Complete Class 11 Physics Notes: Linear Momentum & Collisions (NEET / JEE)
Watch this Live Class https://youtube.com/live/XNXQ_YLdNTA
Welcome to CRACKNEETPhysics! If you are struggling with the complexities of Mechanics, this complete guide to Linear Momentum and Collisions is exactly what you need for your NEET and JEE Mains preparation. In these Class 11 Physics notes, we break down core concepts without the confusing jargon—covering everything from Newton's Second Law in variable mass systems to the Law of Conservation of Momentum and the Coefficient of Restitution (e). Stop losing marks on tricky physics numericals! Read on to learn our timesaving 'axis rotation' trick for 2D explosions, master the physics of elastic vs. inelastic collisions, and tackle high-yield MCQs designed to test your conceptual clarity."
Welcome to the ultimate guide on Linear Momentum and Collisions! Before diving into complex topics like Center of Mass, we must conquer the foundational concept of momentum. These notes break down the theory, problem-solving tricks (like axis rotation), and the real physics behind collisions.
1. Understanding Linear Momentum (p)
What is it? Momentum is a vector quantity that represents the “quantity of motion” contained within a body. It has both magnitude and direction (the direction is the same as the velocity).
Formula: p = m × v (where m = mass, v = velocity)
Conceptual Example: Imagine a very slow-moving truck and a very fast-moving bullet. Even though the bullet is incredibly fast, the truck might have a much higher momentum simply because its mass is so massive. The product of mass and velocity dictates the total impact force!
2. Newton’s Second Law & Variable Mass Systems
We usually learn Newton’s Second Law as F = ma. But the true, most fundamental definition is based on momentum!
The Real Formula: Net Force is the rate of change of momentum over time. F_net = dp / dt
Force as a Transfer Mechanism: Don’t just think of force as a “push or pull.” Think of it as the active mechanism that transfers momentum from one body to another over a brief period (Impulse).
Variable Mass Systems (e.g., Rockets): In real life, mass isn’t always constant.
Rocket Launch: A rocket constantly burns fuel, meaning its mass decreases as it flies.
Leaking Sandbag: A moving trolley with a hole drops sand, changing its mass.
Because mass (m) is changing, we cannot use F = ma. We must use the product rule of derivatives on p = mv: F = m(dv/dt) + v(dm/dt) In these systems, the changing mass (dm/dt) actively contributes to the force and acceleration!
3. The Law of Conservation of Linear Momentum
The Golden Rule: If the net external force on a system is zero (F_ext = 0), then the rate of change of momentum is zero (dp/dt = 0). Therefore, the total initial momentum equals the total final momentum.
P_initial = P_final
Where do we see this?
Collisions: Two billiard balls striking each other.
Explosions: A stationary bomb splitting into three fragments.
Recoil: A gun firing a bullet (the gun jerks backward to cancel the bullet’s forward momentum).
The Origin of Conservation: Why does this law exist? It is deeply rooted in Newton’s Third Law (Every action has an equal and opposite reaction). When two bodies collide, Body 1 applies a force on Body 2, and Body 2 applies an equal and opposite force on Body 1 (F12 = -F21). Because the forces are internal, equal, and opposite, the momentum transferred to Body 1 is perfectly canceled by the momentum transferred to Body 2. The total sum remains completely unchanged!
4. Problem-Solving Hack: 2D Explosions & Axis Rotation
When dealing with explosions or collisions in a 2D plane (X and Y axes), momentum is conserved independently along each axis, provided there is no external force along that specific axis.
Sum of Initial Momentum in X = Sum of Final Momentum in X
Sum of Initial Momentum in Y = Sum of Final Momentum in Y
🔥 The “Rotate the Axes” Trick: Sometimes, examiners give you a 2D collision/explosion diagram with weird, confusing angles. If you use the standard vertical Y-axis and horizontal X-axis, your math will become a nightmare of sines and cosines. The Trick: You are allowed to rotate your X and Y axes! Align your new X-axis directly along the path of one of the moving particles. This immediately sets that particle’s Y-component to zero, eliminating a variable and making the math instantly easier!
5. Types of Collisions & Kinetic Energy
While Momentum is ALWAYS conserved in an isolated system, Kinetic Energy (KE) is not always conserved. Collisions act as a traffic policeman, redistributing the motion among the bodies.
A. Elastic Collisions:
Definition: A collision where there is NO loss of Kinetic Energy.
What happens: The bodies compress slightly like springs upon impact, storing elastic potential energy, but they perfectly bounce back, restoring 100% of that energy into Kinetic Energy. (KE_initial = KE_final).
B. Inelastic Collisions:
Definition: A collision where some Kinetic Energy is lost.
What happens: The mechanical energy is lost to heat, sound, or permanent physical deformation (denting). (KE_initial > KE_final).
C. Perfectly Inelastic Collisions:
Definition: The extreme case of maximum energy loss!
What happens: The two bodies collide and stick together (like throwing a ball of clay at a wall, or a car crashing and hooking into a truck). They move together with a shared final velocity.
6. Coefficient of Restitution (e)
How do we mathematically measure if a collision is elastic, inelastic, or perfectly inelastic? We use the Coefficient of Restitution (e). It is a yardstick to measure how much energy is retained after the crash.
Formula: e = (Relative Velocity of Separation After Collision) / (Relative Velocity of Approach Before Collision)
The Values of ‘e’:
e = 1: Perfectly Elastic (100% energy retained. Velocity of separation equals velocity of approach).
e = 0: Perfectly Inelastic (Maximum energy lost. The bodies stick together, so their relative velocity of separation is exactly 0).
0 < e < 1: Real-world collisions (Partial loss of energy. E.g., a dropping tennis ball doesn’t bounce all the way back to its original height).
(Fun Concept: Kinetic Energy represents “useful, organized motion” capable of doing work, whereas things like heat represent “chaotic, disordered motion” of atoms!)
Multiple Choice Question 1
Car A is moving at 15 m/s and rear-ends Car B, which is moving at 5 m/s in the exact same direction along a straight line. Immediately after the impact, Car B is pushed forward and speeds up to a velocity of 12 m/s, while Car A continues moving forward but slows down to 8 m/s.
Based on the concept of relative velocity, what is the coefficient of restitution (e) for this collision, and how would you classify the nature of this collision?
(A) e = 0.6, Partially Inelastic (B) e = 1.0, Perfectly Elastic (C) e = 0.4, Partially Inelastic (D) e = 0.0, Perfectly Inelastic
Correct Answer: (C) e = 0.4, Partially Inelastic
Detailed Explanation & Step-by-Step Solution:
This question tests the student’s ability to apply the concept of relative velocity to the formula for the coefficient of restitution, rather than just plugging numbers into a memorized equation blindly.
Step 1: Find the Relative Velocity of Approach (Before Collision) Before the crash, Car A is catching up to Car B. The speed at which Car A is closing the gap is their relative velocity.
Velocity of Car A (u₁) = 15 m/s
Velocity of Car B (u₂) = 5 m/s
Relative Velocity of Approach = u₁ - u₂ = 15 - 5 = 10 m/s
Step 2: Find the Relative Velocity of Separation (After Collision) After the crash, Car B is moving faster and pulling away from Car A. The speed at which the gap is widening is their relative velocity of separation.
Velocity of Car A (v₁) = 8 m/s
Velocity of Car B (v₂) = 12 m/s
Relative Velocity of Separation = v₂ - v₁ = 12 - 8 = 4 m/s
Step 3: Calculate the Coefficient of Restitution (e) The coefficient of restitution is simply the ratio of how fast they separate compared to how fast they approached each other.
Formula: e = (Relative Velocity of Separation) / (Relative Velocity of Approach)
e = 4 / 10
e = 0.4
Step 4: Classify the Collision
If e = 1, it is a perfectly elastic collision (no kinetic energy lost).
If e = 0, it is a perfectly inelastic collision (maximum kinetic energy lost, bodies stick together).
Since 0 < e < 1 (our e is 0.4), it is a Partially Inelastic collision. The cars bounced off each other, but some kinetic energy was lost to heat, sound, and the crumpling of the bumpers.
Why the incorrect options are wrong (Examiner Traps):
Trap (A): A student might carelessly divide the final velocities (12 / 15 or 8 / 15) or mix up the subtraction order, failing to use the relative velocities properly.
Trap (B): A student who assumes all collisions in physics problems are ideal will blindly guess perfectly elastic.
Trap (D): The math here is correct (0.4), but the classification is wrong. A “perfectly inelastic” collision requires e = 0, meaning the cars would have locked bumpers and moved together at the exact same final speed.
Multiple Choice Question 2
A small ball of mass ‘m’ and positive charge ‘+q’ is released from rest from a height ‘h’ above a smooth, perfectly rigid horizontal floor. In this space, there is a constant, uniform horizontal electric field ‘E’ pointing to the right. The acceleration due to gravity ‘g’ acts downwards.
If the coefficient of restitution for the collision between the ball and the floor is ‘e’, what is the total kinetic energy of the ball immediately after its first bounce?
(A) e²mgh (B) [ mgh + (q²E²h) / (mg) ] × e² (C) mgh + [ (q²E²h) / (mg) ] × e² (D) e²mgh + (q²E²h) / (mg)
Correct Answer: (D) e²mgh + (q²E²h) / (mg)
Detailed Explanation & Step-by-Step Solution:
This question tests a crucial concept: Perpendicular motions are completely independent of each other. The electric field only affects the horizontal motion, while gravity and the floor’s impact only affect the vertical motion.
Step 1: Find the time taken to hit the floor The ball is released from rest (initial vertical velocity = 0). The only downward acceleration is gravity (g). Using the 2nd equation of motion: s = ut + 1/2 at² h = 0 + 1/2 g t² Time of flight: t = √(2h / g)
Step 2: Find the vertical and horizontal velocities JUST BEFORE impact
Vertical Velocity (v_y): v_y = u + gt = 0 + g × √(2h / g) = √(2gh)
Horizontal Velocity (v_x): The horizontal acceleration is due to the electric force (F = qE). So, a_x = qE / m. v_x = u_x + a_x × t = 0 + (qE / m) × √(2h / g) = (qE / m) × √(2h / g)
Step 3: Apply the Coefficient of Restitution (e) JUST AFTER impact Here is the core physics trick: The floor is smooth (no friction), meaning it can only apply an impulse perpendicular to its surface (vertically).
New Vertical Velocity (v_y’): The floor absorbs some energy, so the vertical velocity reverses and reduces by a factor of ‘e’. v_y’ = e × √(2gh)
New Horizontal Velocity (v_x’): Because there is no friction to slow it down horizontally during the split-second bounce, the horizontal velocity remains completely unchanged! v_x’ = v_x = (qE / m) × √(2h / g)
Step 4: Calculate the New Kinetic Energy Total Kinetic Energy (KE) = 1/2 × m × (Total Velocity)² KE’ = 1/2 × m × [ (v_x’)² + (v_y’)² ] KE’ = 1/2 × m × [ (q²E² / m²) × (2h / g) + e² × (2gh) ]
Now, multiply the (1/2 × m) into the brackets:
First term: 1/2 × m × (q²E² × 2h) / (m²g) = (q²E²h) / (mg)
Second term: 1/2 × m × (e² × 2gh) = e²mgh
Total KE’ = e²mgh + (q²E²h) / (mg)
Trap Analysis (Why students get this wrong):
Trap (A): The student completely forgets about the horizontal electric field and treats this as a standard 1D free-fall collision, calculating only the vertical kinetic energy.
Trap (B): The “Blind Multiplication” trap. The student calculates the total kinetic energy just before impact and multiplies the entire equation by e². They fail to realize that the coefficient of restitution (e) only applies to the velocity component perpendicular to the collision surface!
Trap (C): The reversed trap. The student correctly separates the axes but mistakenly applies the energy loss (e²) to the horizontal motion instead of the vertical motion.
🚀Question 3 Mastering 2D Elastic Collisions: The Maximum Deflection Angle Problem
Question: A particle having a mass of 2 kg and moving with a speed of 6 m/s undergoes an elastic collision with another particle of mass 4 kg moving in the same direction with a speed of 2 m/s. The maximum angle of deviation (deflection) of the 2 kg mass particle will be _________.
Options:
37°
45°
53°
60°
Correct Answer: 3) 53°
🧠 Detailed Step-by-Step Solution & Concept Breakdown
To solve this problem without getting lost in pages of complex 2D momentum equations, we must use the Center of Mass (CM) Reference Frame. This is a powerful technique for advanced JEE and NEET problems.
Step 1: Understand the Goal We need to find the maximum possible angle (θ_max) that the 2 kg particle can deflect from its original straight-line path after bouncing off the 4 kg particle.
Step 2: Find the Velocity of the Center of Mass (V_cm) First, we calculate how fast the “center of mass” of this two-particle system is moving in the laboratory frame.
Formula: V_cm = (m₁u₁ + m₂u₂) / (m₁ + m₂)
m₁ = 2 kg, u₁ = 6 m/s
m₂ = 4 kg, u₂ = 2 m/s
V_cm = (2 × 6 + 4 × 2) / (2 + 4)
V_cm = (12 + 8) / 6 = 20 / 6 = 10/3 m/s
Step 3: Find the Particle’s Velocity IN the Center of Mass Frame (u₁’) Now, imagine you are sitting exactly on the center of mass, moving along with it at 10/3 m/s. How fast does the 2 kg particle appear to be moving toward you before the crash?
Formula: u₁’ = u₁ - V_cm
u₁’ = 6 - (10/3) = (18 - 10) / 3 = 8/3 m/s
Core Theory Check: In a perfectly elastic collision viewed from the CM frame, particles just bounce back or scatter with the exact same speed they approached with. They only change direction. So, after the collision, the speed of the 2 kg particle in the CM frame is still 8/3 m/s!
Step 4: The Maximum Deflection Vector Triangle To convert the velocity back to the real world (Lab Frame), we add the vector V_cm back to the scattered CM velocity vector. Think of it like drawing a circle: The vector V_cm goes straight horizontally. From the tip of V_cm, the scattered velocity (u₁’) can point in any direction, forming a circle of radius u₁’. The maximum angle of deflection (θ_max) in the real world occurs when the final lab velocity vector is exactly tangent to this circle.
A tangent creates a 90° right-angled triangle where:
The hypotenuse is V_cm
The opposite side to our angle is u₁’
Step 5: Apply the Shortcut Formula From that right-angled triangle, we get our master formula:
sin(θ_max) = u₁’ / V_cm
sin(θ_max) = (8/3) / (10/3)
The ‘3’s cancel out, leaving: sin(θ_max) = 8 / 10 = 4/5
Step 6: Find the Angle In physics, a right triangle with sides in the ratio of 3:4:5 has very specific, famous angles.
If sin(θ) = 3/5, the angle is 37°.
If sin(θ) = 4/5, the angle is 53°.
Therefore, the maximum angle of deflection is exactly 53°.
🔥 Exam Tips & Tricks for NEET/JEE Aspirants
1. Memorize the “Max Deflection” Formula! Do not waste time deriving this in the exam hall. If a question asks for the “maximum angle of deviation/deflection” in an elastic collision, instantly write down: 👉 sin(θ_max) = u₁’ / V_cm (Note: This shortcut only works if V_cm > u₁’. If u₁’ > V_cm, the particle can scatter at any angle, up to 180°!).
2. Master the 3-4-5 Magic Triangle! Examiners love the numbers 3, 4, and 5 because they create clean, non-decimal answers. You MUST memorize these two values by heart for mechanics and optics:
sin(37°) = 3/5 (and cos 37° = 4/5)
sin(53°) = 4/5 (and cos 53° = 3/5) If your final calculation reduces to 3/5 or 4/5, you know you are on the right track!
3. Why it happens conceptually: A heavier target (or a target moving in a way that creates a specific V_cm) acts like a partial wall. The 2 kg particle is moving faster (6 m/s) than the CM (3.33 m/s). Because the forward momentum of the system “drags” the final velocity vector forward, the lighter particle cannot bounce straight backward (180°) in the real world; its backward bounce in the CM frame gets overpowered by the forward drag of the whole system. Hence, its scattering is mathematically restricted to a “cone” of a maximum 53°!