Track systems, impulses and relative motion through collisions without confusing energy conservation.
JEE Main · JEE Advanced · NEET UG85 min study guide9 indexed sections
BY THE END, YOU CAN
Locate centres of mass for particles and composite bodies.
Use external impulse to justify momentum conservation by component.
Solve elastic, sticking and restitution-based impacts with energy checks.
Handle explosions, boat motion and staged collision problems.
BEFORE YOU START
Vector components, Newton’s laws and kinetic energy.
Signed velocities and basic integration or graph area.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Choose the system before claiming a conservation law
Momentum is the vector p = mv for a nonrelativistic particle of constant mass. The total momentum P of a system is the vector sum of all particle momenta. Opposite momenta can cancel even when the particles have substantial kinetic energy. A system at zero total momentum is not necessarily motionless or without available energy.
For a fixed collection of particles in an inertial frame, internal Newton-third-law force pairs cancel in the sum, giving dP/dt = F_external. Integrating gives ΔP = J_external. Momentum is conserved over an interval when the net external impulse vanishes, or approximately conserved when that impulse is negligible compared with the momenta being compared.
The choice of boundary matters. In a collision between two carts, their contact forces are internal to the two-cart system but external to either cart separately. Friction from a track is external unless the relevant Earth–track system is included. A wall can exchange a large impulse with a ball even if the wall’s motion is too small to notice.
Conservation can hold along one axis without holding along another. A short collision on a smooth horizontal surface has negligible external horizontal impulse, while vertical forces may constrain motion. Write separate component equations and justify each. “The force acts briefly” alone is not enough: an impulsive external force can be large precisely during that brief time.
Keep the same inertial frame before and after an event. Kinetic energy and momentum values change between frames, but a correctly stated closed-system momentum conservation law holds in every inertial frame. For variable-mass objects such as a rocket, a formula for a fixed collection cannot be applied to the rocket alone without accounting for momentum carried through the boundary.
02 / UNDERSTAND & APPLY
2. Impulse is signed area under a force–time curve
An impulse J = ∫F dt measures the accumulated effect of a force over time. The net impulse equals the vector change in momentum, p_final − p_initial. Its units N s and kg m s⁻¹ are equivalent. A force–time graph provides impulse from signed area; a force–position graph instead provides work. Exchanging these two areas gives the wrong physical quantity.
For an average net force over duration Δt, J = F_average Δt. This average need not equal the peak contact force. A triangular pulse has half the area of a rectangle with the same duration and peak. If positive and negative portions occur, combine their signed areas separately for each force component.
If a ball rebounds, its final velocity can have the opposite sign to its initial velocity. For example, changing from +u to −u gives Δp = −2mu, not zero. Speed change alone does not determine momentum change. In two dimensions, subtract the velocity vectors before taking the magnitude of the impulse.
When estimating contact force during a vertical collision, distinguish contact impulse from net impulse. For upward positive, J_contact − mgΔt = Δp_y if gravity and contact are the only forces. Dropping the weight term is an approximation justified only when its impulse is negligible. Increasing stopping time reduces average force for a specified momentum change, but says nothing by itself about every detail of the peak force.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Quantitative graph. This is net force, not peak force maintained for the full interval. The worked 2.0 kg cart gains 1.5 m/s in the positive direction.
ORIGINAL WORKED EXAMPLE
A triangular force pulse
A 2.0 kg cart initially moves at +1.0 m s⁻¹. Its net horizontal force rises linearly from zero to +12 N and returns to zero during a total 0.50 s. Find impulse and final velocity.
The force–time area is a triangle: J = (1/2)(0.50)(12) = +3.0 N s.
Initial momentum is 2.0 × 1.0 = +2.0 kg m s⁻¹; final momentum is +5.0 kg m s⁻¹.
Final velocity is p/m = 5.0/2.0 = +2.5 m s⁻¹. Using peak force throughout would overestimate the impulse by a factor of two.
Answer +3.0 N s; +2.5 m s⁻¹.
03 / UNDERSTAND & APPLY
3. The centre of mass is a mass-weighted position
For particles of total mass M, the centre-of-mass position is R = Σmᵢrᵢ/M. Compute each Cartesian component with the same masses. The result lies closer to the larger mass in a two-particle system. It is not the unweighted average of positions unless the masses are equal, and it need not lie inside the material itself: a uniform ring has its centre of mass in its empty central space.
For a continuous body, replace the sum by R = (1/M)∫r dm. Choose dm from the actual mass distribution: λ dx for a thin rod, σ dA for a lamina or ρ dV for a volume. A geometrical midpoint is a centre of mass only when the relevant symmetry includes the density distribution. A rod thickening toward one end generally has its centre of mass shifted toward that end.
Symmetry reduces computation. Reflection symmetry about a plane with equal density on reflected points places the centre of mass in that plane. Several intersecting symmetry planes can fix the point completely. For composite objects, use each part’s mass and its own centre of mass. A removed piece can be represented algebraically by a negative contribution relative to the original complete body, while the final remaining mass stays positive.
For two masses on a line, the distances from the centre of mass satisfy m₁d₁ = m₂d₂. This is a geometric consequence of the weighted-position formula, not a requirement that the masses be supported by a physical balancing beam. An origin shift changes all coordinates but leaves the physical location and relative distances consistent.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Positions use a common scale. Circle sizes distinguish masses but do not encode actual object dimensions.
ORIGINAL WORKED EXAMPLE
A composite mass and a removed section
Two point masses, 2 kg at x = 0 and 3 kg at x = 5 m, form one system. Find its centre of mass. Separately, a uniform 6 m rod has its final 2 m section removed; find the remaining centre using subtraction.
For the particles, x_CM = (2 × 0 + 3 × 5)/(2 + 3) = 3 m.
For the rod, let mass per metre be λ. The full rod has mass 6λ and centre x = 3 m; the removed section has mass 2λ and centre x = 5 m.
The remaining centre is [6λ(3) − 2λ(5)]/(6λ − 2λ) = 2 m, agreeing with the midpoint of the remaining interval from 0 to 4 m.
Answer Particle centre: x = 3 m; remaining rod centre: x = 2 m.
04 / UNDERSTAND & APPLY
4. External forces move the centre; internal forces rearrange the parts
For fixed masses, differentiating the weighted position gives V_CM = P/M and A_CM = F_external/M. The centre of mass therefore moves like a particle carrying the whole mass and experiencing the net external force. Internal forces can drastically change shape or relative motion without changing this centre-of-mass acceleration.
If an object explodes while moving in a uniform gravitational field and air resistance is neglected, the centre of mass of all fragments continues on the same projectile trajectory the intact object would have followed. Each individual fragment usually follows a different path. The statement requires keeping all fragments in the system; following only one piece does not preserve the original centre.
For a person walking inside a freely sliding boat with negligible horizontal external force, the combined centre of mass remains fixed if initially at rest. If the person’s displacement relative to the boat is d and masses are m and M, the boat displacement relative to shore is −md/(M + m). The person’s shore displacement is the boat displacement plus d. Confusing relative and shore distances is the common error.
The kinetic energy of a system separates into centre-of-mass translation and motion relative to that centre: K = MV_CM²/2 + K_relative. In a perfectly inelastic collision, the centre-of-mass translational term cannot be removed by internal interactions alone. The relative-motion part can become deformation, heat, sound or other internal energy. This decomposition explains why zero total momentum does not mean zero kinetic energy.
ORIGINAL WORKED EXAMPLE
Walking in a boat
A 60 kg person walks 2.0 m to the right relative to a 90 kg boat initially at rest. Ignore horizontal resistance and take the boat to translate without rotation. Find boat and person displacements relative to shore.
Let the boat move by x. The person moves by x + 2.0 relative to shore.
The combined centre is fixed: 90x + 60(x + 2.0) = 0.
Thus x = −120/150 = −0.80 m; the person’s shore displacement is −0.80 + 2.0 = +1.20 m.
Answer Boat 0.80 m left; person 1.20 m right.
05 / UNDERSTAND & APPLY
5. Momentum conservation does not imply kinetic-energy conservation
A collision is an interaction over a comparatively short interval during which mutual forces change velocities. A perfectly elastic collision conserves total kinetic energy as well as total momentum for an isolated system. An inelastic collision has lower final translational kinetic energy in the usual non-explosive model. A perfectly inelastic collision means the bodies stick together and share one final velocity.
For sticking in one dimension with negligible external impulse, v = (m₁u₁ + m₂u₂)/(m₁ + m₂). Signed velocities are essential. Mechanical kinetic energy lost is not total energy destroyed; it becomes internal energy or other forms. Perfectly inelastic does not mean the final kinetic energy is zero unless the initial total momentum is zero in that frame.
For two initially separated masses, define reduced mass μ = m₁m₂/(m₁ + m₂). The relative kinetic energy in the centre-of-mass frame is μ(u₁ − u₂)²/2. A perfectly inelastic collision removes all this relative translational kinetic energy, leaving only the motion of the shared centre. This provides a fast energy-loss formula without subtracting two large laboratory-frame energy totals.
For an elastic one-dimensional collision, combine momentum with equality of total kinetic energy. Factoring the energy equation and using momentum gives relative separation speed equal to relative approach speed. For equal masses, the velocities exchange. If the second equal mass starts at rest, the first stops and the second leaves with its speed; this is a special case, not the rule for unequal masses.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Smooth horizontal track and negligible external horizontal impulse. The final shared speed is fixed by momentum, not by conserving kinetic energy.
ORIGINAL WORKED EXAMPLE
Sticking carts with opposite velocities
A 2 kg cart moves right at 5 m s⁻¹ and a 3 kg cart moves left at 1 m s⁻¹. They stick on a smooth horizontal track. Find final velocity and kinetic-energy loss.
Choose right positive. Initial momentum is 2(5) + 3(−1) = 7 kg m s⁻¹.
The shared final velocity is 7/5 = +1.4 m s⁻¹.
Initial kinetic energy is (1/2)(2)(25) + (1/2)(3)(1) = 26.5 J. Final kinetic energy is (1/2)(5)(1.4²) = 4.9 J.
The loss is 21.6 J. Independently, μ = 6/5 kg and relative approach speed is 6 m s⁻¹, giving (1/2)(6/5)(36) = 21.6 J.
Answer 1.4 m s⁻¹ right; 21.6 J of translational kinetic energy converted to other forms.
06 / UNDERSTAND & APPLY
6. Restitution describes the normal relative velocity at impact
For a one-dimensional impact where body 1 approaches body 2, coefficient of restitution e is relative separation speed divided by relative approach speed: v₂ − v₁ = e(u₁ − u₂), with a positive approaching difference in this labeling. Use the equation with momentum conservation to determine the two final velocities. In the ordinary passive impact model, 0 ≤ e ≤ 1; e = 1 is elastic and e = 0 has zero relative normal speed immediately after impact.
For general oblique impact, restitution applies along the contact normal, or line of impact, not indiscriminately to the full speed. If surfaces are smooth and the impulse is purely normal, tangential velocity components of each particle do not change during the ideal collision. With friction, tangential impulse and possibly rotation must also be included. Applying the one-dimensional formula to an arbitrary horizontal component can give a wrong answer.
For an isolated one-dimensional two-body impact with this restitution model, loss of translational kinetic energy is ΔK_loss = μ(1 − e²)(u₁ − u₂)²/2. This is zero at e = 1 and greatest at e = 0 for fixed incoming data. Some active or energy-releasing events can have effective e greater than one, outside the ordinary passive-collision assumptions.
A ball bouncing from a stationary massive floor with negligible drag has upward speed e times its incoming downward speed in the vertical model. If it falls from height h and rebounds to height H, H = e²h. The square comes from the energy–speed relation, not from a different definition of restitution. A moving surface requires relative velocities with respect to that surface.
ORIGINAL WORKED EXAMPLE
A partially elastic impact
A 1 kg particle moving at +6 m s⁻¹ collides head-on with a stationary 2 kg particle. External impulse is negligible and e = 0.50. Find final velocities and kinetic-energy loss.
Momentum gives v₁ + 2v₂ = 6.
Restitution gives v₂ − v₁ = 0.50(6 − 0) = 3.
Substitute v₁ = v₂ − 3 to obtain 3v₂ = 9, hence v₂ = +3 and v₁ = 0 m s⁻¹.
Initial energy is 18 J and final energy is (1/2)(2)(9) = 9 J, so the loss is 9 J.
Answer First particle stops; second moves at +3 m s⁻¹; 9 J lost.
07 / UNDERSTAND & APPLY
7. Vector components handle explosions and oblique collisions
In two dimensions, total momentum supplies one equation per independent axis. For two fragments from an initially stationary isolated object, their momenta are equal and opposite, so their velocity magnitudes are inversely proportional to their masses. Three fragments need vector closure; two equal momentum magnitudes need not cancel unless their directions are opposite.
An explosion can increase total kinetic energy by converting stored internal energy. Momentum can still be conserved. It is therefore wrong to infer “inelastic means kinetic energy must always decrease” for every possible rapid interaction. State whether the event is a passive collision or an energy-releasing separation.
For a collision followed by slower motion, solve stages separately. In a ballistic-pendulum idealisation, a projectile sticks in a target during a short collision: conserve the appropriate horizontal momentum, not kinetic energy. During the subsequent swing, if friction is negligible, conserve mechanical energy to relate the common speed to height. Neither conservation statement can simply be applied unchanged across both stages.
An oblique elastic collision with one of two equal masses initially at rest has perpendicular outgoing velocities when both outgoing speeds are nonzero and no other energy channel or external impulse matters. Squaring the momentum-vector equation and comparing kinetic energies gives v₁·v₂ = 0. If one body stops, its direction is undefined, so do not quote a meaningful right angle for that degenerate case.
ORIGINAL WORKED EXAMPLE
Three fragments and vector closure
An object initially at rest breaks into three fragments. A 1 kg fragment moves at (3, 0) m s⁻¹ and a 2 kg fragment at (0, 2) m s⁻¹. The third fragment has mass 1 kg. Find its velocity, neglecting external impulse.
Known momenta are (3, 0) and (0, 4) kg m s⁻¹.
The third momentum must be (−3, −4) so that both component sums vanish.
Dividing by its 1 kg mass gives velocity (−3, −4) m s⁻¹, with speed √(9 + 16) = 5 m s⁻¹. The released internal energy supplies the nonzero final kinetic energy.
Answer (−3, −4) m s⁻¹; speed 5 m s⁻¹.
08 / UNDERSTAND & APPLY
8. Original practice with signed, vector and energy checks
Choose a system and frame first. Show the momentum equation before inserting a collision shortcut, and explicitly identify any kinetic-energy assumption.
ORIGINAL PRACTICE · 1
A 0.20 kg ball changes velocity from +10 to −6 m s⁻¹. Find net impulse.
Reveal answer & reasoning
−3.2 N s.
Δp = m(v − u) = 0.20(−6 − 10). Reversal makes the change larger than the difference of speed magnitudes.
ORIGINAL PRACTICE · 2
The impulse in the previous problem occurs in 0.040 s. Find average net force.
Reveal answer & reasoning
−80 N.
F_average = J/Δt = −3.2/0.040.
ORIGINAL PRACTICE · 3
A force–time rectangle is +4 N for 3 s, followed by −2 N for 2 s. Find total impulse.
Reveal answer & reasoning
+8 N s.
Signed areas add: 4(3) − 2(2) = 8.
ORIGINAL PRACTICE · 4
Masses m and 3m lie at x = 0 and x = 8 m. Find centre of mass.
Reveal answer & reasoning
x = 6 m.
The weighted sum gives [m(0) + 3m(8)]/(4m).
ORIGINAL PRACTICE · 5
A system of mass 5 kg has total momentum (15, −20) kg m s⁻¹. Find centre-of-mass velocity.
Reveal answer & reasoning
(3, −4) m s⁻¹.
V_CM = P/M component by component.
ORIGINAL PRACTICE · 6
Can internal forces accelerate the centre of mass of an isolated fixed-mass system?
Reveal answer & reasoning
No.
Internal pair forces cancel in the total momentum derivative; only net external force changes centre-of-mass momentum.
ORIGINAL PRACTICE · 7
Two equal masses approach at +u and −u and stick. Find final speed and fraction of initial kinetic energy lost.
Reveal answer & reasoning
Zero speed; all initial translational kinetic energy is lost.
Initial total momentum is zero, so common final velocity is zero. Energy becomes internal forms.
ORIGINAL PRACTICE · 8
In a one-dimensional elastic collision between equal masses, initial velocities are +4 and −1 m s⁻¹. Find final velocities.
Reveal answer & reasoning
−1 and +4 m s⁻¹ respectively.
Equal masses exchange velocities in the isolated one-dimensional elastic model.
ORIGINAL PRACTICE · 9
A ball falls from 5.0 m and rebounds to 1.8 m from a stationary floor. Find e, neglecting drag.
Reveal answer & reasoning
0.60.
e = √(H/h) = √(1.8/5.0) = √0.36.
ORIGINAL PRACTICE · 10
Two fragments of masses m and 4m separate from rest. Compare their speeds.
Reveal answer & reasoning
The lighter fragment moves four times faster in the opposite direction.
Equal and opposite momenta require mv = 4mV.
ORIGINAL PRACTICE · 11
Does momentum conservation during a sticking collision justify conserving kinetic energy?
Reveal answer & reasoning
No.
Momentum depends on external impulse; kinetic energy can become deformation, heat and sound.
ORIGINAL PRACTICE · 12
In a smooth-wall oblique impact, which velocity component is affected by the ideal contact impulse?
Reveal answer & reasoning
The normal component; the tangential component is unchanged.
A smooth wall supplies only normal impulse. Restitution is applied to relative normal velocities.
ORIGINAL PRACTICE · 13
A 2 kg system has total momentum zero but two parts moving oppositely. Must its kinetic energy vanish?
Reveal answer & reasoning
No.
Zero momentum removes centre-of-mass translation, but relative motion can carry positive kinetic energy.
ORIGINAL PRACTICE · 14
A 0.010 kg projectile embeds in a 0.990 kg target initially at rest. Immediately after impact they move at 2.0 m s⁻¹ horizontally. Find projectile speed if external horizontal impulse is negligible.
Reveal answer & reasoning
200 m s⁻¹.
0.010u = (0.010 + 0.990)(2.0), giving u = 200. Kinetic energy is not conserved during embedding.
09 / UNDERSTAND & APPLY
9. A short audit prevents inappropriate conservation
Write the system boundary, positive directions and time interval beside the first equation. Ask which external impulses are negligible, rather than whether the interaction simply “looks like a collision.” Then distinguish elastic, partially inelastic, sticking and explosive events. If two stages have different forces or energy losses, put a separate model on each stage.
Finally test the result: total momentum components should match, passive-collision energy loss should not be negative under 0 ≤ e ≤ 1, and the outgoing bodies should not continue to approach along the contact normal after a completed simple impact. A centre of mass of positive point masses should lie in their convex hull. Such checks often expose a sign or relative-distance error immediately.
Impulse is a vector and comes from force–time area.
Centre of mass uses mass weights, including density for continuous bodies.
Momentum conservation requires negligible net external impulse for the selected system.
Kinetic-energy conservation is an extra condition, not a consequence of momentum conservation.
Restitution refers to relative normal velocities.
Use distinct collision and post-collision stages when their assumptions differ.
KEEP THE BIG IDEAS
Quick revision
Internal forces rearrange a system but do not change its total momentum in isolation.
Centre-of-mass translation and relative motion carry different parts of kinetic energy.
Collision type supplies the additional equation beyond momentum conservation.
The relevant frame, system boundary and impact normal must be stated.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.