Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This is Part 1 of Laws of Motion, focusing purely on Newton’s Laws, Momentum, and frictionless systems for CBSE Class 11 Physics.
Key Concepts
Laws of Motion (No Friction)
The foundations of classical mechanics
1
Linear Momentum
Formula
The quantity of motion contained in a body. It is a vector quantity, defined as the product of mass and velocity.
$$\vec{p} = m\vec{v}$$
2
Newton’s Second Law
Formula
The rate of change of linear momentum of a body is directly proportional to the applied external force and takes place in the direction of the force. This is the real, fundamental law of motion.
$$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = m\vec{a} \quad (\text{if mass is constant})$$
3
Impulse & Impulse-Momentum Theorem
Formula
A large force acting for a very short time is an impulsive force. Impulse ($J$) is the total effect of this force. The theorem states that Impulse equals the change in momentum.
$$\vec{J} = \vec{F}_{\text{avg}} \Delta t = \int \vec{F} dt = \Delta \vec{p} = \vec{p}_f – \vec{p}_i$$
4
Newton’s Third Law
Definition
To every action, there is always an equal and opposite reaction. Forces always occur in pairs. Action and reaction act on different bodies, which is why they do not cancel each other out!
$$\vec{F}_{AB} = -\vec{F}_{BA}$$
5
Conservation of Linear Momentum
Formula
If the net external force on a system is zero, the total linear momentum of the system remains constant, regardless of the internal forces acting between the particles.
$$\text{If } \vec{F}_{\text{ext}} = 0 \implies \vec{p}_{\text{initial}} = \vec{p}_{\text{final}}$$
6
Equilibrium & Lami’s Theorem
Formula
A particle is in equilibrium if the net external force is zero ($\sum \vec{F} = 0$). If three concurrent forces keep a body in equilibrium, each force is proportional to the sine of the angle between the other two.
$$\frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma}$$
7
Apparent Weight in an Elevator (Lift)
Case Formulas
Your true weight is $mg$. Your apparent weight is the Normal Reaction ($N$) from the floor. It changes when the lift accelerates.
$$\text{Accelerating UP: } W_{\text{app}} = m(g + a)$$
$$\text{Accelerating DOWN: } W_{\text{app}} = m(g – a)$$
$$\text{Free Fall: } W_{\text{app}} = m(g – g) = 0$$
8
Connected Bodies (Atwood Machine)
Formula
Two masses $m_1$ and $m_2$ ($m_1 > m_2$) connected by a light string over a frictionless pulley. The heavier mass accelerates down, the lighter accelerates up with the same magnitude $a$.
$$a = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)g \quad | \quad T = \left(\frac{2 m_1 m_2}{m_1 + m_2}\right)g$$
Concept Deep Dive
01
The Art of the Free Body Diagram (FBD)
Isolate to calculateA Free Body Diagram is the most crucial tool in Mechanics. To draw one:
1. Imagine the object is completely isolated from its surroundings.
2. Draw a dot or a box representing the object.
3. Draw vectors for all the forces acting ON the object (Gravity, Tension, Normal, Applied force). Never include the forces the object exerts on other things.
4. Set up an X-Y coordinate system (often aligning one axis with the direction of acceleration).
5. Write $\sum F_x = m a_x$ and $\sum F_y = m a_y$.
1. Imagine the object is completely isolated from its surroundings.
2. Draw a dot or a box representing the object.
3. Draw vectors for all the forces acting ON the object (Gravity, Tension, Normal, Applied force). Never include the forces the object exerts on other things.
4. Set up an X-Y coordinate system (often aligning one axis with the direction of acceleration).
5. Write $\sum F_x = m a_x$ and $\sum F_y = m a_y$.
02
Variable Mass Systems (Rocket Propulsion)
When $F=ma$ failsStudents memorize $F = ma$, but Newton’s original law was $F = \frac{d(mv)}{dt}$. Using the product rule, this expands to $F = m\frac{dv}{dt} + v\frac{dm}{dt}$.
For a rocket blasting into space, the mass $m$ is constantly decreasing as fuel burns! Therefore, the second term becomes the dominant source of force, called Thrust. The thrust force on a rocket is the relative velocity of the exhaust gas multiplied by the rate at which mass is ejected.
For a rocket blasting into space, the mass $m$ is constantly decreasing as fuel burns! Therefore, the second term becomes the dominant source of force, called Thrust. The thrust force on a rocket is the relative velocity of the exhaust gas multiplied by the rate at which mass is ejected.
$$F_{\text{thrust}} = v_{\text{rel}} \frac{dm}{dt}$$
Compare & Contrast
✗ Internal Forces
- Forces exerted by particles of a system on each other.
- Occur in action-reaction pairs that completely cancel out.
- Cannot change the total momentum of the system.
- Example: The expanding gases pushing on a bomb casing just before it explodes.
✓ External Forces
- Forces exerted on the system by an external agent.
- Do not have their reaction pair inside the system.
- Can change the total momentum and accelerate the center of mass.
- Example: Gravity pulling the bomb down toward the earth.
Common Mistakes to Avoid
Mistake 1
Cancelling Action and Reaction: Since action and reaction are equal and opposite, students often think they cancel out, resulting in zero net force. They DO NOT cancel because they act on two different bodies. A horse pulls a cart (Action on cart), the cart pulls the horse (Reaction on horse). If you draw an FBD of just the cart, only one of these forces is present!
Mistake 2
Normal Force Always Equals Weight: Many students automatically write $N = mg$. This is only true for a flat, horizontal surface with no vertical acceleration or other vertical forces. On an inclined plane, $N = mg \cos \theta$. In an accelerating lift, $N = m(g \pm a)$. Always use $\sum F_y = ma_y$ to find $N$.
Mistake 3
Misinterpreting “Constant Velocity”: If a problem states a block is moving at a constant velocity, it implies acceleration is zero. Therefore, by Newton’s Second Law, the Net Force is ZERO ($\sum F = 0$). Students often try to calculate a positive net force just because the object is moving.
Exam Tips
Tip 1
For problems involving a gun firing bullets: If $n$ bullets each of mass $m$ are fired per second with velocity $v$, the average force required to hold the gun steady is simply $F = nmv$. (Because $F = dp/dt$, and $p = nmv$ is the momentum lost per second).
Tip 2
The “System” Shortcut: If multiple blocks are connected by strings and pulled by a force $F$ on a smooth surface, treat all the blocks as a single “System” to find the common acceleration quickly: $a = \frac{F_{\text{net}}}{\text{Total Mass}}$. Then, use FBDs of individual blocks to find the tension in specific strings.
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Laws of Motion · CBSE Exam1
A machine gun fires a bullet of mass $40 \text{ g}$ with a velocity of $1200 \text{ m/s}$. The man holding it can exert a maximum force of $144 \text{ N}$ on the gun. How many bullets can he fire per second at the most? [2 marks]
Answer
$3 \text{ bullets/second}$
📝
Explanation
Given: Mass of bullet $m = 40 \text{ g} = 0.04 \text{ kg}$, Velocity $v = 1200 \text{ m/s}$, Max Force $F = 144 \text{ N}$.
Let $n$ be the number of bullets fired per second.
Force = Change in momentum per second = $n \times m \times v$.
$144 = n \times 0.04 \times 1200$
$144 = n \times 48 \implies n = \frac{144}{48} = 3$.
2
A mass of $10 \text{ kg}$ is suspended by a rope of length $2 \text{ m}$ from the ceiling. A force of $50 \text{ N}$ in the horizontal direction is applied at the midpoint of the rope. What is the angle the rope makes with the vertical in equilibrium? ($g = 10 \text{ m/s}^2$) [3 marks]
Answer
$\theta = \tan^{-1}(0.5)$ or $\approx 26.5^\circ$
📝
Explanation
Let the tension in the upper half of the rope be $T_1$, and it makes an angle $\theta$ with the vertical.
The tension in the lower half of the rope $T_2$ simply supports the weight: $T_2 = mg = 10 \times 10 = 100 \text{ N}$.
Considering the equilibrium at the midpoint:
Vertical forces balance: $T_1 \cos \theta = T_2 = 100 \text{ N}$ (Eq. 1)
Horizontal forces balance: $T_1 \sin \theta = F_{\text{applied}} = 50 \text{ N}$ (Eq. 2)
Dividing Eq. 2 by Eq. 1:
$\frac{T_1 \sin \theta}{T_1 \cos \theta} = \frac{50}{100} \implies \tan \theta = 0.5 \implies \theta = \tan^{-1}(0.5)$.
3
Two masses $8 \text{ kg}$ and $12 \text{ kg}$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released. ($g = 10 \text{ m/s}^2$) [3 marks]
Answer
$a = 2 \text{ m/s}^2$, $T = 96 \text{ N}$
📝
Explanation
Given: $m_1 = 12 \text{ kg}$, $m_2 = 8 \text{ kg}$.
Using the standard Atwood machine formulas:
Acceleration $a = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)g = \left(\frac{12 – 8}{12 + 8}\right) \times 10 = \left(\frac{4}{20}\right) \times 10 = 2 \text{ m/s}^2$.
Tension $T = \left(\frac{2 m_1 m_2}{m_1 + m_2}\right)g = \left(\frac{2 \times 12 \times 8}{20}\right) \times 10 = \frac{192}{20} \times 10 = 96 \text{ N}$.
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