PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 11 · Chapter 3 · JEE Main

Motion in a Plane

Chapter-wise practice and revision for JEE Main: concise notes on vectors, projectiles, circular motion and relative velocity, concept figures, and 50 selected previous-year JEE questions (49 JEE Main or AIEEE, 1 IIT JEE) covering the major concepts and question patterns, each with a step-by-step solution.

50 selected questions5 topics7 concept figuresPreview · first 3 of 50 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 138 previous-year JEE questions in the PiTheory bank

Projectile58
Vectors38
Circular16
2D motion14
Relative10

Contents

  1. 1Vectors: addition, resolution and productsQ1–Q13
  2. 2Motion in a plane with constant accelerationQ14–Q19
  3. 3Projectile motionQ20–Q38
  4. 4Uniform circular motionQ39–Q44
  5. 5Relative velocity in two dimensionsQ45–Q50
  6. ✓Last-day checklist
01

Topic 1 · Q1–Q13

Vectors: addition, resolution and products

38 of the 138 JEE PYQs in the bank test this as their main idea

Look first

ABR = A + BθαR² = A² + B² + 2AB cos θ, tan α = B sin θ / (A + B cos θ)
Parallelogram law: the resultant is the diagonal from the common tail.
B cos θABθarea = |A × B| = AB sin θA·B = A (B cos θ)
Dot product uses the projection; cross product is the area.

Quick notes

$$R^2=A^2+B^2+2AB\cos\theta,\qquad \tan\alpha=\frac{B\sin\theta}{A+B\cos\theta}$$

Resultant of $\vec A$ and $\vec B$ at angle $\theta$; $\alpha$ is measured from $\vec A$. For $|\vec A-\vec B|$ replace $\cos\theta$ by $-\cos\theta$. Always $|A-B|\le R\le A+B$.

Given conditionConclusion
Equal magnitudes $A$$|\vec A+\vec B|=2A\cos\dfrac\theta2$, $|\vec A-\vec B|=2A\sin\dfrac\theta2$
$|\vec A+\vec B|=n|\vec A-\vec B|$, equal magnitudes$\cos\theta=\dfrac{n^2-1}{n^2+1}$
$|\vec A+\vec B|=|\vec A-\vec B|$$\theta=90^\circ$
Resultant perpendicular to $\vec A$$\cos\theta=-\dfrac AB$, and $R^2=B^2-A^2$
$\vec A\cdot\vec B=|\vec A\times\vec B|$$\theta=45^\circ$
$|\vec P+\vec Q|=|\vec P|$$2\vec P\cdot\vec Q=-Q^2$, so $(2\vec P+\vec Q)\perp\vec Q$
$$\vec A\cdot\vec B=AB\cos\theta=A_xB_x+A_yB_y+A_zB_z,\qquad |\vec A\times\vec B|=AB\sin\theta$$

Dot product zero means perpendicular. Component of $\vec A$ along $\vec B$: $\dfrac{\vec A\cdot\vec B}{B}$; as a vector, $(\vec A\cdot\hat B)\hat B$. Cross product is perpendicular to both vectors; its magnitude is the area of the parallelogram they form (half for the triangle).

$$\vec A\times\vec B=\begin{vmatrix}\hat i&\hat j&\hat k\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix},\qquad \vec A\cdot(\vec B\times\vec C)=\begin{vmatrix}A_x&A_y&A_z\\B_x&B_y&B_z\\C_x&C_y&C_z\end{vmatrix}$$

$\hat i\times\hat j=\hat k$, $\hat j\times\hat k=\hat i$, $\hat k\times\hat i=\hat j$; $\vec A\times\vec A=\vec0$; $\vec B\times\vec A=-\vec A\times\vec B$. The triple product is the volume of the box on the three vectors; zero means the vectors are coplanar.

  • Unit vector $\hat A=\dfrac{\vec A}{A}$; direction cosines satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$. Parallel (collinear) vectors have proportional components.
  • Vectors forming a closed polygon (head to tail) add to zero. Vectors from the centre of a regular polygon to its vertices add to zero, so vectors from one vertex to all the others add to $n\overrightarrow{AO}$.
  • Torque $\vec\tau=\vec r\times\vec F$ is perpendicular to both $\vec r$ and $\vec F$: $\vec r\cdot\vec\tau=\vec F\cdot\vec\tau=0$.
  • Expand dot products like algebra: $|\vec A+\vec B|^2=A^2+B^2+2\vec A\cdot\vec B$.
Traps. $\vec A\times\vec A$ is the zero vector, not a number. Angle between vectors is taken tail to tail. The $\hat j$ term in a determinant expansion has a minus sign.
More standard resultsResult
Sum of magnitudes $S$, resultant perpendicular to the smaller$F_2^2-F_1^2=R^2$
Angle between body diagonals of a cube$\cos^{-1}\tfrac13$
Vector of magnitude $|\vec A|$ along $\vec B$$|\vec A|\,\hat B$
Area of a triangle with sides $\vec A$, $\vec B$$\tfrac12|\vec A\times\vec B|$

Practice: first 3 of 13 questions

Q1JEE Main 2024 (31 Jan, Shift 2)Easy

If two vectors $\vec A$ and $\vec B$ having equal magnitude $R$ are inclined at an angle $\theta$, then

  1. A$|\vec A - \vec B| = \sqrt2R\sin\dfrac\theta2$
  2. B$|\vec A + \vec B| = 2R\sin\dfrac\theta2$
  3. C$|\vec A + \vec B| = 2R\cos\dfrac\theta2$
  4. D$|\vec A - \vec B| = 2R\cos\dfrac\theta2$
Answer and solution

Answer: Option C: $|\vec A + \vec B| = 2R\cos\dfrac\theta2$

Concept. The magnitude of a sum or difference of two vectors comes from the law of cosines: $|\vec A\pm\vec B|^2=A^2+B^2\pm2AB\cos\theta$. With equal magnitudes the half-angle identities turn the square roots into neat sines and cosines.

Step 1. Magnitude of the sum. Put $A=B=R$: $$|\vec A+\vec B|^2=R^2+R^2+2R^2\cos\theta=2R^2(1+\cos\theta)$$

Step 2. Use the half-angle identity. Since $1+\cos\theta=2\cos^2\dfrac\theta2$, $$|\vec A+\vec B|^2=4R^2\cos^2\frac\theta2\ \Rightarrow\ |\vec A+\vec B|=2R\cos\frac\theta2$$

Step 3. Magnitude of the difference (to test the other options). Here the cosine term changes sign and $1-\cos\theta=2\sin^2\dfrac\theta2$: $$|\vec A-\vec B|^2=2R^2(1-\cos\theta)=4R^2\sin^2\frac\theta2\ \Rightarrow\ |\vec A-\vec B|=2R\sin\frac\theta2$$ So options A, B and D are all wrong.

Answer. Option C. Quick check: at $\theta=0$ the sum is $2R$ and the difference is $0$, which both formulas give.

Point to note. Equal magnitudes: sum $2R\cos\frac\theta2$, difference $2R\sin\frac\theta2$.

Q2JEE Main 2019 (10 Jan, Shift 2)Easy

Two forces $P$ and $Q$, of magnitude $2F$ and $3F$, respectively, are at an angle $\theta$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Then, the angle $\theta$ is:

  1. A$120^\circ$
  2. B$60^\circ$
  3. C$30^\circ$
  4. D$90^\circ$
Answer and solution

Answer: Option A: $120^\circ$

Concept. The resultant of two forces at angle $\theta$ is given by $R^2=P^2+Q^2+2PQ\cos\theta$. We can write this law twice, once before and once after $Q$ is doubled, and the unknown resultant $R$ then cancels out, leaving only $\theta$.

Step 1. Before doubling. With $P=2F$, $Q=3F$: $$R^2=4F^2+9F^2+12F^2\cos\theta=13F^2+12F^2\cos\theta$$

Step 2. After doubling $Q$. Now $Q=6F$ and the resultant is $2R$: $$4R^2=4F^2+36F^2+24F^2\cos\theta=40F^2+24F^2\cos\theta$$

Step 3. Eliminate $R$. Multiply the first equation by 4 and set it equal to the second: $$52F^2+48F^2\cos\theta=40F^2+24F^2\cos\theta\ \Rightarrow\ 24\cos\theta=-12\ \Rightarrow\ \cos\theta=-\frac12$$

Answer. $\theta=120^\circ$ (Option A). Check: at $120^\circ$, $R^2=13-6=7$ and $(2R)^2=40-12=28=4\times7$.

Point to note. Write $R^2$ twice (before and after doubling) and eliminate $R$.

Q3JEE Main 2024 (4 Apr, Shift 1)Medium

Two forces $\vec F_1$ and $\vec F_2$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $\vec F_1$ and $\vec F_2$ is $\cos^{-1}\left(\dfrac1n\right)$. The value of $|n|$ is ______.

Numerical answer type

Answer and solution

Answer: 6

Concept. Use the resultant formula $R^2=F_1^2+F_2^2+2F_1F_2\cos\theta$. The condition "resultant equals the larger force" makes the larger force's square cancel from both sides.

Step 1. Name the forces. Let the smaller force be $F$ and the larger $3F$. The resultant is $3F$.

Step 2. Apply the resultant formula. $$(3F)^2=F^2+(3F)^2+2(F)(3F)\cos\theta$$

Step 3. Simplify. The $9F^2$ terms cancel: $$0=F^2+6F^2\cos\theta\ \Rightarrow\ \cos\theta=-\frac16$$

Step 4. Compare with the given form. $\cos\theta=\dfrac1n$ gives $n=-6$.

Answer. $|n|=6$. The angle is obtuse: a resultant can only equal the larger force if the smaller force partly opposes it.

Point to note. Resultant equal to the larger force: $A^2+2AB\cos\theta=0$.

47 more questions and 4 more topics in the full chapter

  1. 2Motion in a plane with constant acceleration6 questions with solutions
  2. 3Projectile motion19 questions with solutions
  3. 4Uniform circular motion6 questions with solutions
  4. 5Relative velocity in two dimensions6 questions with solutions
  5. ✓Last-day checklist

The complete Motion in a Plane chapter is in the Class 11 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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