Class 11 Physics · NCERT chapter 6 · 8 topics · free to read
Work
Work done by a force is W = Fd — the component of force along the displacement times the displacement. Work can be positive, negative, or zero depending on . For variable forces, work is the area under the F - x graph. NEET loves sign-of-work questions and work done by gravity, normal force, and friction.
W = Fd = F d | W = _ x_1 ^ x_2 F(x) dx
Work by a Constant Force
Work is the dot product of force and displacement vectors. The angle between them decides whether work is positive, negative, or zero.
Work done by a constant force F over a displacement d : W = F d = Fd , where is the angle between F and d . Only the component of force along the displacement does work. The perpendicular component does zero work.
W = Fd = F d
Work is a scalar quantity (dot product of two vectors) — it has magnitude but no direction
Work depends on three things: magnitude of force, magnitude of displacement, and angle between them
No displacement no work, regardless of how large the force is
Positive, Negative & Zero Work
Positive work means force aids the motion, negative work means force opposes it, and zero work means force is perpendicular to motion. This sign convention is a NEET staple.
The sign of work comes from . When 0° < 90° , > 0 → positive work (force has a component along displacement). When 90° < 180° , < 0 → negative work (force has a component opposite to displacement). When = 90° , 90° = 0 → zero work.
Force component along motion
Force component opposite to motion
Speeds up the body (or assists motion)
Slows down the body (or opposes motion)
Gravity on a falling body
Gravity on a rising body
Applied force pulling a box forward
Friction on a sliding box
Energy transferred TO the body
Energy transferred FROM the body
Zero work ( = 90° ): Normal force on a block sliding on a horizontal surface, centripetal force in circular motion, magnetic force on a moving charge
Positive work: Gravity on a freely falling object ( = 0° )
Negative work: Gravity on a ball thrown upward ( = 180° ), friction on a sliding body ( = 180° )
The TOTAL work by all forces = algebraic sum of work done by each force individually
Quick recall
A ball is thrown vertically upward. What is the sign of work done by gravity during the upward journey?
Negative. Gravity acts downward, displacement is upward → = 180° → W = mgh 180° = -mgh . Gravity opposes the motion, so it does negative work and the ball slows down.
Work by a Variable Force
When force varies with position, you can't just multiply F and d. Instead, work is the integral (area under the curve) of the force-displacement graph. This is how spring force problems are solved.
When force changes with position, split the displacement into tiny intervals dx where force is approximately constant. Work in each interval: dW = F(x) dx . Total work = sum of all these tiny contributions = integral.
W = _ x_1 ^ x_2 F(x) dx = Area under the F-x graph
For a constant force, the F - x graph is a horizontal line → area = rectangle = F d (recovers W = Fd )
For a spring ( F = kx ), the graph is a straight line through origin → area = triangle = 1 2 kx^2
Area above the x-axis = positive work; area below = negative work
This is the most general definition of work — W = Fd is just a special case for constant force
Quick recall
A force F = 3x^2 N acts on a body. Find work done in moving it from x = 0 to x = 2 m.
W = _0^2 3x^2 dx = [x^3 ]_0^2 = 8 - 0 = 8 J .
Work Done by Specific Forces
Gravity, friction, and normal force appear in almost every NEET mechanics problem. Knowing their work formulas by heart saves time and prevents sign errors.
Three forces dominate NEET problems. Each has a quick formula for work done:
Gravity: W_g = mgh (falling) or W_g = -mgh (rising). Only depends on vertical height change, NOT the path taken
Friction (kinetic): W_f = -f_k d = - _k N d . Always negative because friction opposes motion
Normal force (on horizontal surface): W_N = 0 . Normal force is perpendicular to displacement, so 90° = 0
Tension in an inextensible string (constant length): Work by tension on the system = 0 (internal force, no net displacement)
W_ gravity = mgh | W_ friction = - _k N d | W_ normal = 0
Kinetic Energy & Work-Energy Theorem
Kinetic energy is the energy a body possesses due to its motion: KE = 1 2 mv^2 . The work-energy theorem states that the net work done on a body equals its change in kinetic energy. This theorem is a powerful shortcut for NEET — it lets you skip force and acceleration entirely.
KE = 1 2 mv^2 | W_ net = KE = 1 2 mv_f^2 - 1 2 mv_i^2
Kinetic Energy
Any moving object has kinetic energy proportional to its mass and the square of its speed. The v^2 dependence means doubling speed quadruples the energy — which is why high-speed crashes are so devastating.
Kinetic energy is the energy a body has because it is moving: KE = 1 2 mv^2 . It is always non-negative (mass is positive, v^2 is non-negative). It depends on the square of velocity — so doubling speed means 4 the kinetic energy.
KE = 1 2 mv^2
KE is a scalar — always positive or zero, never negative
SI unit: joule (J). Dimensions: [ML^2T^ -2 ] — same as work
KE v^2 : doubling speed → 4 KE; tripling speed → 9 KE
KE m : doubling mass → 2 KE (linear, not quadratic)
KE depends on the frame of reference — a person sitting in a moving train has KE relative to the ground but zero KE relative to the train
Work-Energy Theorem
The total work done by ALL forces on a body equals the change in its kinetic energy. This bypasses the need to find acceleration — go directly from forces to energy.
The work-energy theorem states: the net work done by all forces acting on a body equals the change in its kinetic energy. W_ net = KE = KE_f - KE_i = 1 2 mv_f^2 - 1 2 mv_i^2 .
W_ net = 1 2 mv_f^2 - 1 2 mv_i^2
W_ net includes work done by ALL forces: gravity, friction, applied force, normal force, spring force — everything
Positive net work → KE increases → body speeds up
Negative net work → KE decreases → body slows down
Zero net work → KE unchanged → speed remains constant
The theorem is valid in both inertial and non-inertial frames (if pseudo forces are included)
Quick recall
A 2 kg block initially at rest is pushed by a 10 N force over 5 m on a frictionless surface. Find the final speed.
W_ net = F d = 10 5 = 50 J. Using work-energy theorem: 50 = 1 2 (2)v_f^2 - 0 , so v_f^2 = 50 → v_f = 50 = 5 2 7.07 m/s .
KE-Momentum Relation
The relation KE = p^2/2m connects kinetic energy and momentum. This shortcut is tested repeatedly in NEET — especially in comparison problems.
Since p = mv , we get v = p/m . Substituting into KE = 1 2 mv^2 :
KE = p^2 2m p = 2m KE
Same KE, different masses: Heavier body has MORE momentum ( p = 2mKE , so p m )
Same momentum, different masses: Lighter body has MORE KE ( KE = p^2/2m , so KE 1/m )
Same force, same time: Both get same impulse → same p . Lighter body gets more KE
These three comparison cases cover ~90% of NEET questions on this topic
Quick recall
Two bodies of mass m and 4m have equal kinetic energy. What is the ratio of their momenta?
p = 2mKE . Ratio: p_1 p_2 = m 4m = 1 4 = 1:2 . The heavier body has twice the momentum.
Potential Energy
Potential energy is stored energy due to position or configuration. Gravitational PE = mgh (height-based) and spring PE = 1 2 kx^2 (deformation-based) are the two forms tested in NEET. The choice of reference point is arbitrary but must be consistent.
U_g = mgh | U_s = 1 2 kx^2
Gravitational PE
When you lift an object to a height h , you store energy mgh in the Earth-object system. This energy is released when the object falls. The reference point for h is your choice — only changes in PE matter physically.
Gravitational potential energy of a body of mass m at height h above a chosen reference level: U_g = mgh . This is the work done by you (against gravity) to lift the body from the reference level to height h , OR equivalently, the work gravity will do if the body falls back down.
U_g = mgh (valid near Earth's surface, where g is approximately constant)
PE is a property of the system (Earth + object), not just the object alone
Only changes in PE ( U ) are physically meaningful — not absolute values
Reference point (where h = 0 ) is arbitrary: ground, table top, bottom of incline — your choice
U_g = mgh assumes constant g — valid near Earth's surface. For large heights, use U = - GMm r
PE can be positive, negative, or zero depending on your reference point
Spring PE
A spring compressed or stretched by x from its natural length stores PE = 1 2 kx^2 . This comes from integrating the spring force F = -kx (Hooke's law). The x^2 means compression and extension store the same energy.
A spring with spring constant k , displaced by x from its natural (equilibrium) length, stores elastic PE: U_s = 1 2 kx^2 . This is derived from the work done against the spring force F = -kx (Hooke's law). Since force varies linearly with x , we integrate: W = _0^x kx' dx' = 1 2 kx^2 .
Spring PE is always non-negative ( x^2 0 ) — both compression and extension store energy
k = spring constant (N/m) — measures stiffness. Larger k → stiffer spring → more PE for same x
At natural length ( x = 0 ): PE = 0 (the natural length is the built-in reference point)
PE x^2 : doubling the stretch → 4 the stored energy
The graph of U_s vs x is a parabola opening upward with minimum at x = 0
Quick recall
A spring with k = 200 N/m is compressed by 0.1 m. How much PE is stored?
U_s = 1 2 (200)(0.1)^2 = 1 2 (200)(0.01) = 1 J .
Relationship Between PE and Force
Force is the negative gradient of potential energy. This connects the PE function to the force at every point and is key to understanding equilibrium and potential energy curves.
For a conservative force in one dimension, the force is related to PE by: F(x) = - dU dx . The force points in the direction of decreasing PE — objects naturally move from high PE to low PE (like a ball rolling downhill).
F(x) = - dU dx
The negative sign means force acts toward lower PE — nature prefers minimum energy
At points where dU/dx = 0 : force = 0 → these are equilibrium positions
Stable equilibrium: PE is minimum (valley). d^2U/dx^2 > 0 . Small push → returns to equilibrium
Unstable equilibrium: PE is maximum (hilltop). d^2U/dx^2 < 0 . Small push → moves away
Neutral equilibrium: PE is constant. d^2U/dx^2 = 0 . Small push → stays in new position
Quick recall
Given U(x) = 5x^2 - 20x + 10 J, find the equilibrium position and check if it's stable.
F = -dU/dx = -(10x - 20) = 0 → x = 2 m. Check: d^2U/dx^2 = 10 > 0 → minimum → stable equilibrium.
Conservation of Mechanical Energy
When only conservative forces do work, the total mechanical energy E = KE + PE remains constant. Energy transforms between KE and PE but the sum never changes. This is the most powerful problem-solving tool in mechanics — it converts complex force/acceleration problems into simple algebra.
E = KE + PE = constant | 1 2 mv_1^2 + U_1 = 1 2 mv_2^2 + U_2
The Conservation Principle
If the only forces doing work are conservative (gravity, spring), then KE and PE trade back and forth while their sum stays fixed. This is conservation of mechanical energy.
Total mechanical energy E = KE + PE . If only conservative forces (gravity, spring) do work on the system, then E is constant: KE_i + PE_i = KE_f + PE_f . Energy is not created or destroyed — it just transforms between kinetic and potential forms.
1 2 mv_i^2 + U_i = 1 2 mv_f^2 + U_f
Conservation applies ONLY when non-conservative forces (friction, air resistance) do zero work
Normal force and tension (inextensible string) do zero work — they don't break conservation
If friction is present: KE_i + PE_i = KE_f + PE_f + |W_ friction | — energy is 'lost' to heat
The reference point for PE doesn't matter — as long as you use the SAME reference at both points
Conservation of energy is ALWAYS true (First Law of Thermodynamics). Conservation of MECHANICAL energy is the special case with no friction.
Classic Applications
Conservation of energy turns seemingly complex problems into simple algebra. These four setups cover the vast majority of NEET energy conservation questions.
Setup 1: Free fall from height h . At top: KE = 0 , PE = mgh . At bottom: KE = 1 2 mv^2 , PE = 0 . Conservation: mgh = 1 2 mv^2 → v = 2gh . Notice: mass cancels! All objects fall with the same speed (same height, no air resistance).
Setup 2: Pendulum. At extreme position: v = 0 , h = h_ . At lowest point: v = v_ , h = 0 . Conservation: mgh_ = 1 2 mv_ ^2 → v_ = 2gh_ .
Setup 3: Spring-mass system. At maximum compression x : KE = 0 , PE = 1 2 kx^2 . At natural length: KE = 1 2 mv^2 , PE = 0 . Conservation: 1 2 kx^2 = 1 2 mv^2 → v = x k/m .
Setup 4: Block on frictionless incline. Height dropped = h . Conservation: mgh = 1 2 mv^2 → v = 2gh . Same answer as free fall — the angle of the incline doesn't matter! Only the vertical height change matters.
Free fall: v = 2gh — independent of mass (Galileo's discovery!)
Pendulum: v_ at the lowest point; height = L - L = L(1 - ) for angle
Spring: speed is maximum at natural length, zero at maximum compression/extension
Incline: only VERTICAL height matters, not the length of the incline or the angle
Quick recall
A pendulum of length 1 m is released from 60° to the vertical. Find speed at the lowest point. ( g = 10 m/s ^2 )
h = L(1 - 60°) = 1(1 - 0.5) = 0.5 m. v = 2gh = 2(10)(0.5) = 10 3.16 m/s .
Energy Conservation with Friction
When friction is present, mechanical energy is NOT conserved — some converts to heat. The modified energy equation includes friction's work. This is how NEET tests 'real-world' energy problems.
When non-conservative forces (friction, air drag) act, mechanical energy decreases. The 'lost' energy becomes heat, sound, or deformation. The modified conservation equation:
W_ friction = _k N d is always positive in this equation (energy REMOVED from the system)
Total energy is STILL conserved — but mechanical energy converts to thermal energy
If friction is present, the body reaches the bottom with LESS speed than 2gh
On a rough incline of length l : mgh = 1 2 mv^2 + _k mg l
Quick recall
A 5 kg block slides down a 3 m high rough incline ( 30° , _k = 0.2 ). Find speed at the bottom. ( g = 10 m/s ^2 )
Length of incline: l = h/ 30° = 3/0.5 = 6 m. Friction work: _k mg 30° l = 0.2(5)(10)(0.866)(6) = 51.96 J. Energy equation: mgh = 1 2 mv^2 + 51.96 → 150 = 2.5v^2 + 51.96 → v^2 = 39.2 → v 6.26 m/s .
Conservative & Non-Conservative Forces
A conservative force does work that depends only on the initial and final positions — not the path. Gravity and spring force are conservative; friction and air resistance are non-conservative. The distinction determines whether mechanical energy is conserved or 'lost' to heat.
F _ cons d r = 0 | W_ cons = - U
Conservative Forces
A force is conservative if the work done by it depends only on the initial and final positions, not on the path taken. Equivalently, the work done in any closed loop is zero. A potential energy can be defined only for conservative forces.
A force is conservative if: (1) Work done depends ONLY on initial and final positions (not the path), OR equivalently (2) Work done in a complete closed loop is zero. Both conditions are equivalent — if one is true, the other must be true.
W_ conservative = - U = -(U_f - U_i) | F _ cons d r = 0
Examples of conservative forces: gravitational force, spring force (elastic), electrostatic force
For conservative forces, a potential energy function U can be defined: W = - U
Work done in a closed loop = 0 (go from A to B by any path, return to A — net work is zero)
Conservative forces depend only on position — they don't depend on velocity or time
Quick recall
You lift a 2 kg book from the floor to a shelf (2 m high) via three different paths: straight up, along a ramp, and through a spiral staircase. Compare work done by gravity in all three.
Same in all three: W_g = -mgh = -(2)(10)(2) = -40 J. Gravity is conservative — only the height difference matters, not the path.
Non-Conservative Forces
Friction and air resistance are non-conservative — the work they do depends on the path taken. Longer path = more energy lost to heat. No potential energy can be defined for these forces.
A force is non-conservative if the work done by it depends on the path taken. Equivalently, the work done in a closed loop is NOT zero. Friction is the classic example — push a box in a circle back to its starting point, and friction does negative work throughout (you've lost energy to heat even though displacement is zero).
Work is path-independent
Work is path-dependent
Work in closed loop = 0
Work in closed loop ≠ 0
PE can be defined ( W = - U )
PE cannot be defined
Gravity, spring, electrostatic
Friction, air resistance, viscous drag
Mechanical energy is conserved
Mechanical energy is NOT conserved
Friction always does negative work regardless of direction — so in a closed loop, work by friction is negative (not zero)
Longer path = more distance = more negative work by friction = more energy lost to heat
Non-conservative forces convert mechanical energy into other forms (heat, sound, deformation)
Total energy is still conserved (First Law of Thermodynamics) — but MECHANICAL energy decreases
Power
Power is the rate of doing work: P = W/t (average) or P = F v (instantaneous). It tells you how FAST energy is being delivered or consumed. SI unit is watt (W). NEET tests unit conversions (HP, kW), and P = Fv applications for vehicles climbing hills.
P = W t = dW dt = F v
Average & Instantaneous Power
Average power is total work divided by total time. Instantaneous power is the rate of doing work at a specific instant. The formula P = Fv is the workhorse for NEET power problems.
Average power: P_ avg = W t — total work done divided by total time taken. Instantaneous power: P = dW dt — the rate of doing work at a specific moment. Since dW = F d s and d s /dt = v , we get the extremely useful form: P = F v = Fv .
P_ avg = W t | P_ inst = dW dt = F v = Fv
SI unit: watt (W) = J/s = kg m ^2 /s ^3
1 horsepower (HP) = 746 W 750 W (used for engines and motors)
1 kilowatt-hour (kWh) = 3.6 10^6 J = 3.6 MJ (unit of ENERGY, not power — used in electricity bills)
Power is a scalar (dot product of F and v )
Dimensions: [P] = [ML^2T^ -3 ]
P = Fv Applications
The formula P = Fv connects engine power to the force and speed of a vehicle. At constant power, force and velocity are inversely related — this explains why cars slow down on hills and why low gear gives more force.
For a vehicle moving at velocity v with engine force F (in the direction of motion): P = Fv . At constant power P : F = P/v . This means as speed increases, the available force decreases — and vice versa. This is why vehicles shift to lower gears (lower v , higher F ) on steep hills.
P = Fv F = P v (at constant power)
At constant power, maximum speed is reached when driving force = resistive force (net force = 0, no acceleration)
Maximum speed: v_ = P/(resistive force)
Lift/elevator: Power to lift at constant v : P = mgv (force = weight, velocity = constant speed of lift)
Pump: Power to pump water to height h : P = mgh t = V g h / t (where V/t = volume flow rate)
Quick recall
A pump lifts 200 kg of water per minute to a height of 10 m. Find the power of the pump. ( g = 10 m/s ^2 )
W = mgh = 200 10 10 = 20 , 000 J. Time = 60 s. P = W/t = 20 , 000/60 = 333.3 W .
Quick recall
A car of power 50 kW faces a constant resistance of 1000 N. Find its maximum speed.
At max speed, engine force = resistance (no acceleration). P = Fv_ → v_ = P/F = 50 , 000/1000 = 50 m/s .
Collisions in 1D
Collisions are classified by whether kinetic energy is conserved: elastic (KE conserved), inelastic (KE not conserved), and perfectly inelastic (bodies stick together). Momentum is ALWAYS conserved in all collisions. NEET heavily tests velocity formulas for elastic collisions, coefficient of restitution, and energy loss calculations.
All collisions conserve momentum. The difference is in kinetic energy: elastic collisions conserve it, inelastic don't, and perfectly inelastic (where bodies stick) lose the maximum possible KE.
In any collision, momentum is always conserved (Newton's third law ensures internal forces cancel). The classification depends on what happens to kinetic energy:
Elastic collision: Both momentum AND KE are conserved. Bodies bounce off. Examples: billiard balls (approximately), atomic collisions
Inelastic collision: Momentum conserved, KE NOT conserved (some KE → heat, sound, deformation). Most real-world collisions are inelastic
Perfectly inelastic collision: Momentum conserved, bodies STICK together after collision ( v_1 = v_2 = v_ common ). Maximum possible KE loss
In ALL three types, momentum is conserved. Never doubt this — it's Newton's third law at work
KE conserved
KE NOT conserved (maximum loss)
Bodies separate after collision
Bodies stick together
e = 1
e = 0
Two equations (momentum + KE)
One equation (momentum only)
Billiard balls, atomic collisions
Bullet embedding in block, clay balls
Elastic Collision Formulas
For a 1D elastic collision, combining momentum and KE conservation gives direct formulas for final velocities. Special cases (equal masses, one at rest, heavy-light) are NEET favorites.
For elastic collision between mass m_1 (velocity u_1 ) and mass m_2 (velocity u_2 , often u_2 = 0 ): apply both momentum conservation and KE conservation simultaneously. The resulting velocity formulas are:
In a perfectly inelastic collision, the two bodies move as one unit after the collision. Only momentum conservation is used (KE is not conserved). The common velocity formula and energy loss formula are NEET staples.
After a perfectly inelastic collision, both bodies move with a common velocity v . Using momentum conservation: m_1 u_1 + m_2 u_2 = (m_1 + m_2)v .
v = m_1 u_1 + m_2 u_2 m_1 + m_2
The KE lost in a perfectly inelastic collision (with u_2 = 0 ):
Common velocity v is always LESS than u_1 (the faster body slows down)
The fraction of KE lost = m_2 m_1 + m_2 (when u_2 = 0 )
Bullet-block problem: Bullet (mass m , speed u ) embeds in block (mass M , at rest). v = mu m+M . KE lost = M m+M 1 2 mu^2
If m_1 = m_2 : half the initial KE is lost to deformation/heat
Perfectly inelastic collision has the MAXIMUM possible energy loss for given masses and velocities
Quick recall
A 2 kg ball moving at 3 m/s collides head-on with a 1 kg ball moving at 6 m/s in the opposite direction. They stick together. Find the common velocity.
Taking rightward as positive: m_1u_1 + m_2u_2 = (m_1 + m_2)v . 2(3) + 1(-6) = 3v → 6 - 6 = 3v → v = 0 m/s . They come to rest!
Coefficient of Restitution
The coefficient of restitution e is the ratio of relative velocity of separation to relative velocity of approach. e = 1 for elastic, e = 0 for perfectly inelastic, and 0 < e < 1 for real-world collisions.
The coefficient of restitution e measures how 'bouncy' a collision is. It is defined as the ratio of the relative speed of separation to the relative speed of approach along the line of impact.
e = |v_2 - v_1| |u_1 - u_2| = relative speed of separation relative speed of approach
e = 1 : Perfectly elastic — relative speed is fully restored after collision
e = 0 : Perfectly inelastic — bodies stick, no separation
e is dimensionless and depends on the materials of colliding bodies
For a ball dropped from height h_1 bouncing to height h_2 : e = h_2/h_1
e = h_2 h_1 (ball dropped from h_1, bounces to h_2)
Quick recall
A ball dropped from 5 m bounces to 3.2 m. Find the coefficient of restitution.
e = h_2/h_1 = 3.2/5 = 0.64 = 0.8 .
Collisions in 2D
In oblique (2D) collisions, momentum is conserved independently along both x and y directions. The collision is analyzed by resolving velocities into components along and perpendicular to the line of impact. NEET tests glancing collisions of equal masses (perpendicular scattering) and basic 2D momentum conservation.
In a 2D collision, the total momentum is a vector. Conservation of momentum gives two independent equations — one for the x-component and one for the y-component. This is enough to solve many NEET problems.
In a 2D collision, momentum conservation is a vector equation: m_1 u _1 + m_2 u _2 = m_1 v _1 + m_2 v _2 . This gives two independent scalar equations — one for x and one for y.
x: m_1 u_ 1x + m_2 u_ 2x = m_1 v_ 1x + m_2 v_ 2x
y: m_1 u_ 1y + m_2 u_ 2y = m_1 v_ 1y + m_2 v_ 2y
Choose your x-axis along the initial direction of the incoming body (simplifies u_ 1y = 0 , u_ 2x = u_ 2y = 0 if target is at rest)
This gives two equations. With KE conservation (elastic), you get a third equation — enough for most problems
The coefficient of restitution applies ONLY along the line of impact (the line joining centers at contact), NOT perpendicular to it
Components perpendicular to the line of impact are unchanged in the collision
Equal Mass Elastic Scattering
When two equal masses undergo an elastic collision (with one initially at rest), the two bodies always scatter at right angles ( 90° ) to each other. This is a famous result that NEET tests directly.
Key result: In an elastic 2D collision between equal masses (one at rest), the angle between the final velocity vectors of the two bodies is always 90° . This is the '90-degree scattering rule' — one of the most elegant results in collision physics.
_1 + _2 = 90° (elastic, equal masses, one at rest)
Proof sketch: Momentum conservation (vector): m u = m v _1 + m v _2 → u = v _1 + v _2 . Squaring: u^2 = v_1^2 + v_2^2 + 2v_1 v_2 (where = angle between v _1 and v _2 ). KE conservation: u^2 = v_1^2 + v_2^2 . Comparing: 2v_1 v_2 = 0 . Since v_1, v_2 0 : = 0 → = 90° .
This result holds ONLY for elastic collisions with equal masses and one body initially at rest
The angle between the scattered bodies is exactly 90° regardless of the impact parameter (glancing or near head-on)
For a head-on elastic collision ( _1 = 0° ): one ball stops, the other moves forward (1D equal-mass result)
For unequal masses, the angle is NOT 90° — it depends on the mass ratio
Quick recall
A ball moving at 5 m/s hits an identical stationary ball. After elastic collision, one moves at 30° to the original direction. Find the direction of the other.
_1 + _2 = 90° → _2 = 90° - 30° = 60° to the original direction (on the opposite side).