Kinematics Previous Year Questions — JEE Main Physics
38 past-year Kinematics questions from JEE Main (Physics) — page 6.
Q55 (2024)
A vector has magnitude same as that of $\vec{\mathrm{A}} = 3\hat{\mathrm{i}} + 4\hat{\mathrm{j}}$ and is parallel to $\vec{\mathrm{B}} = 4\hat{\mathrm{i}} + 3\hat{\mathrm{j}}$. The $x$ and $y$ components of this vector in first quadrant are $x$ and $3$ respectively where $x = $ ________
### Related Formula
$$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$$
$$\vec{N} = |\vec{A}| \hat{B}$$
### Core Logic
We need to find a new vector $\vec{N}$ that has the magnitude of $\vec{A}$ and the direction of $\vec{B}$.
Magnitude of $\vec{A}$: $|\vec{A}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
Unit vector in the direction of $\vec{B}$: $\hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{4\hat{i} + 3\hat{j}}{\sqrt{4^2 + 3^2}} = \frac{4\hat{i} + 3\hat{j}}{5}$.
### Step 1: Construct the Vector
$$\vec{N} = |\vec{A}| \hat{B} = 5 \left( \frac{4\hat{i} + 3\hat{j}}{5} \right)$$
$$\vec{N} = 4\hat{i} + 3\hat{j}$$
### Step 2: Match Components
The $x$ and $y$ components are given as $x$ and $3$.
From $\vec{N} = 4\hat{i} + 3\hat{j}$, we see the $x$-component is $4$.
Therefore, $x = 4$.
### Pattern Recognition
Constructing a vector matching magnitude and direction is a simple scalar multiplication of the desired magnitude by the target direction's unit vector.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Motion in a Plane
Q56 (2024)
The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t + 1) \mathrm{~s}$ are $125 \mathrm{~m}$ and $50 \mathrm{~m / s}$, respectively. The distance travelled by the particle in $(t + 2)^{\mathrm{th}} \mathrm{~s}$ is $\_ \_ \_ \_ \_ \mathrm{m}$.
### Related Formula
$v = u + at$
$$s = ut + \frac{1}{2}at^2$$
$$S_{n^{\text{th}}} = u + \frac{a}{2}(2n - 1)$$
### Core Logic
Let the velocity at time $t$ be $u$. The time interval $\Delta t = (t+1) - t = 1 \mathrm{~s}$. The increase in velocity over 1 second is exactly the acceleration $a$. The displacement in that 1-second interval acts as the $(t+1)^{\text{th}}$ second displacement equation.
### Step 1: Determine Acceleration
Increase in velocity $\Delta v = 50 \mathrm{~m/s}$ in $1 \mathrm{~s}$.
$v = u + at$
$$u + 50 = u + a(1) \Rightarrow a = 50 \mathrm{~m/s^2}$$
### Step 2: Determine Velocity 'u' at time t
Displacement in the 1-second interval from $t$ to $t+1$ is $125 \mathrm{~m}$.
Using $s = ut' + \frac{1}{2}at'^2$ where $t' = 1 \mathrm{~s}$:
$$125 = u(1) + \frac{1}{2}a(1)^2$$
$$125 = u + \frac{50}{2}$$
$$125 = u + 25 \Rightarrow u = 100 \mathrm{~m/s}$$
### Step 3: Distance in the next second
We need the distance travelled in the $(t+2)^{\text{th}}$ second, which corresponds to the 1-second interval starting with an initial velocity equal to the velocity at $t+1$.
Alternatively, we can use the $n^{\text{th}}$ second formula directly by re-indexing. The velocity at start of this interval is $u_{new} = u + a = 100 + 50 = 150 \mathrm{~m/s}$.
Distance $S_{1\text{st}}$ using new parameters:
$$s = 150(1) + \frac{1}{2}(50)(1)^2 = 150 + 25 = 175 \mathrm{~m}$$
Or using PDF logic from base $u$ with $n=2$ (since $n=1$ was $t+1^{\text{th}}$):
$$S_{2^{\text{nd}}} = u + \frac{a}{2}[2n - 1] = 100 + 25[4 - 1] = 100 + 75 = 175 \mathrm{~m}$$
### Pattern Recognition
For 1-second interval mechanics, $\Delta v$ directly yields $a$. The displacement equation simplifies gracefully to $s = u_{start} + a/2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Q44 (2024)
If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at an angle $\theta$, then
- $|\vec{A} - \vec{B}| = \sqrt{2}R \sin \left(\frac{\theta}{2}\right)$
- $|\vec{A} + \vec{B}| = 2R \sin \left(\frac{\theta}{2}\right)$
- $|\vec{A} + \vec{B}| = 2R \cos \left(\frac{\theta}{2}\right)$
- $|\vec{A} - \vec{B}| = 2R \cos \left(\frac{\theta}{2}\right)$
### Related Formula
The magnitude of the resultant vector is given by:
$$|\vec{R}_{res}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$$
### Core Logic
Let $|\vec{A}| = |\vec{B}| = R$.
Then for vector addition:
$$|\vec{A} + \vec{B}| = \sqrt{R^2 + R^2 + 2R^2 \cos \theta}$$
### Step 1: Simplify Addition Form
$$|\vec{A} + \vec{B}| = \sqrt{2R^2 (1 + \cos \theta)}$$
Using the trigonometric identity $1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right)$:
$$|\vec{A} + \vec{B}| = \sqrt{2R^2 \times 2 \cos^2 \left(\frac{\theta}{2}\right)} = 2R \cos \left(\frac{\theta}{2}\right)$$
### Step 2: Cross-check Subtraction Form
For subtraction:
$$|\vec{A} - \vec{B}| = \sqrt{R^2 + R^2 - 2R^2 \cos \theta}$$
$$|\vec{A} - \vec{B}| = \sqrt{2R^2 (1 - \cos \theta)} = \sqrt{2R^2 \times 2 \sin^2 \left(\frac{\theta}{2}\right)} = 2R \sin \left(\frac{\theta}{2}\right)$$
Checking options, only $|\vec{A} + \vec{B}| = 2R \cos \left(\frac{\theta}{2}\right)$ is correctly paired in the choice list.
### Pattern Recognition
Standard geometry shortcut: Addition of two equal vectors yields a cosine half-angle dependency. Subtraction yields a sine half-angle dependency. $(+ \to \cos)$, $(- \to \sin)$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Motion in a Plane
Q2005 (2024)
A body starts falling freely from height $H$ hits an inclined plane in its path at height $h$. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of $\frac{H}{h}$ for which the body will take the maximum time to reach the ground is
### Related Formula
$$t = \sqrt{\frac{2d}{g}}$$
### Core Logic
{{SOL_IMG1}}
The body falls freely from height $H$ to height $h$. The distance fallen is $(H - h)$.
Time taken to fall this distance:
$$t_1 = \sqrt{\frac{2(H - h)}{g}}$$
After elastic impact, the vertical velocity becomes zero (the entire velocity is directed horizontally). From height $h$, it now acts as a horizontal projectile.
The time taken to reach the ground vertically from height $h$ is:
$$t_2 = \sqrt{\frac{2h}{g}}$$
Total time of flight $T = t_1 + t_2$:
$$T = \sqrt{\frac{2(H - h)}{g}} + \sqrt{\frac{2h}{g}}$$
### Step 2: Maximizing Time
To find the maximum time $T$, differentiate $T$ with respect to $h$ and equate to zero:
$$\frac{dT}{dh} = \sqrt{\frac{2}{g}} \left( \frac{-1}{2\sqrt{H - h}} + \frac{1}{2\sqrt{h}} \right) = 0$$
$$\frac{1}{2\sqrt{h}} = \frac{1}{2\sqrt{H - h}}$$
$$\sqrt{H - h} = \sqrt{h}$$
Squaring both sides:
$H - h = h$
$H = 2h$
$$\frac{H}{h} = 2$$
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinematics
Q33 (2024)
The relation between time '$t$' and distance '$x$' is $t = \alpha x^2 + \beta x$, where $\alpha$ and $\beta$ are constants. The relation between acceleration ($a$) and velocity ($v$) is:
- $a = -2\alpha v^3$
- $a = -5\alpha v^5$
- $a = -3\alpha v^2$
- $a = -4\alpha v^4$
### Related Formula
$$v = \frac{dx}{dt}$$
$$a = \frac{dv}{dt} = v\frac{dv}{dx}$$
### Step 1: Differentiate with respect to time
Given the equation:
$$t = \alpha x^2 + \beta x$$
Differentiating with respect to time $t$:
$$\frac{dt}{dt} = \frac{d}{dt}(\alpha x^2 + \beta x)$$
$$1 = 2\alpha x \frac{dx}{dt} + \beta \frac{dx}{dt}$$
$$1 = (2\alpha x + \beta) v$$
$$v = (2\alpha x + \beta)^{-1}$$
### Step 2: Calculate Acceleration
Now, acceleration $a = \frac{dv}{dt}$. Differentiating $v$ with respect to time $t$:
$$a = \frac{d}{dt} \left[ (2\alpha x + \beta)^{-1} \right]$$
$$a = -1(2\alpha x + \beta)^{-2} \cdot \frac{d}{dt}(2\alpha x + \beta)$$
$$a = -(2\alpha x + \beta)^{-2} \cdot (2\alpha) \frac{dx}{dt}$$
Substitute $v$ and $(2\alpha x + \beta)^{-2} = v^2$:
$$a = -(v^2) \cdot (2\alpha) \cdot v$$
$$a = -2\alpha v^3$$
### Pattern Recognition
Standard kinematic shortcut: Whenever $t = Ax^2 + Bx$, $v = (2Ax+B)^{-1}$ and $a = -2A v^3$. Memorizing this directly saves derivation time during the exam.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
