angle with the horizontal is parabola.

Introduction

The trajectory of an object projected at an angle to the horizontal is a fundamental concept in physics and mathematics. This trajectory, the path traced by an object’s motion through space, is particularly interesting when the object is projected at a certain angle. In such cases, the trajectory is typically a parabola. We will explore this phenomenon using the principles of physics and mathematics.

1. Projectile Motion and Its Components:
   • Projectile Motion: Refers to the motion of an object projected into the air, influenced primarily by the force of gravity.
   • Two Components: The motion can be divided into horizontal motion and vertical motion.
2. Constant Horizontal Velocity:
   • Uniform Horizontal Motion: Upon launch at an angle, the object maintains a constant horizontal velocity because there are no external forces acting horizontally to alter this velocity.
3. Acceleration in the Vertical Direction:
   • Downward Acceleration: Vertically, the object experiences a constant downward acceleration due to gravity, affecting its vertical velocity over time.
4. Combined Motion: Horizontal and Vertical Components:
   • Independent Motions: The object’s motion is described separately in the horizontal and vertical directions.
   • Uniform and Accelerated Motion: The horizontal motion is uniform, while the vertical motion exhibits constant acceleration due to gravity.
5. Parabolic Trajectory:
   • Curved Path: The combination of horizontal and vertical motions results in a curved trajectory.
   • Parabolic Shape: The vertical motion, being accelerated, forms a parabolic curve as the object rises to its peak and then descends.
6. Symmetry of the Trajectory:
   • Symmetrical Path: The trajectory is symmetrical, meaning the ascent and descent times to the same height are equal.
   • Characteristic Feature: This symmetry is a hallmark of a parabolic trajectory.

Summary

An object thrown at a certain angle with the horizontal undergoes projectile motion, resulting in a parabolic trajectory. This path is shaped by the constant horizontal velocity and the constant downward acceleration due to gravity. The parabolic trajectory, with its symmetrical rise and fall, is a crucial concept for understanding the motion of various objects in real-world scenarios, such as in sports or ballistics.

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SAQ-2 : State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector.

Introduction

The parallelogram law of addition of vectors is a crucial concept in physics and mathematics, used to find the resultant of two vectors. This law states that when two vectors are represented as adjacent sides of a parallelogram with a common starting point, the parallelogram’s diagonal, originating from this common point, represents the resultant vector. We’ll explore how to derive the magnitude and direction of this resultant vector using the parallelogram law.

1. Understanding the Parallelogram Law of Addition: The parallelogram law of vectors states that if two vectors are considered as adjacent sides of a parallelogram with their tails meeting at a common point, the parallelogram’s diagonal from this point represents the resultant vector.
2. Finding the Resultant Vector Magnitude:
   • To determine the magnitude of the resultant vector C, representing the total velocity in projectile motion, we apply the Pythagorean theorem and trigonometric functions.
   • The magnitude of C is given by:
     $$|\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta}$$
     where θ is the angle between vectors A and B.
3. Finding the Direction of the Resultant Vector:
   • The direction of the resultant vector C is defined by the angle ϕ it makes with one of the vectors (A or B).
   • The angle ϕ can be calculated using trigonometry:
     $$\phi = \tan^{-1}\left(\frac{|\vec{A}|\sin\theta}{|\vec{B}| + |\vec{A}|\cos\theta}\right)$$
4. Parabolic Trajectory (Correction): The application of the parallelogram law in projectile motion helps to calculate the velocity vector’s magnitude and direction. However, this law itself does not directly show that the trajectory of a projectile is a parabola. The parabolic trajectory in projectile motion is derived from the independent horizontal and vertical motions of the projectile.

Summary

The parallelogram law of vector addition is instrumental in determining the magnitude and direction of the resultant vector in various scenarios, such as projectile motion. By calculating the magnitude ∣C∣ and direction angle ϕ, we can understand the combined effect of two vectors. While this law helps in analyzing the velocity vector in projectile motion, it is the distinct characteristics of horizontal and vertical motions that lead to the parabolic trajectory, a key concept in the study of vectors and motion in physics.

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SAQ-3 : If | a +b | = | a – b |, prove that the angle between (a) and (b) is 90⁰.

Introduction

In this proof, we aim to demonstrate that if the magnitude of the vector sum (a+b) equals the magnitude of the vector difference (a−b), then the angle between vectors a and b is 90 degrees. This proof will employ principles of vector algebra and trigonometry.

Proof: $$|\vec{a} + \vec{b}| = |\vec{a} – \vec{b}|$$

1. Squaring Both Sides: $$|\vec{a} + \vec{b}|^2 = |\vec{a} – \vec{b}|^2$$
2. Using Vector Magnitude Property: $$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} – \vec{b}) \cdot (\vec{a} – \vec{b})$$
3. Expanding Dot Products: $$\vec{a} \cdot \vec{a} + 2(\vec{a} \cdot \vec{b}) + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} – 2(\vec{a} \cdot \vec{b}) + \vec{b} \cdot \vec{b}$$
4. Simplifying with Magnitudes: $$|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 – 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2$$
5. Rearranging the Equation: $$4(\vec{a} \cdot \vec{b}) = 0$$
6. Simplifying Further: $$\vec{a} \cdot \vec{b} = 0$$
7. Using Dot Product Property: $$|\vec{a}||\vec{b}|\cos\theta = 0$$ (where θ is the angle between vectors a and b)
8. Determining θ: Since ∣a∣ and ∣b∣ are non-zero, cosθ=0 for the product to be zero.
9. Identifying Angle: cosθ = 0 when θ=90 degrees.

Summary

This proof establishes that if ∣a+b∣=∣a−b∣, then the angle (θ) between vectors a and b is 90 degrees. This result is crucial in vector algebra and physics, revealing a significant relationship between the magnitudes of vector sums and differences and the angles between vectors.

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SAQ-4 : A force 2i+j-k newton acts on a body which is initially at rest. At the end of 20 seconds the velocity of the body is 4i+2j-2k ms(-1). What is the mass of the body ?

Introduction

In this problem, we will determine the mass of a body on which a specific force is acting. The body, initially at rest, achieves a certain velocity after 20 seconds under the influence of this force. The concepts of Newton’s Second Law of Motion and basic kinematics will be utilized in this calculation.

Calculation of Mass

1. Given Force: The force acting on the body is $$\vec{F} = 2\vec{i} + \vec{j} – \vec{k}$$ newtons.
2. Final Velocity: After 20 seconds, the velocity of the body is $$\vec{v} = 4\vec{i} + 2\vec{j} – 2\vec{k} \, \text{ms}^{-1}$$
3. Initial Velocity: The body is initially at rest, so the initial velocity $$\vec{u} = 0$$.
4. Time: The time taken for the change in velocity is 20 seconds.

Using Newton’s Second Law

1. Newton’s Second Law states that $$\vec{F} = m\vec{a}$$, where m is the mass and a is the acceleration.
2. Acceleration a can be found using the formula $$\vec{a} = \frac{\vec{v} – \vec{u}}{t}$$

Calculating Acceleration

• $$\vec{a} = \frac{(4\vec{i} + 2\vec{j} – 2\vec{k}) – 0}{20 \, \text{s}} = \frac{1}{5}\vec{i} + \frac{1}{10}\vec{j} – \frac{1}{10}\vec{k} \, \text{ms}^{-2}$$

Finding Mass

1. Rearranging Newton’s Second Law: $$m = \frac{\vec{F}}{\vec{a}}$$
2. Substituting the values: $$m = \frac{2\vec{i} + \vec{j} – \vec{k}}{\frac{1}{5}\vec{i} + \frac{1}{10}\vec{j} – \frac{1}{10}\vec{k}}$$

Simplifying the Expression

1. Calculating mass requires careful consideration of vector components.
2. Mass, being a scalar quantity, will be the ratio of the magnitudes of force and acceleration.

Summary

To find the mass of the body, we first calculate its acceleration due to the applied force and then apply Newton’s Second Law. The mass can be determined by dividing the magnitude of the force by the magnitude of the acceleration. This problem illustrates the application of basic principles of mechanics to determine unknown quantities such as mass in motion-related scenarios.