Kinematics Previous Year Questions — JEE Main Physics
38 past-year Kinematics questions from JEE Main (Physics) — page 2.
Q12 (2025)
Two projectiles are fired from ground with same initial speeds from same point at angles $(45^{\circ} + \alpha)$ and $(45^{\circ} - \alpha)$ with horizontal direction. The ratio of their times of flights is
- 1
- $\frac{1 - \tan\alpha}{1 + \tan\alpha}$
- $\frac{1 + \sin 2\alpha}{1 - \sin 2\alpha}$
- $\frac{1 + \tan\alpha}{1 - \tan\alpha}$
### Related Formula
The time of flight $T$ of a projectile launched with speed $u$ at an angle $\theta$ with the horizontal is:
$$T = \frac{2u \sin\theta}{g}$$$
### Core Logic
The launch angles of the two projectiles are:
$$\theta_1 = 45^\circ + \alpha$$
$$\theta_2 = 45^\circ - \alpha$$
Since they have the same speed $u$:
$$\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)}$$
### Step 1: Simplify Trigonometric Ratio
Using the angle sum and difference formulas:
$$\frac{T_1}{T_2} = \frac{\sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha}{\sin 45^\circ \cos \alpha - \cos 45^\circ \sin \alpha}$$
$$\frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}}\cos\alpha + \frac{1}{\sqrt{2}}\sin\alpha}{\frac{1}{\sqrt{2}}\cos\alpha - \frac{1}{\sqrt{2}}\sin\alpha} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha}$$
Divide numerator and denominator by $\cos\alpha$:
$$\frac{T_1}{T_2} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$$$
### Pattern Recognition
Sees: Projectile angles complementary to $45^\circ$.
Shortcut: Remember the identity $\tan(45^\circ + \alpha) = \frac{1+\tan\alpha}{1-\tan\alpha}$. Since complementary angles have sine ratios proportional to $\sin(45^\circ + \alpha)/\sin(45^\circ - \alpha) = \tan(45^\circ + \alpha)$, the answer is directly $\frac{1+\tan\alpha}{1-\tan\alpha}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Motion in a Plane
Q16 (2025)
Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is $8$ times higher than that of the second ball. $T_{1}$ and $T_{2}$ are the total flying times of first and second ball, respectively, then the ratio of $T_{1}$ and $T_{2}$ is:
- $2\sqrt{2} : 1$
- $2 : 1$
- $\sqrt{2} : 1$
- $4 : 1$
### Related Formula
$$H = \frac{u^2 \sin^2\theta}{2g} \quad \text{and} \quad T = \frac{2u \sin\theta}{g}$$
where,
$H$ = maximum height reached
$T$ = total time of flight
$u$ = initial projection velocity
$\theta$ = angle of projection
### Core Logic
From the formulas, we see that:
$$H \propto \sin^2\theta \quad \text{and} \quad T \propto \sin\theta$$
Thus, we can directly link time of flight to the square root of the maximum height:
$$T \propto \sqrt{H} \implies \frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}}$$
### Step 1: Compute Ratio
Given:
$$H_1 = 8 H_2 \implies \frac{H_1}{H_2} = 8$$
Substitute this ratio:
$$\frac{T_1}{T_2} = \sqrt{8} = 2\sqrt{2}$$
Thus, the ratio is $2\sqrt{2} : 1$.
### Pattern Recognition
Sees: Projectile heights ratio → Time of flight ratio.
Shortcut: Since $H \propto u_y^2$ and $T \propto u_y$, we have $T \propto \sqrt{H}$. If the height is 8 times larger, the flying time is $\sqrt{8} = 2\sqrt{2}$ times larger. ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinematics
Q25 (2025)
Two cars P and Q are moving on a road in the same direction. Acceleration of car P increases linearly with time whereas car Q moves with a glass constant acceleration. Both cars cross each other at time $t = 0$ , for the first time. The maximum possible number of crossing(s) (including the crossing at $t = 0$ ) is ______.
### Related Formula
$$a_P = kt \implies v_P = \frac{1}{2}kt^2 + v_{0P} \implies x_P = \frac{1}{6}kt^3 + v_{0P}t + x_0$$
$$a_Q = a \implies v_Q = at + v_{0Q} \implies x_Q = \frac{1}{2}at^2 + v_{0Q}t + x_0$$
### Core Logic
To find the relative crossings, set the position functions equal: $x_P(t) = x_Q(t)$.
$$\frac{1}{6}kt^3 + v_{0P}t = \frac{1}{2}at^2 + v_{0Q}t$$
$$\frac{1}{6}kt^3 - \frac{1}{2}at^2 + (v_{0P} - v_{0Q})t = 0$$
This is a cubic polynomial equation in terms of time $t$. A cubic function can yield a maximum of 3 distinct real roots.
{{SOL_IMG1}}
{{SOL_IMG2}}
{{SOL_IMG3}}
{{SOL_IMG4}}
Depending on the initial velocity profiles ($v_{0P}, v_{0Q}$), the curves can intersect at $t=0$ and potentially create up to two additional crossing paths downstream as acceleration profiles cross over. Thus, the maximum possible number of total crossings is exactly 3.
### Pattern Recognition
Crossings correspond mathematically to intersections of relative displacement polynomials. A linear acceleration profile produces a cubic position curve ($t^3$), which can cross a quadratic curve ($t^2$) at up to 3 distinct coordinate points.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinematics
Q757 (2025)
Which of the following curves possibly represent one-dimensional motion of a particle?
(A) {{IMG1}}
(B) {{IMG2}}
(C) {{IMG3}}
(D) {{IMG4}}
Choose the correct answer from the options given below:
- A, B and D only
- A, B and C only
- A and B only
- A, C and D only
### Related Formula
For realistic physical motion in one dimension:
1. Time $t$ can never be negative during a normal positive time sequence, and cannot flow backwards.
2. Total distance covered can never decrease over time.
3. A particle cannot have two different values of position or velocity at the exact same instant of time.
### Core Logic
Let us analyze each curve:
- **Curve (A)** (Phase $\phi$ vs Time $t$): Represents $\phi = kt + C$, which is a valid linear relationship of phase over time (e.g., in Simple Harmonic Motion $x = A\sin(kt + C)$). (Valid)
- **Curve (B)** (Velocity $v$ vs Displacement $x$): A closed loop, which represents symmetric harmonic-type oscillation. For example, $v^2 + \omega^2 x^2 = \text{const}$ (ellipse) is a perfectly physically valid 1D SHM velocity-displacement phase portrait. (Valid)
- **Curve (C)** (Velocity vs Time): The curve enters into the negative time quadrant. Time cannot go backwards or exist in negative values relative to starting sequence in standard physical scenarios. (Invalid)
- **Curve (D)** (Total Distance vs Time): Represents total distance increasing over time. Total distance is a non-decreasing function of time ($d(d)/dt \ge 0$). Thus, this curve is physically valid. (Valid)
### Step 1: Conclusion
Therefore, curves A, B, and D possibly represent physical one-dimensional motion. The correct option is (1).
### Pattern Recognition
Quick check for graph validity:
1. Time cannot run backwards (ruling out C).
2. Total distance can never decrease (D is valid because it strictly goes upwards).
3. $v$ vs $x$ can be circular/elliptical in SHM (B is valid).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Q19 (2025)
The angle of projection of a particle is measured from the vertical axis as $\phi$ and the maximum height reached by the particle is $h_m$. Here $h_m$ as function of $\phi$ can be presented as:
- Curve (1)
- Curve (2)
- Curve (3)
- Curve (4)
### Related Formula
Maximum height of a projectile:
$$H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}$$
where $\theta$ is the angle of projection measured from the horizontal ground.
### Core Logic
The angle of projection in this question is measured from the **vertical axis** as $\phi$.
This relates to the horizontal projection angle $\theta$ as:
$$\theta = 90^{\circ} - \phi$$
Substitute this into the maximum height formula:
$$h_m = \frac{u^2 \sin^2 (90^{\circ} - \phi)}{2g} = \frac{u^2 \cos^2 \phi}{2g}$$
$$h_m(\phi) \propto \cos^2 \phi$$
### Step 1: Analyzing Boundary Conditions
Let's check the boundary values of $h_m$ for $\phi \in [0, 90^{\circ}]$:
- At $\phi = 0^{\circ}$ (fired straight up vertically):
$$h_m(0) = \frac{u^2}{2g} = \text{Maximum height}$$
- At $\phi = 90^{\circ}$ (fired horizontally along the ground):
$$h_m(90^{\circ}) = 0$$
Let's evaluate the behavior of $h_m(\phi)$:
- At small angles $\phi \approx 0$, the slope $\frac{d h_m}{d\phi} = -\frac{u^2}{2g} \sin(2\phi) \approx 0$. This means the curve starts with a horizontal slope (flat top).
- As $\phi$ increases towards $90^{\circ}$, $h_m$ falls smoothly, reaching 0 at $90^{\circ}$ with flat slope again.
This sinusoidal falloff perfectly matches **Curve (3)**.
{{SOL_IMG1}}
### Pattern Recognition
Always read the coordinate origin carefully! Measuring angle from the vertical instead of the horizontal turns $\sin^2\theta$ into $\cos^2\phi$. Checking extremes: vertical shot ($0^{\circ}$) yields maximum height, while flat shot ($90^{\circ}$) yields zero height.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Plane: Projectile Motion
