Collisions (1D and 2D) Collisions (1D and 2D) When two bodies hit or push each other for a short time, we call it a collision. During that brief contact, very large forces act, but only for a tiny interval, so the quantities that matter are momentum and impulse. In an isolated system (no net external impulse), total momentum is conserved in every collision. Kinetic energy, however, may or may not be conserved: perfectly elastic collisions conserve it, perfectly inelastic collisions do not, and the general case lies in between. To quantify how “bouncy” a collision is, we define the coefficient of restitution e as the ratio of relative speed after to relative speed before impact, measured along the line of impact (the common normal at the contact). In 1D head-on collisions, all motion lies along a single axis, so we combine momentum conservation with e to determine final speeds. In 2D, we decompose velocities into components normal and tangential to the line of impact: the normal components change according to e , while tangential components remain unchanged if the surfaces are smooth (no friction). This component method makes oblique collisions predictable, from billiard-ball deflections to balls bouncing off a wall. Mastering sign conventions, directions, and which components to apply e to are the keys to scoring collision questions correctly. remember Think of a cricket ball hitting a concrete wall: it squashes a bit, then springs back. Along the normal (perpendicular) direction to the wall, it reverses with some loss captured by e . Along the tangential direction (parallel to the wall), its speed barely changes if the wall is smooth. A short-duration interaction between bodies where large contact forces act and momentum may be exchanged. Collision The common normal at the point of contact. Only the components of velocity along this line enter Newton’s restitution law. Line of impact Ratio of relative speed of separation to relative speed of approach along the line of impact; 0 e 1 for passive materials. Coefficient of restitution (e) A collision in which both momentum and kinetic energy are conserved (effectively e=1 for head-on impact). Elastic collision Momentum is conserved but kinetic energy is not. For a perfectly inelastic collision, bodies stick together ( e=0 for head-on impact). Inelastic collision A 1D collision where initial velocities are collinear with the line of impact. Head-on collision Oblique collision A 2D collision where initial velocities have components both along and perpendicular to the line of impact. Change in momentum caused by a force acting over a short time, p = F ,dt . Impulse Sign convention and axes: In 1D, choose the positive axis along the initial motion of object 1. Keep directions consistent in all equations. In 2D, first identify the line of impact at the instant of contact. Resolve each velocity into (i) normal component along the line of impact and (ii) tangential component perpendicular to it. Apply the restitution relation and momentum conservation only to the normal components when surfaces are smooth; the tangential components remain unchanged (no frictional impulse). Recombine components to obtain final velocities. 1D head-on: u 1,u 2 are initial velocities, v 1,v 2 are final velocities, measured along the line of impact. Directions with sign matter. Newton’s Law of Restitution Use the restitution formula only along the line of impact. In 2D, replace u and v by their normal components to the contact surface. tip Short contact time; external impulses (e.g., weight) are negligible compared to contact impulse. Interaction force acts predominantly along the line of impact (smooth contact). Deformation and restitution phases are symmetric up to a material-dependent factor captured by e. Newton’s law of restitution: e equals the ratio of relative separation speed to relative approach speed along the line of impact. e = v 2 - v 1 u 1 - u 2 Relative speed of approach along the line of impact (1D sign convention). Relative speed of separation along the line of impact. By definition, 0 e 1 for passive materials. With a consistent sign convention, the absolute values can be dropped to get the working 1D formula. Physical meaning of e: During contact, bodies compress and then attempt to regain shape. If no mechanical energy is lost to heat, sound, or permanent deformation, the separation speed equals the approach speed and e=1 . If maximum energy is lost (stick together), the separation speed is zero and e=0 . Most real collisions have 0<e<1 . Note that e is primarily a property of the contacting materials and geometry and can depend on impact speed; within NEET scope we treat it as a given constant for a collision. Momentum Conservation (Vector Form) Valid for the system of colliding bodies when external impulse during contact is negligible. For isolated systems, the total momentum before and after a collision remains constant, provided external forces are negligible. When is momentum conserved? During the short collision, the contact impulse is huge, but external forces like gravity act for the same short time, giving a tiny external impulse. If the ground is smooth and the interval is tiny, we can neglect external impulse and apply momentum conservation to the system. Caution: If a fixed wall is involved, model the wall as an external agent that can supply impulse; then you do not conserve the two-body momentum (ball + wall) unless you include the wall or Earth as part of the system. 1D head-on analysis: For masses m 1 and m 2 with initial velocities u 1 and u 2 and final velocities v 1 and v 2 , write two equations along the line of impact. Momentum conservation gives m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 . The restitution relation gives e = v 2 - v 1 u 1 - u 2 . Solving these yields v 1 and v 2 for any e [0,1] . For e=1 we obtain the elastic case; for e=0 the perfectly inelastic case where bodies stick and share a common final speed. v 1 = (m 1 - m 2)u 1 + 2 m 2 u 2 m 1 + m 2 , v 2 = 2 m 1 u 1 + (m 2 - m 1)u 2 m 1 + m 2 Isolated two-body system during collision (no net external impulse). Head-on (1D) motion along a single axis. Elastic collision: kinetic energy and momentum are conserved. Final velocities in a 1D perfectly elastic collision Momentum conservation. Kinetic energy conservation. Solve the pair of equations. Symmetric expression for v 2 . Useful limit checks: if m 1 m 2 , v 1 u 1 , small deflection; if m 1=m 2 and u 2=0 , they exchange speeds. Elastic 1D Results This equation determines the final speed of the first mass after a perfectly elastic, one-dimensional collision. Interpretation checks: If m 1=m 2 in a head-on elastic hit with u 2=0 , the incoming body stops and the target moves with u 1 . If m 1 m 2 (a truck hits a ping-pong ball elastically), the truck’s speed barely changes, while the light ball can rebound with roughly twice the truck’s speed relative to the truck. These limits are sanity checks for your final formulas. neet-alert Sign trap: In e= v 2-v 1 u 1-u 2 , use algebraic velocities along the chosen positive axis. Don’t insert magnitudes or flip signs mid-way. Another trap: ‘distance in the nth second’ analog mistake—here it’s the relative speed between bodies before/after, not each body’s speed alone. Energy loss in inelastic collisions: Combine momentum conservation with restitution to show that the kinetic energy lost depends on the relative speed before impact. Introducing reduced mass = m 1 m 2 m 1+m 2 , the total kinetic energy decreases by an amount proportional to (1-e 2) and to the square of the initial relative speed of approach. This result is powerful for quick percentage-loss questions without solving for individual final speeds. Momentum conservation. Restitution relation. Define kinetic energy loss. Algebraic rearrangement using center-of-mass identities (or by elimination). Substitute v 2 - v 1 = e(u 1 - u 2) . Kinetic energy loss in a general 1D collision with restitution e Two-body, 1D head-on collision. No external impulse; momentum is conserved. Restitution coefficient e relates normal components. K i - K f = 1 2 (1 - e 2)(u 1 - u 2) 2 Zero loss for e=1 ; maximum loss for e=0 (perfectly inelastic). Energy Loss with Restitution This formula applies to two-body collisions where the interaction is inelastic, causing a loss of mechanical energy. Perfectly inelastic collision (stick together): The final common speed is the momentum-weighted average, v com = m 1 u 1 + m 2 u 2 m 1+m 2 . The kinetic energy loss is the maximum possible for given masses and initial relative speed. This model fits soft clay lumps colliding or a bullet embedding in a block (ignoring heat and deformation details). Applies when the bodies stick together after impact ( e=0 for head-on). Common Speed (Perfectly Inelastic) This equation determines the common final speed of two masses after they undergo a perfectly inelastic collision. m/s Use elastic 1D formulas or momentum plus KE conservation. Final speeds v 1 and v 2 . Equal masses collide elastically; one initially at rest. easy Classic outcome: exchange of speeds for equal masses. With m 1=m 2 and u 2=0 , the incident ball stops. The target leaves with the incident speed. masses: m 1=m 2=0.20 , kg initial speeds: u 1=4.0 , m/s , u 2=0 elastic collision: e=1 Interpretation: With equal masses in an elastic head-on collision, the moving one transfers all its kinetic energy and momentum to the other. This is why a moving billiard ball can stop dead after striking a similar stationary ball straight on. Any deviation from perfect elasticity or head-on alignment will change this clean exchange. Final speeds v 1 and v 2 . 1D inelastic collision with given e between unequal masses. m/s Use momentum conservation and e= v 2-v 1 u 1-u 2 . masses: m 1=2.0 , kg , m 2=1.0 , kg initial speeds along +x: u 1=5.0 , m/s , u 2=1.0 , m/s coefficient of restitution: e=0.60 medium Standard restitution application. Total momentum equals 11 , kg ,m/s . Relative speed relation. From restitution. From momentum. Algebra. Final speed of mass m 2 . Applies to two-dimensional collisions between objects where the collision surface is smooth (no friction) and the initial and final velociti Check reasonableness: The lighter body ( 1 , kg ) speeds up from 1.0 to about 5.3 , m/s , while the heavier one slows from 5.0 to about 2.9 , m/s . The relative speed after is v 2-v 1 2.4 , m/s , which is 0.60 times the approach speed 4.0 , m/s , consistent with e=0.60 . Oblique collisions: For smooth spherical bodies, the contact force has no tangential component, so the tangential velocity components (perpendicular to the line of impact) are unchanged by the collision. Only the normal components change, as per momentum conservation (along normal) and restitution. This yields a clean algorithm: resolve, apply, and recombine to get the final direction and speed. Component Relations (Smooth Contact) Tangential components unchanged; normal components follow Newton’s restitution law. If contact is rough, an impulsive friction may act and change tangential components too. That is beyond NEET’s standard scope. For walls and floors modeled as smooth and fixed, only the ball’s normal component flips in sign and shrinks in magnitude by factor e , while tangential remains the same. This directly gives rebound angles for a smooth bounce from a wall. Speed before impact: u=10 , m/s Angle of incidence with the normal: =30 Coefficient of restitution (wall–ball): e=0.80 Smooth wall (no tangential impulse) hard Normal component toward the wall. Tangential component along the wall. Magnitude after rebound; direction reverses, so normal component is outward. Smooth wall: tangential unchanged. Rebound speed. Angle with the normal after collision. Oblique wall bounce using restitution. Ball strikes a smooth vertical wall obliquely and rebounds. Rebound speed v and angle with the normal . m/s, degree Resolve along normal (perpendicular to wall) and tangential (parallel to wall). Equal-mass elastic scattering at 90°: In an elastic collision where a moving sphere hits an identical stationary sphere, the two final velocity vectors are at right angles to each other, provided the impact is not head-on (i.e., it is a glancing collision). This famous 90° rule follows from conserving momentum and kinetic energy in 2D and appears frequently in worked examples and lab demonstrations with billiard balls. v 1 v 2 For m 1=m 2 and elastic collision with target initially at rest, the final velocities are perpendicular: v 1 v 2=0 . Two identical masses m 1=m 2=m . Target initially at rest: u 2= 0 . Elastic collision in 2D; smooth contact. Vector momentum conservation. Kinetic energy conservation. Expand the dot product. Therefore, the angle between v 1 and v 2 is 90 . y-direction Vector triangle for equal-mass elastic scattering: u1 = v1 + v2 with right angle between v1 and v2. Momentum triangle closes with a right angle between v1 and v2 for identical masses in an elastic glancing hit. control u1 v1 derived derived v2 x-direction custom Origin u1 x Initial velocity u1 u1 y v1 y Final v1 v1 x v1 y + v2 y v1 + v2 = u1 v1 x + v2 x Choose a consistent axis (1D) or identify the line of impact (2D). Resolve velocities into normal and tangential components. Write momentum conservation along the line of impact (and overall vector conservation if no external impulse). Apply Newton’s law of restitution to normal components: e= v 2n -v 1n u 1n -u 2n . Use tangential conditions: for smooth contact, v 1t =u 1t and v 2t =u 2t . Solve simultaneously for unknown components; then recombine to get magnitudes and directions. Check limits ( e 1 , e 0 , equal-mass cases) and units. Verify that relative speed after equals e times the relative speed before (normal components). Steps to solve collision problems Quick checks and heuristics If one mass is much larger than the other, the heavy mass changes speed very little. For equal masses and elastic collision, velocities are exchanged in 1D; in 2D, the final velocities are perpendicular if the target was at rest. Perfectly inelastic: immediately write v com =(m 1 u 1 + m 2 u 2)/(m 1+m 2) . Wall bounce (smooth): flip only the normal component and scale by e ; keep tangential unchanged. Always track directions with signs; do not plug magnitudes into restitution. Collision Types Type Momentum Conserved? KE Conserved? Restitution ( e ) Physical Outcome Elastic e=1 bounces free, Inelastic e is in-between, and Perfectly Inelastic e=0 sticks together like glue. Elastic Collision Yes Yes e = 1 Bodies bounce back without loss of total KE or permanent deformation. Inelastic Collision Yes No 0 < e < 1 Bodies separate after collision but some KE is converted to heat or sound. Perfectly Inelastic Collision Yes No (Max Loss) e = 0 Bodies stick together and move with a common velocity V c after impact. Super-elastic Collision Yes No (KE Increases) e > 1 Internal potential energy is released, increasing the total KE of the system. collision types remember Impulse acts along the line of impact for smooth contact. That’s why only normal velocity components are altered in such collisions. Center-of-mass (COM) frame trick: In elastic collisions, speeds measured in the COM frame are unchanged; velocities reverse along the line of impact. Compute in COM, then transform back. tip Center-of-mass frame: Let V cm = m 1 u 1 + m 2 u 2 m 1+m 2 . Define pre-collision velocities in COM frame as u i' = u i - V cm . For an elastic collision, | v i'|=| u i'| and the relative velocity reverses along the line of impact. After finding v i' in the COM frame, add back V cm to obtain lab-frame velocities v i= v i' + V cm . This method often reduces algebra and clarifies geometry. Compute in COM frame, then transform back to lab frame. COM Transformations Determines the average spatial location of the system, which serves as the reference point for analyzing relative velocities during collisions. In the COM frame for elastic collisions, the two bodies approach and separate with equal and opposite velocities (same magnitudes). Geometrically, u 1' + u 2' = 0 and v 1' + v 2' = 0 . Therefore, a collision acts like a mirror reflection of the relative velocity about the tangential direction at impact. This picture helps explain the 90° scattering for equal masses when one was initially at rest. Equal masses, elastic glancing collision; find final speeds using the COM frame. Possible | v 1| and | v 2| and the angle between them. m/s Use V cm = u 1 2 =(2,0) , equal masses. masses: m 1=m 2=m initial: u 1=(4,0) , m/s , u 2=(0,0) elastic, smooth contact medium Speeds in COM are 2 m/s and opposite. Elastic: magnitudes unchanged. Perpendicular in lab as well (from earlier derivation). One-parameter family depending on the contact geometry; the angle between v 1 and v 2 is 90 . Geometry of equal-mass scattering. Not true. Momentum is always conserved in an isolated system, but kinetic energy is conserved only for elastic collisions. Inelastic collisions lose kinetic energy to heat, sound, or deformation. If momentum is conserved, kinetic energy must also be conserved. e primarily depends on material properties and contact geometry; masses influence momentum exchange but do not directly set e. The coefficient of restitution e depends on masses of the bodies. Restitution applies only to the components along the line of impact (normal direction). Tangential components are unchanged for smooth contact. In 2D, apply the restitution formula to the full speeds. Restitution quantifies bounciness: e is the ratio of relative separation speed to relative approach speed along the line of impact. This formula applies to the calculation of rebound velocity components when an object collides with a smooth, rigid wall, assuming the colli Boundaries of applicability: The simple restitution model assumes short contact, negligible rotation, and smooth contact (no impulsive friction). It applies to the normal components of velocity. If impacts cause significant spin, if contact is rough, or if the bodies deform plastically beyond the elastic range, more advanced models are needed. In exam problems, unless stated otherwise, treat contact as smooth and use Newton’s restitution with momentum conservation. Common mistake: Applying e to speed magnitudes in 2D. Only use normal components with proper signs. Also, do not mix lab-frame and COM-frame velocities in the same equation set. neet-alert Ball–wall formulae (smooth wall): If the wall’s inward normal is taken as +n, then before impact u n>0 , u t arbitrary. After impact, v n = e ,u n but along −n (outward), and v t=u t . In vector form with unit normal n and tangent t , v = (-e ,u n) n + u t , t . This compact rule quickly gives rebound speeds and angles. Smooth Wall Bounce (Components) Angles measured with respect to the wall’s normal: incidence , rebound . Successive bounces with the floor: Modeling the floor as a smooth horizontal plane with restitution e , the vertical speed right after each bounce is multiplied by e relative to the speed just before. For a ball dropped from height h , the first rebound height is e 2 h (since v 2 h ). Each subsequent peak height is multiplied by e 2 again. This geometric progression idea can save time in multi-bounce questions. Edge cases and limits: If the relative approach speed is zero ( u 1=u 2 in 1D), there is no impact (no compression), so the restitution formula is not used; the bodies continue with the same speeds unless other forces act. As e 1 , results approach the elastic formulas; as e 0 , the normal relative speed after impact vanishes. For r 0 duration impacts with huge force, impulse remains finite and sets the momentum change: that’s why impulse methods are more natural than force-time details for collisions. Worked check with energy loss formula: In the medium example ( e=0.60 ), = 2 1 2+1 = 2 3 , kg . Relative approach speed is (u 1-u 2)=4.0 , m/s . Predicted loss is 1 2 (1-e 2)(u 1-u 2) 2= 1 2 2 3 (1-0.36) 16 3.41 , J . If you compute K i-K f from the solved v 1,v 2 , you will get the same number. This is a great cross-check during exams. Practical note on materials and speed: Real materials have a restitution that can vary with impact speed and temperature. At low speeds, rubber can have high e , while at very high speeds it may lose more energy and show lower e . Metals can appear quite elastic for modest speeds. In NEET questions, e is usually a given constant for the event; treat it as such unless told otherwise. Multiple-object or sequential collisions: When three or more bodies collide in sequence, apply conservation and restitution pairwise at each contact event in correct time order. Between contacts, use free motion. The reduced-mass and relative-speed viewpoint still helps: compute the effective two-body interactions one collision at a time, updating velocities after each event. Rotational effects (beyond scope): Off-center impacts can generate spins; then angular momentum about the contact point and impulsive friction matter. NEET typically avoids this. If a question mentions ‘smooth spheres’ or ‘smooth wall’, interpret that as no tangential impulse and hence no change in tangential components or spins. Checking directions after recombination: After solving for normal and tangential components, always reconstruct the vector and verify if magnitudes and directions make sense. For example, a wall rebound should point away from the wall; a reduction in the normal component should increase the rebound angle with the normal compared to incidence if e<1 . Momentum bookkeeping tip: For a two-body 1D collision, it is often efficient to solve using the pair (m 1+m 2) and the reduced mass with the relative speed. Momentum conservation gives the COM motion, while restitution fixes the relative motion. Combining them yields the individual final velocities with minimal algebra. Be careful with which angle is given: many problems state the angle with the surface, not the normal. For wall bounces, using the wrong reference (normal vs surface) flips tan relations and leads to wrong rebound angles. neet-alert Time of contact vs impulse: Even though the contact force is huge, the time is tiny, so J= F ,dt is finite and sets the momentum change. Problems rarely ask for contact time or force profiles; instead, use conservation and restitution to sidestep the unknown force-time curve. This is exactly why momentum methods dominate collision analysis in mechanics. Line of impact Common normal at contact; only along this line does e apply. e = (relative separation speed)/(relative approach speed) along the line of impact. Coefficient of restitution Elastic collision Momentum and kinetic energy conserved. Inelastic collision Momentum conserved; kinetic energy decreases. Perfectly inelastic Bodies stick; maximum kinetic energy loss. Impulse Change in momentum from a short, large force action. Reduced mass = m 1 m 2 m 1+m 2 used to express relative-motion energy. Key terms recap