Kinematics Questions for NEET Previous Year Question Papers

Kinematics NEET Questions 2027 – Chapter Importance and Exam Weightage

The topics of kinematics in the NEET Physics paper test more than just the application of formulae. Some possible topics include interpretation of the motion graph, relation between velocity and acceleration, projectile motion, and equations of motion in certain scenarios. Based on the pattern of previous years, some of the topics which have been tested and which students need to practice well for NEET 2027 include equations of motion, speed and velocity, projectile motion, uniform circular motion, and Kinematics graphs.

Motion graphs and vector-related problems are difficult if the student is not aware of what exactly the slope or the area in the graph means and how to resolve a vector. Practising Kinematics NEET PYQs will therefore help in knowing how concepts are used in the NEET Physics paper. Analysis of recent exams will thus be helpful in identifying common concepts and formats of questions.

The unit comprises the major concepts given below:

NEET Kinematics PYQ Analysis: Frequency of Questions

To help students understand the importance of Kinematics in NEET, we have compiled PYQs from 2020-2026, categorising them by sub-topic, difficulty level, and type of question.

Year-wise Distribution of PYQs (2020-2026)

Key Insights from the NEET Kinematics PYQ Analysis

• Equations of Motion and Projectile Motion are the most frequently asked concepts in NEET Physics.
• The difficulty level is moderate to high, with most questions being of medium difficulty.
• Conceptual and graphical questions dominate the unit.
• Relative Velocity and Vector Analysis are recurring topics in NEET PYQs.

High‑Yield Kinematics PYQs with Solutions for NEET 2027 Preparation

Here are some of the most important PYQs from Kinematics, with detailed solutions.

Ques 1: If the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors, the angle between these vectors is:

Option 1: 0^o

Option 2: 90o

Option 3: 45^o

Option 4: 180^o

Difficulty level: Easy

Answer:

Represents the law of parallelogram vector addition:

$|\vec{A} - \vec{B}| = |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 - 2AB\cos\theta}$

Now, using the given values:

$A^2 + B^2 - 2AB\cos\theta = A^2 + B^2 + 2AB\cos(90^\circ)$

Since $\cos(90^\circ) = 0$, we get:

$2AB\cos(90^\circ) = 0$

Therefore,

$|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2}$

Hence, the vectors are perpendicular:

$\angle AB = 90^\circ$

Hence, the answer is option (2).

Ques 2: A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 x 10^-4 J by the end of the second revolution after the beginning of the motion?

Option 1: 0.1 m/s²

Option 2: 0.15 m/s²

Option 3: 0.18 m/s²

Option 4: 0.2 m/s²

Difficulty level: Medium

Answer:

Given:

Mass of particle, $m = 10\,\text{g} = 10 \times 10^{-3}\,\text{kg}$

Radius of circle, $r = 6.4\,\text{cm} = 6.4 \times 10^{-2}\,\text{m}$

Kinetic energy after 2 revolutions, $KE = 8 \times 10^{-4}\,\text{J}$

Let the constant tangential acceleration be $a_t$.

Step 1: Total distance travelled in 2 revolutions

$s = 2 \times 2\pi r = 4\pi r$

$s = 4 \times 3.14 \times 6.4 \times 10^{-2} = 8.0384\,\text{m}$

Step 2: Using the equation of motion

$v^2 = 2a_t s$

Using

$\frac{1}{2}mv^2 = \frac{1}{2}m(2a_t s) = KE$

Therefore,

$KE = ma_t s$

$\Rightarrow a_t = \frac{KE}{ms}$

$a_t = \frac{8 \times 10^{-4}}{(10 \times 10^{-3}) \times 8.0384}$

$= \frac{8 \times 10^{-4}}{8.0384 \times 10^{-2}}$

$\approx 0.0995 \approx 0.1\,\text{m/s}^2$

Final Answer: Option 1) $0.1\,\text{m/s}^2$

Alternate verification

Given,

$KE = \frac{1}{2}mV^2 = 8 \times 10^{-4}\,\text{J}$

Then,

$V^2 = \frac{2 \times KE}{m}$

$= \frac{2 \times 8 \times 10^{-4}}{10 \times 10^{-3}} = 0.16$

Now, using

$V^2 = 2a_t s$

$\Rightarrow a_t = \frac{V^2}{2s}$

$= \frac{0.16}{2 \times 8.0384}$

$\approx 0.1\,\text{m/s}^2$

Hence, the answer is option 1.

Ques 3: A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is: (g = 10 m/s)

Option 1: 360 m

Option 2: 340 m

Option 3: 320 m

Option 4: 300 m

Difficulty level: Medium

Answer:

Given:

Initial velocity, $u = 20\,\text{m/s}$ (downward)

Final velocity, $v = 80\,\text{m/s}$

Acceleration due to gravity, $g = 10\,\text{m/s}^2$

Let the height of the tower be $h$.

Using the kinematic equation

$v^2 = u^2 + 2gh$

Substitute the values

$(80)^2 = (20)^2 + 2 \cdot 10 \cdot h$

$\Rightarrow 6400 = 400 + 20h$

$\Rightarrow 20h = 6000$

$\Rightarrow h = \frac{6000}{20} = 300\,\text{m}$

Final Answer: Option 4) $300\,\text{m}$

Alternate method

$v^2 - u^2 = 2as$

$(80)^2 - (20)^2 = 2 \times 10 \times h$

$h = \frac{6400 - 400}{20}$

$= \frac{6000}{20} = 300\,\text{m}$

Hence, the answer is option 4.

Ques 4: A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval t=n-1 to t=n.

Then, the ratio $S_n/S_{n+1}$ is

A) $\frac{2n-1}{2n}$

B) $\frac{2n-1}{2n+1}$

C) $\frac{2n+1}{2n-1}$

D) $\frac{2n}{2n-1}$

Difficulty level: Medium

Answer:

From $t = 0$ to $t = n - 1$,

$S_1 = 0 + \frac{1}{2}a(n - 1)^2$

From $t = 0$ to $t = n$,

$S_2 = 0 + \frac{1}{2}an^2$

From $t = 0$ to $t = n + 1$,

$S_3 = 0 + \frac{1}{2}a(n + 1)^2$

Therefore,

$S_n = S_2 - S_1$

$= \frac{1}{2}a\left[n^2 - (n^2 - 2n + 1)\right]$

$= \frac{1}{2}a(2n - 1)$

Similarly,

$S_{n+1} = S_3 - S_2$

$= \frac{1}{2}a\left[n^2 + 2n + 1 - n^2\right]$

$= \frac{1}{2}a(2n + 1)$

Hence,

$\frac{S_n}{S_{n+1}} = \frac{\frac{1}{2}a(2n - 1)}{\frac{1}{2}a(2n + 1)}$

$\therefore \frac{S_n}{S_{n+1}} = \frac{2n - 1}{2n + 1}$

Hence, the answer is option (2).

Ques 5: A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution.

If this particle were projected with the same speed at an angle ′θ′ to the horizontal, the maximum height attained by it equals 4 R . The angle of projection,θ, is then given by :

Option 1: $\theta = \cos^{-1}\left(\frac{gT^2\pi}{2R}\right)$

Option 2: $\theta = \cos^{-1}\left(\frac{\pi^2 R g T^2}{2}\right)$

Option 3: $\theta = \sin^{-1}\left(\pi^2 R g T^2\right)$

Option 4: $\theta = \sin^{-1}\left(\frac{2gT^2\pi}{2R}\right)$

Difficulty level: Medium

Given:

$T = \frac{2\pi R}{V}$

and

$V = \frac{2\pi R}{T}$

Range:

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

$\Rightarrow u^2 = \frac{Rg}{\sin 2\theta}$

Height:

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

Substituting $V = \frac{2\pi R}{T}$,

$H = \frac{\left(\frac{2\pi R}{T}\right)^2\sin^2\theta}{2g}$

Therefore,

$\sin^2\theta = \frac{2gHT^2}{4\pi^2R^2}$

$\Rightarrow \sin\theta = \sqrt{\frac{2gHT^2}{4\pi^2R^2}}$

Hence,

$\theta = \sin^{-1}\left(\sqrt{\frac{2gHT^2}{4\pi^2R^2}}\right)$

Ques 6: The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second :

Option 1: 1: 4: 9: 16

Option 2: 1: 3: 5: 7

Option 3: 1: 1: 1: 1

Option 4: 1: 2: 3: 4

Difficulty level: Medium

Answer:

The initial speed of the body is zero, i.e. $u = 0$.

Distance travelled in the $n^\text{th}$ second is given by

$S_n = u + \frac{a}{2}(2n - 1)$

Since $u = 0$,

$S_n = \frac{a}{2}(2n - 1)$

Distance travelled in the 1st second

$S_1 = \frac{a}{2}(2 \times 1 - 1) = \frac{a}{2}$

Distance travelled in the 2nd second

$S_2 = \frac{a}{2}(2 \times 2 - 1) = \frac{3a}{2}$

Distance travelled in the 3rd second

$S_3 = \frac{a}{2}(2 \times 3 - 1) = \frac{5a}{2}$

Distance travelled in the 4th second

$S_4 = \frac{a}{2}(2 \times 4 - 1) = \frac{7a}{2}$

Therefore,

$S_1 : S_2 : S_3 : S_4 = 1 : 3 : 5 : 7$

Hence, according to Galileo's law, the distances travelled by a freely falling body in successive seconds are in the ratio of odd numbers

$1 : 3 : 5 : 7$

Hence, the answer is option (2).

Ques 7: A ball is projected with a velocity of $10\,\text{m/s}$ at an angle of $60^\circ$ with the vertical direction. Its speed at the highest point of its trajectory will be

Option 1: $5\sqrt{3}\,\text{m/s}$

Option 2: $5\,\text{m/s}$

Option 3: $10\,\text{m/s}$

Option 4: Zero

Difficulty level: Medium

Answer:

The ball is projected at an angle of $60^\circ$ with the vertical. Therefore, its angle with the horizontal is

$\theta = 90^\circ - 60^\circ = 30^\circ$

At the highest point of the trajectory, the vertical component of velocity becomes zero, while the horizontal component remains unchanged.

Therefore, the speed at the highest point is

$v = u\cos\theta$

$= 10\cos30^\circ$

$= 10 \times \frac{\sqrt{3}}{2}$

$= 5\sqrt{3}\,\text{m/s}$

Hence, the correct answer is Option 1: $5\sqrt{3}\,\text{m/s}$.

NEET 2027 Last-Minute Kinematics Concept Revision: Key Formulas & Exam Tips

Here are some important Physics Formulas which serve as a base for Kinematics:

• Equations of Motion:

  $v = u + at$

  $s = ut + \frac{1}{2}at^2$

  $v^2 = u^2 + 2as$

• Projectile Motion:

  Time of flight: $T = \frac{2u\sin\theta}{g}$

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

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

Graphical Analysis:

• The slope of a position-time graph gives velocity.

• The slope of a velocity-time graph is acceleration.

• Relative Velocity

Quick Tips to Solve Kinematics Questions for NEET 2027
Motion graph analysis

Practice position-time, velocity-time, and acceleration-time graphs.
Learn vectors
Revise vector components and relative velocity for projectiles and motions.
Be careful with calculations
Verify units, signs, and directions before finalising your answer.
Solve PYQs
Solve recent PYQs of Kinematics in NEET, including NEET 2026.
Take mock tests
Take NEET mock tests to increase your accuracy.
Maintain notes
Keep a small list of formulas and mistakes for revision.

In NEET 2027, develop conceptual understanding and PYQ/mock test-solving skills for accurate and fast solutions to Kinematics problems.