NEET Physics Lecture Notes: Vector Operations and Applications

Course Overview and Target Goals

• Course Details:

  • Program: YAKEEN NEET 2.0 2027

  • Subject: Physics

  • Lecture Number: 05

  • Topic: Vector

  • Faculty: Manish Raj (MR Sir)

• Academic Performance Targets (NEET Aspirants):

  • Biology: $330+$ to $360$ marks.

  • Chemistry: $130$ to $170$ marks.

  • Physics: $130$ to $170$ marks.

  • Goal for Top 100 Rank: Total score target of $690+$ marks.

  • Goal for Top 500 Rank: Total score target of $670+$ marks.

• Study Requirements:

  1. Regular attendance of classes.

  2. Completion of all Daily Practice Problems (DPP).

  3. Thorough revision of notes.

  4. Solving advanced problems (JEE Advanced/HC Verma) for high-level preparation.

Today's Learning Goals

• Applications of Vectors in Physics.

• Introduction and practice with Dot Product and Cross Product of vectors.

Vector Properties and Unit Vector Calculations

• Finding Unknowns in a Unit Vector:

  • Example Problem: If \vec{A} = 0.4hat{i} + 0.3hat{j} + \betahat{k} is a unit vector, find the value of $\beta$.

  • Concept: The magnitude of a unit vector is always $1$. Therefore, |vec{A}| = 1.

  • Calculation Steps:

    1. Formula for magnitude: |vec{A}| = sqrt{(0.4)^2 + (0.3)^2 + \beta^2} = 1

    2. Square both sides: $0.16 + 0.09 + \beta^2 = 1$

    3. Combine constants: $0.25 + \beta^2 = 1$

    4. Isolate unknown: $\beta^2 = 1 - 0.25 = 0.75$

    5. Final Result: \beta = sqrt{0.75}

• Constructing a Vector along another Vector:

  • Problem: Find vector $\vec{C}$ such that its magnitude is equal to $\vec{A}$ and its direction is along $\vec{B}$, where \vec{A} = 2hat{i} + hat{j} + 2hat{k} and \vec{B} = hat{i} + hat{j} + 2hat{k}.

  • Analysis:

    • The vector $\vec{C}$ is defined as |vec{C}|hat{C}.

    • Magnitude: |vec{C}| = |vec{A}| = sqrt{2^2 + 1^2 + 2^2} = sqrt{9} = 3.

    • Direction: \hat{C} = hat{B} = frac{vec{B}}{|vec{B}|} = frac{hat{i} + hat{j} + 2hat{k}}{sqrt{1^2 + 1^2 + 2^2}} = frac{hat{i} + hat{j} + 2hat{k}}{sqrt{6}}.

  • Result: \vec{C} = 3 times \left( frac{hat{i} + hat{j} + 2hat{k}}{sqrt{6}} right).

Vector Mechanics: Force and Acceleration

• Deriving Force and Acceleration Vectors:

  • Scenario: A force of magnitude 20,N acts along the vector \vec{A} = 2hat{i} + 3hat{j} - sqrt{3}hat{k}. Find the force vector and the acceleration of a 5,kg object.

  • Force Vector Calculation:

    • hat{A} = frac{2hat{i} + 3hat{j} - sqrt{3}hat{k}}{sqrt{2^2 + 3^2 + (-sqrt{3})^2}} = frac{2hat{i} + 3hat{j} - sqrt{3}hat{k}}{sqrt{4 + 9 + 3}} = frac{2hat{i} + 3hat{j} - sqrt{3}hat{k}}{4}.

    • vec{F} = |vec{F}|hat{A} = 20 times \left( frac{2hat{i} + 3hat{j} - sqrt{3}hat{k}}{4} right) = 5(2hat{i} + 3hat{j} - sqrt{3}hat{k}).

  • Acceleration Calculation:

    • Using Newton's Second Law: \vec{a} = frac{vec{F}}{m} = frac{5(2hat{i} + 3hat{j} - sqrt{3}hat{k})}{5} = 2hat{i} + 3hat{j} - sqrt{3}hat{k},m/s^2.

Resultant Force and Equilibrium

• Total Force on Point Objects:

  • Problem: Forces 2hat{i} + 2hat{j}, -2hat{i} - 2hat{j}, and -4hat{i} act on an object at rest. Determine the direction of motion.

  • Analysis: Sum the vectors: \vec{F}_{net} = (2 - 2 - 4)hat{i} + (2 - 2)hat{j} = -4hat{i}.

  • Conclusion: Since the net force is in the negative x direction, the object moves along the negative x-axis.

• Minimum Additional Force for Vertical Resultant:

  • Scenario: Three forces (including components like 4,N and 2,N at angles) act on a body. The goal is to have the resultant force directed only along the y-axis.

  • Requirement: To have the resultant only along y-axis, the net force in the x-direction must be zero ($F_{net, x} = 0$).

  • Calculation: If there's an unbalanced component \vec{F}_x = 2 cos(60^{circ}) = 1,N, an additional force of 1,N in the opposite direction is required.

Vectors in Electromagnetism and Gravity

• Electrostatic Force (Equilateral Triangle):

  • For three charges $+q$ at the vertices of an equilateral triangle with side length $a$, the force on any one charge due to the other two is calculated using the superposition of two force vectors, each of magnitude F = frac{kq^2}{a^2}, separated by an angle of 60^{circ}.

  • F_{net} = sqrt{F^2 + F^2 + 2F^2 cos(60^{circ})} = sqrt{3}F = frac{sqrt{3}kq^2}{a^2}.

• Force with Opposing Charges:

  • If one charge is $-q$ and the other is $+q$ (forming a 120^{circ} angle of attraction/repulsion components at the third vertex), the magnitude of the net force is:

  • F_{net} = sqrt{F^2 + F^2 + 2F^2 cos(120^{circ})} = F = frac{kq^2}{a^2}.

• Gravitational Superposition:

  • Principle of Null Point: For a symmetric distribution (triangle, square, etc.) of identical masses $m$, the net gravitational force on a mass $M$ placed exactly at the centroid is zero ($F_{net} = 0$).

  • Proof: The vectors cancel out due to symmetry (angles of 120^{circ} in a triangle cancel when summed).

Fundamental Concepts and Logic Checks

• True/False Statements:

  • (A) Resultant of two vectors must be in the plane of the two vectors: True.

  • (B) Minimum number of unequal vectors to give zero resultant is 2: False (must be at least 3 for unequal vectors).

  • (C) Minimum number of vectors (equal) to give zero resultant is 2: True (provided they are opposite).

  • (D) Minimum number of unequal vectors to give zero resultant is 3: True (if they form a triangle).

  • (E) Minimum number of non-coplanar vectors to give zero resultant is 4: True.

Kinematics: Change in Velocity (Delta vec{v},)

• The Difference between Velocity and Speed:

  • Scalar (Speed) changes only by changing magnitude.

  • Vector (Velocity) changes if the magnitude changes, the direction changes, or both change.

• Specific Cases of Velocity Change (Initial speed $u$):

  1. Rebound (Elastic Collision): A ball hits a wall and rebounds with the same speed.

     • vec{v}_i = -uhat{i}

     • vec{v}_f = +uhat{i}

     • Delta vec{v} = vec{v}_f - vec{v}_i = 2uhat{i}; Magnitude |Delta vec{v}| = 2u.

  2. Stopping: Ball hits and stops.

     • Delta vec{v} = 0 - (-uhat{i}) = +uhat{i}; Magnitude $u$.

  3. Perpendicular Deflection: Ball deflects at 90^{circ} with same speed.

     • vec{v}_i = -uhat{i}, vec{v}_f = uhat{j}.

     • Delta vec{v} = uhat{j} + uhat{i}.

     • |Delta vec{v}| = sqrt{u^2 + u^2} = sqrt{2}u.

  4. Circular Path: Object at diametrically opposite points.

     • vec{v}_i = vhat{j}, vec{v}_f = -vhat{j}.

     • Change in speed: $|v| - |v| = 0$.

     • Change in velocity: -vhat{j} - vhat{j} = -2vhat{j}; Magnitude $2v$.

Position and Displacement Vectors

• Position Vector (vec{r}): Reference from the origin $(0,0,0)$.

  • Relative position of Pinky (P) w.r.t Ramlal (R): \vec{r}_{PR} = vec{r}_P - vec{r}_R.

• Calculating Position Differences:

  • Find the position of an object at $(4, 3, 9)$ relative to a point $(1, 7, 4)$.

  • vec{r}_{rel} = (4-1)hat{i} + (3-7)hat{j} + (9-4)hat{k} = 3hat{i} - 4hat{j} + 5hat{k}.

  • Magnitude |vec{r}_{rel}| = sqrt{3^2 + (-4)^2 + 5^2} = sqrt{9+16+25} = sqrt{50} = 5sqrt{2},m.

• Displacement (vec{S}):

  • Defined as the change in position: \vec{S} = vec{r}_{final} - vec{r}_{initial}.

  • Example: Student moves 50,m East, then 20,m North, then 40sqrt{2},m South-East.

    • vec{d}_1 = 50hat{i}

    • vec{d}_2 = 20hat{j}

    • vec{d}_3 = 40sqrt{2} times ( cos(45^{circ})hat{i} - sin(45^{circ})hat{j} ) = 40hat{i} - 40hat{j}.

    • vec{S}_{net} = (50+40)hat{i} + (20-40)hat{j} = 90hat{i} - 20hat{j}.

• Distance vs Displacement:

  • Distance is the sum of magnitudes of individual path segments: $|d_1| + |d_2| + |d_3|$.

  • Displacement is the vector sum: \vec{d}_1 + vec{d}_2 + vec{d}_3.