Laws of Motion Previous Year Questions — JEE Main Physics
34 past-year Laws of Motion questions from JEE Main (Physics) — page 4.
Q45 (2024)
A body of mass $1000\text{ kg}$ is moving horizontally with a velocity $6\text{ m/s}$. If $200\text{ kg}$ extra mass is added, the final velocity (in m/s) is:
- $6$
- $2$
- $3$
- $5$
### Related Formula
$$m_1 v_1 = m_2 v_2$$
### Core Logic
Since there is no external horizontal force acting on the body, linear momentum along the horizontal axis is conserved.
Initial mass $m_1 = 1000\text{ kg}$, initial velocity $v_1 = 6\text{ m/s}$.
Final mass $m_2 = 1000 + 200 = 1200\text{ kg}$.
### Step 1: Balance conservation equation
$$1000 \times 6 = 1200 \times v_2$$
$$6000 = 1200 v_2 \implies v_2 = \frac{6000}{1200} = 5\text{ m/s}$$
### Pattern Recognition
Inelastic mass addition transitions are classic momentum-balance equations, scaling velocity inversely with total expanded mass profiles.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q39 (2024)
A block of mass $100 \mathrm{~kg}$ slides over a distance of $10 \mathrm{~m}$ on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4, then the work done against friction (in J) is:
- 4200
- 3900
- 4000
- 4500
### Related Formula
Kinetic frictional force ($f$) on a flat surface:
$$f = \mu N = \mu m g$$
Work done against friction ($W_{\text{against}}$):
$$W_{\text{against}} = f \cdot s = \mu m g s$$
### Core Logic
Given values:
$$m = 100 \mathrm{~kg}, \quad s = 10 \mathrm{~m}, \quad \mu = 0.4$$
Take $g = 10 \mathrm{~m/s^2}$.
### Step 1: Calculate Friction Force
The normal force on a horizontal surface is:
$$N = mg = 100 \times 10 = 1000 \mathrm{~N}$$
The force of friction is:
$$f = \mu N = 0.4 \times 1000 = 400 \mathrm{~N}$$
### Step 2: Calculate Work Done
The work done against friction is:
$$W_{\text{against}} = f \cdot s = 400 \mathrm{~N} \times 10 \mathrm{~m} = 4000 \mathrm{~J}$$
Therefore, the work done is $4000 \mathrm{~J}$.
### Pattern Recognition
Work done *by* friction is negative ($-4000 \mathrm{~J}$) because the frictional force opposes displacement. Work done *against* friction is positive ($+4000 \mathrm{~J}$) because it represents the external energy that must be spent to sustain slide.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q1727 (2024)
Three blocks $\mathrm{A}$, $\mathrm{B}$ and $\mathrm{C}$ are pulled on a horizontal smooth surface by a force of $80\mathrm{N}$ as shown in figure
{{IMG1}}
The tensions $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ in the string are respectively:
- $40 \mathrm{N}, 64 \mathrm{N}$
- $60 \mathrm{N}, 80 \mathrm{N}$
- $88 \mathrm{N}, 96 \mathrm{N}$
- $80 \mathrm{N}, 100 \mathrm{N}$
### Related Formula
$$a = \frac{F_{\text{net}}}{m_{\text{total}}}$$
$$T = m_{\text{behind}} \times a$$
### Core Logic
Since the surface is smooth, the entire system of three blocks moves with a common acceleration $a$.
$$a_{\mathrm{A}} = a_{\mathrm{B}} = a_{\mathrm{C}} = \frac{F}{m_{\mathrm{A}} + m_{\mathrm{B}} + m_{\mathrm{C}}}$$
### Step 1: Calculate Common Acceleration
$$a = \frac{80}{5 + 3 + 2} = \frac{80}{10} = 8 \mathrm{~m/s}^2$$
### Step 2: Calculate Tensions
{{SOL_IMG1}}
$\mathrm{T}_1$ pulls only block $\mathrm{A}$ ($5 \mathrm{~kg}$):
$$\mathrm{T}_1 = m_{\mathrm{A}} \times a = 5 \times 8 = 40 \mathrm{~N}$$
{{SOL_IMG2}}
$\mathrm{T}_2$ pulls both blocks $\mathrm{A}$ and $\mathrm{B}$ ($5 \mathrm{~kg} + 3 \mathrm{~kg} = 8 \mathrm{~kg}$):
From free body diagram of block B:
$$\mathrm{T}_2 - \mathrm{T}_1 = m_{\mathrm{B}} \times a$$
$$\mathrm{T}_2 - 40 = 3 \times 8 = 24 \implies \mathrm{T}_2 = 64 \mathrm{~N}$$
### Pattern Recognition
Tension at any point in a train of connected accelerating bodies (without friction) is directly proportional to the total mass being pulled *behind* that point.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q32 (2024)
A block of mass $1 \mathrm{~kg}$ is pushed up a surface inclined to horizontal at an angle of $60^{\circ}$ by a force of $10 \mathrm{~N}$ parallel to the inclined surface as shown in figure. When the block is pushed up by $10 \mathrm{~m}$ along inclined surface, the work done against frictional force is: $\left[\mathrm{g} = 10 \mathrm{~m} / \mathrm{s}^{2}\right]$
{{IMG1}}
- $5\sqrt{3} \mathrm{~J}$
- $5 \mathrm{~J}$
- $5 \times 10^{3} \mathrm{~J}$
- $10 \mathrm{~J}$
### Related Formula
$$W_f = f_k \cdot d$$
$f_k = \mu N$
$$N = mg \cos\theta$$
### Core Logic
The work done against the frictional force is the product of the kinetic friction force and the displacement along the plane.
The normal force $N$ on the block is given by $N = mg \cos\theta$, where $\theta = 60^{\circ}$.
### Step 1: Calculate Normal and Frictional Force
Given:
$\mu = 0.1$
$m = 1 \mathrm{~kg}$
$\theta = 60^{\circ}$
$d = 10 \mathrm{~m}$
Normal force $N = 1 \times 10 \times \cos(60^{\circ}) = 10 \times \frac{1}{2} = 5 \mathrm{~N}$.
Frictional force $f_k = \mu N = 0.1 \times 5 = 0.5 \mathrm{~N}$.
### Step 2: Work Done Against Friction
Work done against frictional force $= f_k \times d$
$$W = 0.5 \mathrm{~N} \times 10 \mathrm{~m} = 5 \mathrm{~J}$$
### Pattern Recognition
Whenever asked for "work done against friction", simply compute $\mu mg \cos\theta \times d$. The applied force ($10 \mathrm{~N}$) is irrelevant to the friction calculation itself since it is parallel to the plane.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Work, Energy and Power
Q39 (2024)
A block of mass $m$ is placed on a surface having vertical cross section given by $y = x^2 / 4$. If coefficient of friction is $0.5$, the maximum height above the ground at which block can be placed without slipping is:
- $1 / 4 \mathrm{~m}$
- $1 / 2 \mathrm{~m}$
- $1 / 6 \mathrm{~m}$
- $1 / 3 \mathrm{~m}$
### Related Formula
$$\tan \theta = \mu$$
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \tan \theta$$
### Core Logic
For a block to remain stationary on a rough surface without slipping, the maximum slope of the surface it can rest on is determined by the angle of repose.
$$\tan \theta \le \mu$$
The slope of the given parabolic curve at any point $(x,y)$ is $\frac{\mathrm{d}y}{\mathrm{d}x}$.
### Step 1: Evaluate the Slope
Given curve equation:
$$y = \frac{x^2}{4}$$
Differentiating with respect to $x$:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x}{4} = \frac{x}{2}$$
### Step 2: Find Maximum Valid Position
At the critical point of slipping, slope equals $\mu$:
$$\frac{x}{2} = \mu$$
Given $\mu = 0.5 = \frac{1}{2}$, we have:
$$\frac{x}{2} = \frac{1}{2} \implies x = 1$$
### Step 3: Calculate Maximum Height
Substitute $x = 1$ back into the curve equation to find the maximum height $y$:
$$y = \frac{(1)^2}{4} = \frac{1}{4} \mathrm{~m}$$
### Pattern Recognition
When asked for maximum height on a curve $y=f(x)$ without slipping, set $\frac{\mathrm{d}y}{\mathrm{d}x} = \mu$, solve for $x$, and plug it back into the original equation to find $y$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
