Kinematics Previous Year Questions — JEE Main Physics
38 past-year Kinematics questions from JEE Main (Physics) — page 5.
Q51 (2024)
A particle starts from the origin at $t = 0$ with a velocity $5\hat{i}\text{ m/s}$ and moves in the x-y plane under the action of a force which produces a constant acceleration of $(3\hat{i} + 2\hat{j})\text{ m/s}^{2}$.
If the x-coordinate of the particle at that instant is $84\text{ m}$, then the speed of the particle at this time is $\sqrt{a}\text{ m/s}$. The value of $a$ is ______.
### Related Formula
$$v_x^2 - u_x^2 = 2 a_x x$$
$$v_y = u_y + a_y t$$
### Core Logic
Analyze the motion along the x-axis first to find the final x-velocity component ($u_x = 5\text{ m/s}$, $a_x = 3\text{ m/s}^2$, $x = 84\text{ m}$):
$$v_x^2 - 5^2 = 2(3)(84)$$
$$v_x^2 - 25 = 504 \implies v_x^2 = 529 \implies v_x = 23\text{ m/s}$$
### Step 1: Compute time interval
Using the velocity relation along the x-axis:
$$v_x = u_x + a_x t \implies 23 = 5 + 3t \implies 3t = 18 \implies t = 6\text{ s}$$
### Step 2: Evaluate y-axis velocity component
Since $u_y = 0$ and $a_y = 2\text{ m/s}^2$, calculate $v_y$ at $t = 6\text{ s}$:
$$v_y = 0 + 2(6) = 12\text{ m/s}$$
### Step 3: Total net speed calculation
$$v^2 = v_x^2 + v_y^2 = 23^2 + 12^2 = 529 + 144 = 673$$
$$v = \sqrt{673}\text{ m/s}$$
Comparing this with $\sqrt{a}$ yields $a = 673$.
### Pattern Recognition
Splitting coordinates explicitly isolates vector calculation lines cleanly, optimizing equation selections relative to independent components.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
Q1642 (2024)
A ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \mathrm{~m}$ high and $0.1 \mathrm{~m}$ wide. The minimum velocity $u$ with which that ball just hits the step 5 of the stairway will be $\sqrt{x} \mathrm{~ms}^{-1}$ where $x =$ ________ [use $g = 10 \mathrm{~m/s}^2$].
### Related Formula
For a projectile fired horizontally from height $h$ with speed $u$:
$$\text{Horizontal Range } R = u \cdot t$$
$$\text{Vertical Displacement } h = \frac{1}{2} g t^2$$
### Core Logic
To just clear step 4 and land on the tread of step 5, the flight trajectory must pass beyond the outer corner boundary vertex of the $4^{\text{th}}$ step.
Therefore, net horizontal distance required to cross 4 complete steps is:
$$R = 4 \times 0.1 \mathrm{~m} = 0.4 \mathrm{~m}$$
Similarly, net vertical fall matching the top of the $4^{\text{th}}$ step level is:
$$h = 4 \times 0.1 \mathrm{~m} = 0.4 \mathrm{~m}$$
{{SOL_IMG1}}
### Step 1: Determine Time of Flight
Using the vertical kinematic equation:
$$0.4 = \frac{1}{2} \times 10 \times t^2 \implies 0.4 = 5 t^2$$
$$t^2 = \frac{0.4}{5} = 0.08 \mathrm{~s^2}$$
### Step 2: Determine Velocity
Using the horizontal path equation:
$$R = u \cdot t \implies R^2 = u^2 \cdot t^2$$
Substituting range and time values:
$$(0.4)^2 = u^2 \times 0.08$$
$$0.16 = u^2 \times 0.08$$
$$u^2 = \frac{0.16}{0.08} = 2 \implies u = \sqrt{2} \mathrm{~m/s}$$
### Step 3: Extract x
Matching the parameter form $u = \sqrt{x}$, we find:
$x = 2$
### Pattern Recognition
To clear the $n$-th step, the projectile must safely pass the outer point of step $(n-1)$. Treat the geometric coordinates of corners as bounding constraints ($x = (n-1)w$, $y = (n-1)h$) to configure kinematics instantaneously.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Plane
Q36 (2024)
A body starts moving from rest with constant acceleration covers displacement $S_{1}$ in first $(p - 1)$ seconds and $S_{2}$ in first p seconds. The displacement $S_{1} + S_{2}$ will be made in time:
- $(2 p + 1) \mathrm{~s}$
- $\sqrt{(2p^2 - 2p + 1)} \mathrm{~s}$
- $(2p - 1) \mathrm{~s}$
- $(2p^2 - 2p + 1) \mathrm{~s}$
### Related Formula
For a body starting from rest ($u = 0$) with constant acceleration $a$, displacement $S$ in time $t$ is:
$$S = \frac{1}{2} a t^2$$
### Core Logic
Let the constant acceleration be $a$.
* Displacement $S_1$ covered in the first $(p - 1)$ seconds:
$$S_1 = \frac{1}{2} a (p - 1)^2$$
* Displacement $S_2$ covered in the first $p$ seconds:
$$S_2 = \frac{1}{2} a p^2$$
### Step 1: Express Net Displacement
Let the time taken to achieve a displacement of $S_1 + S_2$ be $t$.
$$S_1 + S_2 = \frac{1}{2} a t^2$$
Substituting the expressions of $S_1$ and $S_2$:
$$\frac{1}{2} a (p - 1)^2 + \frac{1}{2} a p^2 = \frac{1}{2} a t^2$$
### Step 2: Solve for Time
Dividing the entire equation by $\frac{1}{2}a$:
$$(p - 1)^2 + p^2 = t^2$$
$$p^2 - 2p + 1 + p^2 = t^2$$
$$t^2 = 2p^2 - 2p + 1$$
$$t = \sqrt{2p^2 - 2p + 1} \mathrm{~s}$$
### Pattern Recognition
When dealing with equations of motion from rest, notice that displacement scales quadratically with time ($S \propto t^2$). This means if displacement sums up ($S_t = S_1 + S_2$), the corresponding times will add in quadrature: $t = \sqrt{t_1^2 + t_2^2}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Q41 (2024)
If the radius of curvature of the path of two particles of same mass are in the ratio $3:4$, then in order to have constant centripetal force, their velocities will be in the ratio of:
- $\sqrt{3}: 2$
- $1: \sqrt{3}$
- $\sqrt{3}: 1$
- $2: \sqrt{3}$
### Related Formula
The centripetal force ($F$) acting on a particle of mass $m$ moving with velocity $v$ in a path of radius $r$ is given by:
$$F = \frac{m v^2}{r}$$
### Core Logic
Given that the two particles have the same mass ($m_1 = m_2$) and the ratio of their radii of curvature is:
$$\frac{r_1}{r_2} = \frac{3}{4}$$
To maintain a constant (equal) centripetal force ($F_1 = F_2$):
### Step 1: Relate Velocity to Radius
$$\frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2}$$
Since $m_1 = m_2$, the equation simplifies to:
$$\frac{v_1^2}{r_1} = \frac{v_2^2}{r_2} \implies \frac{v_1^2}{v_2^2} = \frac{r_1}{r_2}$$
$$\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}}$$
### Step 2: Calculate the Ratio
Substituting the given ratio of radii:
$$\frac{v_1}{v_2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$
Therefore, the ratio of their velocities is $\sqrt{3}:2$.
### Pattern Recognition
For problems involving steady values under constraint variations, establish the proportionality relation first. Here, $F \propto \frac{v^2}{r} \implies v \propto \sqrt{r}$ when $F$ and $m$ are held constant.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Plane
Q1730 (2024)
Projectiles $\mathrm{A}$ and $\mathrm{B}$ are thrown at angles of $45^{\circ}$ and $60^{\circ}$ with vertical respectively from top of a $400 \mathrm{~m}$ high tower. If their ranges and times of flight are same, the ratio of their speeds of projection $\mathrm{v_A}: \mathrm{v_B}$ is:
- $1: \sqrt{3}$
- $\sqrt{2}:1$
- $1:2$
- $1:\sqrt{2}$
### Core Logic
{{SOL_IMG1}}
For two projectiles launched from the same height to have the same time of flight ($T$), their vertical components of velocity must be equal.
Since the angles given are with the *vertical*, the vertical components are $v_A \cos(45^{\circ})$ and $v_B \cos(60^{\circ})$.
If $T_A = T_B$, then $v_{Ay} = v_{By}$.
$$v_A \cos(45^{\circ}) = v_B \cos(60^{\circ})$$
$$v_A \left(\frac{1}{\sqrt{2}}\right) = v_B \left(\frac{1}{2}\right)$$
$$\frac{v_A}{v_B} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$
### Step 1: Check Inconsistency
For the ranges to be the same while having the same time of flight, their horizontal components of velocity must also be equal: $v_{Ax} = v_{Bx}$.
$$v_A \sin(45^{\circ}) = v_B \sin(60^{\circ})$$
$$v_A \left(\frac{1}{\sqrt{2}}\right) = v_B \left(\frac{\sqrt{3}}{2}\right)$$
$$\frac{v_A}{v_B} = \sqrt{\frac{3}{2}}$$
This yields a contradiction. It is impossible for both the ranges and the times of flight to be simultaneously equal for different angles of projection from a tower.
### Step 2: Conclusion
The question contains inconsistent data and is technically a Bonus question. However, if one arbitrarily equates only the time of flight (or if NTA intended a different scenario), the ratio $\frac{v_A}{v_B} = \frac{1}{\sqrt{2}}$ matches option (4), which was the officially provided key before corrections.
### Pattern Recognition
Be wary of over-constrained physics problems. If a question specifies both range AND time of flight are identical for two different angles, check if the math yields a contradiction. NTA often accepts the result of one partial constraint
By NTA 4
BY Rankbit (Bonus).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Plane
