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Motion in a Plane — NEET & JEE MCQs with answers
Two- and three-dimensional motion is described using vectors — scalars have magnitude only, vectors have both magnitude and direction and obey the triangle/parallelogram law of addition. The chapter applies vector kinematics to derive equations for projectile motion and uniform circular motion, introducing centripetal acceleration.
Topics in this chapter
- 3.1 Introduction
- 3.2 Scalars and Vectors
- 3.3 Multiplication of Vectors by Real Numbers
- 3.4 Addition and Subtraction of Vectors — Graphical Method
- 3.5 Resolution of Vectors
- 3.6 Vector Addition — Analytical Method
- 3.7 Motion in a Plane
- 3.8 Motion in a Plane with Constant Acceleration
- 3.9 Projectile Motion
- 3.10 Uniform Circular Motion
Free practice questions
A projectile is launched at $30\ \text{m s}^{-1}$ at $90^\circ$ (straight up), $g = 10\ \text{m s}^{-2}$. The horizontal range is:
- A. $45\ \text{m}$
- B. $30\ \text{m}$
- C. $0\ \text{m}$
- D. $90\ \text{m}$
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Answer: C — $R = v_0^2\sin 2\theta_0/g$. With $\theta_0 = 90^\circ$, $\sin 180^\circ = 0$, so $R = 0$. A vertically thrown object has no horizontal displacement.
NCERT Class 11 Physics · Ch 3 · 3.9 Projectile Motion
For the analytical addition of two vectors, the y-component of the resultant $\vec{R} = \vec{A} + \vec{B}$ is:
- A. $R_y = A_x + B_x$
- B. $R_y = A_y - B_y$
- C. $R_y = A_y + B_y$
- D. $R_y = A_y B_y$
Show answer
Answer: C — Per Eq. (3.21), $R_y = A_y + B_y$.
NCERT Class 11 Physics · Ch 3 · 3.6 Vector Addition - Analytical Method
The subtraction of a vector $\vec{B}$ from a vector $\vec{A}$ is defined as
- A. $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$
- B. $\vec{A} - \vec{B} = \vec{A} + \vec{B}$
- C. $\vec{A} - \vec{B} = -\vec{A} + \vec{B}$
- D. $\vec{A} - \vec{B} = \vec{B} + (-\vec{A})$
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Answer: A — NCERT: 'We define the difference of two vectors A and B as the sum of two vectors A and -B : A - B = A + (-B).'
NCERT Class 11 Physics · Ch 3 · 3.4 Addition and Subtraction of Vectors - Graphical Method
According to NCERT, a vector quantity obeys
- A. the triangle law of addition or equivalently the parallelogram law of addition
- B. only the rules for combining scalars
- C. no law of addition at all
- D. only the rules of ordinary algebra
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Answer: A — NCERT: a vector quantity has both magnitude and direction and obeys the triangle law of addition or equivalently the parallelogram law of addition.
NCERT Class 11 Physics · Ch 3 · 3.2 Scalars and Vectors
A vector $\mathbf{A}$ of magnitude $10\ \text{units}$ is multiplied by a real number $\lambda$. If the resultant vector has magnitude $25\ \text{units}$ and points opposite to $\mathbf{A}$, then $\lambda$ equals:
- A. $+0.4$
- B. $+2.5$
- C. $-2.5$
- D. $-15$
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Answer: C — Opposite direction means $\lambda$ is negative. Magnitude $= |\lambda| \times |\mathbf{A}|$, so $|\lambda| = 25 / 10 = 2.5$. Since direction is reversed, $\lambda = -2.5$.
NCERT Class 11 Physics · Ch 3 · 3.3 Multiplication of Vectors by Real Numbers
A vector is written as $\vec{A} = 5\hat{i} + 12\hat{j}$. Its magnitude $A = \sqrt{A_x^2 + A_y^2}$ is:
- A. $17$ units
- B. $13$ units
- C. $169$ units
- D. $7$ units
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Answer: B — $A_x = 5$, $A_y = 12$, so $A = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ units (Eq. 3.14).
NCERT Class 11 Physics · Ch 3 · 3.5 Resolution of Vectors
A particle starts from origin with $\vec{v}_0 = (0\,\hat{i}+10\,\hat{j})\ \text{m/s}$ and constant acceleration $(2.0\,\hat{i}+0\,\hat{j})\ \text{m/s}^2$. What is its $x$-coordinate when $y = 50\ \text{m}$?
- A. $25\ \text{m}$
- B. $50\ \text{m}$
- C. $5.0\ \text{m}$
- D. $10\ \text{m}$
Show answer
Answer: A — $y = (10)t = 50 \Rightarrow t = 5.0\ \text{s}$ ($a_y=0$). Then $x = \tfrac{1}{2}(2.0)(5.0)^2 = 25\ \text{m}$ (independent x-motion).
NCERT Class 11 Physics · Ch 3 · 3.8 Motion in a Plane with Constant Acceleration
The position of a particle is $\mathbf{r} = (3.0t)\hat{i} + (2.0t^2)\hat{j}\,\text{m}$. What is the magnitude of its velocity at $t = 1.0\,\text{s}$?
- A. 5.0 m/s
- B. 4.0 m/s
- C. 3.0 m/s
- D. 7.0 m/s
Show answer
Answer: A — $v_x = d(3.0t)/dt = 3.0\,\text{m/s}$; $v_y = d(2.0t^2)/dt = 4.0t$. At $t = 1.0\,\text{s}$, $v_y = 4.0\,\text{m/s}$. Magnitude $= \sqrt{3.0^2 + 4.0^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0\,\text{m/s}$.
NCERT Class 11 Physics · Ch 3 · 3.7 Motion in a Plane
A particle moves uniformly in a circle of radius $R = 0.25\,\text{m}$ at $\omega = 6.28\,\text{rad s}^{-1}$. Its linear speed is
- A. $25.12\,\text{m s}^{-1}$
- B. $0.04\,\text{m s}^{-1}$
- C. $3.14\,\text{m s}^{-1}$
- D. $1.57\,\text{m s}^{-1}$
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Answer: D — Using $v = \omega R = 6.28 \times 0.25 = 1.57\,\text{m s}^{-1}$.
NCERT Class 11 Physics · Ch 3 · 3.10 Uniform Circular Motion
The title of the chapter introduced in section 3.1 is:
- A. Motion in a Plane
- B. Motion in a Straight Line
- C. Laws of Motion
- D. Kinematics in One Dimension
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Answer: A — The chapter is titled 'MOTION IN A PLANE', as given at the start of the chapter.
NCERT Class 11 Physics · Ch 3 · 3.1 Introduction
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