Kinematics Questions for NEET Previous Year Question Papers

Kinematics Questions for NEET: We are well aware that the topic of Kinematics is a fundamental part of the NEET Physics syllabus, covering concepts like motion in a straight line, motion in a plane, velocity, acceleration, equations of motion, projectile motion, and relative velocity. Mastering these concepts is crucial, as they build the foundation for solving advanced mechanics problems in NEET.

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This Story also Contains

1. Importance of Kinematics in NEET Syllabus
2. PYQ Analysis: Frequency of Questions in the Last 10 Years
3. Top Kinematics PYQs with Solutions
4. Kinematics Concept Summary: Key Formulas & Exam Tips
5. Final Tips for Mastering Kinematics in NEET

This article presents NEET PYQs Kinematics from the last 10 years, a chapter-wise breakdown of questions, difficulty level analysis, and high-yield questions with solutions. Additionally, we offer a topic-wise weightage analysis and strategic preparation tips for cracking the NEET 2025 exam.

Importance of Kinematics in NEET Syllabus

This unit contributes approximately 8-10% of the NEET Physics section, making it a highly scoring topic. Questions in this topic range from basic formula-based problems to conceptual and graphical interpretations, such as position-time, velocity-time, and acceleration-time graphs.

Many students struggle with understanding motion graphs, vector decomposition, and applying equations of motion in different scenarios. However, solving Previous Year Questions (PYQs) effectively familiarises students with exam patterns, enhances problem-solving skills, and improves accuracy.

The unit comprises the major concepts given below:

PYQ Analysis: Frequency of Questions in the Last 10 Years

To help students understand the importance of Kinematics in NEET, we have compiled PYQs from 2015 to 2024, categorising them by sub-topic, difficulty level, and type of question.

Year-wise Distribution of PYQs (2015-2024)

Key Insights from the PYQ Analysis

• Equations of Motion and Projectile Motion are the most frequently asked topics.

• 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.

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• How to Crack NEET 2025 in First Attempt

• How To Prepare For NEET At Home

• NEET Difficulty Level

Top Kinematics PYQs with Solutions

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: 0o

Option 2: 90o

Option 3: 45o

Option 4: 180o

Difficulty level: Easy

Answer:

Represents the law of parallelogram Vector Addition

$\begin{array}{rl}& |\overrightarrow{A}-\overrightarrow{B}|=|\overrightarrow{A}+\overrightarrow{B}|\\ & A^2+B^2-2AB\cos \theta =A^2+B^2+2AB\cos \theta \\ & 4AB\cos \theta =0\\ & \cos \theta =\frac{0}{4AB}=0={90}^{\circ }\end{array}$

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:

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:

Option 1: 360 m

Option 2: 340 m

Option 3: 320 m

Option 4: 300 m

Difficulty level: Medium

Answer:

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 $\mathrm{t}=\mathrm{n}$-1 to $\mathrm{t}=\mathrm{n}$.

Then, the ratio $\frac{S_n}{S_{n+1}}$ is:

Option 1: $\frac{2n-1}{2n}$

Option 2: $\frac{2n-1}{2n+1}$

Option 3: $\frac{2_n+1}{2n-1}$

Option 4: $\frac{2n}{2n-1}$

Difficulty level: Medium

Answer:

$\begin{array}{rl}& \text{ From }t=0\text{ to }t=n-1\\ & S_1=0+\frac{1}{2}a(n-1{)}^2\\ & \text{ From }t=0\text{ to }t=n\\ & S_2=0+\frac{1}{2}a(n{)}^2\\ & \text{ From }t=0\text{ to }t=n+1\\ & S_3=0+\frac{1}{2}a(n+1{)}^2\\ & S_n=S_2-S_1\\ & S_n=\frac{1}{2}a[n^2-(n^2-2n+1)]\\ & S_{n+1}=S_3-S_2=\frac{1}{2}a[n^2+2n+1-n^2]\\ & \frac{S_n}{S_n+1}=\frac{(2n-1)}{(2n+1)}\end{array}$

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 ${}^{\prime }{\theta }^{\prime }$ to the horizontal, the maximum height attained by it equals 4 R . The angle of projection,$\theta$, is then given by :

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

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

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

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

Difficulty level: Medium

Answer: $\begin{array}{rl}& T=\frac{2\pi R}{V};V=\frac{2\pi R}{T}\\ & H=4R=\frac{u^2{\sin }^2\theta }{2g}\\ & u=v\\ & H=\frac{V^2{\sin }^2\theta }{2g}\\ & {\sin }^2=\frac{2gh}{v^2}=\frac{2gH}{4{\pi }^2{R}^2\frac{T^2}{}}\end{array}$

$\begin{array}{rl}& \sin \theta ={\left(\frac{2gH{T}^2}{4{\pi }^2{R}^2}\right)}^{\frac{1}{2}}=\left(\frac{2g(4g)T^2}{4{\pi }^2{R}^2}\right)\frac{1}{2}\\ & \sin \theta =\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)\frac{1}{2}\end{array}$

Hence, the answer is option (4).

Ques 6: The ratio of the distances traveled by a freely falling body in the 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 nth second is given by Snth=u+a/2(2n−1)

⟹ Snth=(a/2)(2n−1) (∵u=0)

So, distance travelled in the first second, S1=(a/2)(2×1−1)=a/2

So, distance travelled in 2nd second, S2=(a/2)(2×2−1)=3a/2

So, distance travelled in 3rd second, S3=(a/2)(2×3−1)=5a/2

So, distance travelled in 4th second, S4=(a/2)(2×4−1)=7a/2

⟹ S1:S2:S3:S4=1:3:5:7

By Galileo's ratio, the distance travelled by a freely falling body is in the ratio 1: 3: 5: 7
(odd number)

Hence, the answer is option (2).

Ques 7: A ball is projected with a velocity $10{\mathrm{ms}}^{-1}$ at an angle of ${60}^{\circ }$ with the vertical direction. Its speed at the highest point of its trajectory will be:

Option 1: $5\sqrt{3}{\mathrm{ms}}^{-1}$

Option 2: $5{\mathrm{ms}}^{-1}$

Option 3: $10{\mathrm{ms}}^{-1}$

Option 4: Zero

Difficulty level: Medium

Answer:

At the topmost position, the speed of the projectile is $\mathrm{u}\cos \theta =10\times \cos {30}^{\circ }=5\sqrt{3}\frac{\mathrm{m}}{\mathrm{s}}$

Hence, the answer is option 1.

Kinematics Concept Summary: Key Formulas & Exam Tips

Important Formulas

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

• Equations of Motion: v=u+at, s=ut+12at2, v2=u2+2as, v = u + at,

• Projectile Motion: Time of flight=2usin⁡θg, Max height=u2sin⁡2θ2g, Range=u2sin⁡2θg

Graphical Analysis:

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

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

• Relative Velocity

Final Tips for Mastering Kinematics in NEET

• Understand motion graphs deeply, as they appear frequently in PYQs.

• Master vectors because this helps in relative motion & projectile problems.

• Avoid calculation errors & unit conversion mistakes as these cost valuable marks.

• Practice PYQs & mock tests as they help build accuracy & confidence.

Thus, with consistent practice and conceptual clarity, Kinematics can become one of your strongest topics in NEET Physics. Keep solving PYQs, focus on tricky concepts, and revise formulas regularly. Practice, Practice, Practice!

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