AP Exams (Advanced Placement) · Physics (JEE & NEET)
Collisions in One and Two Dimensions
Impulse, coefficient of restitution, elastic and inelastic collisions, 1D head-on results, oblique/2D collisions and COM-frame methods.
The standalone collision topic expands beyond the thin work-energy treatment: impulse, restitution, 1D results, oblique impacts and COM-frame thinking.
- AP Exams (Advanced Placement)
- Hard level
- 5 concepts
- 5 practice questions
1Impulse is momentum transfer
A collision force is large but short-lived, so the useful quantity is impulse: \vec J=\int\vec F\,dt=\Delta\vec p. If the same momentum change happens in a longer time, the average force is smaller. This is why padding and crumple zones reduce injury without changing the required momentum change much.
Figure. Impulse is the area under the force–time pulse — equal to Δp. Spread the same momentum change over a longer contact and the peak force falls.
How it works
- Find momentum changeUse final momentum minus initial momentum as a vector.
- Divide by timeAverage force is impulse over contact time.
- Use directionImpulse points in the direction of momentum change.
Average force on rebound
A 0.20\text{ kg} ball moving right at 10\text{ m s}^{-1} rebounds left at 6\text{ m s}^{-1} in 0.04\text{ s}.
- \Delta p=m(v-u)=0.20(-6-10)-3.2\text{ N s}
- F_{avg}=\Delta p/\Delta t-80\text{ N}
Pro tip. The sign says the impulse is leftward. The magnitude is 80 N.
Doubling collision time for the same momentum change makes average force
- Double
- Half
- Unchanged
F_{avg}=\Delta p/\Delta t.
2Restitution lives on the line of impact
The coefficient of restitution compares relative speeds along the line of impact only. For a head-on collision this is the whole motion. For an oblique collision, first resolve velocity into normal and tangential components; the normal component follows restitution, while the tangential component is unchanged if the contact is smooth.
Figure. Restitution compares only the normal components along the line of impact. The tangential piece rides along unchanged on a smooth wall; e multiplies the outgoing normal speed.
How it works
- Draw normalThe normal/common line of impact is where the impulse acts.
- Resolve velocitySplit into normal and tangential components.
- Apply e only normalTangential velocity changes only if friction gives tangential impulse.
Ball off a smooth wall
A ball has u_n=6 m/s into a wall and u_t=8 m/s along it. With e=0.5, find final components.
- v_t8\text{ m s}^{-1}
- v_n-0.5\times6=-3\text{ m s}^{-1}
- Final speed\sqrt{8^2+3^2}=8.54\text{ m s}^{-1}
Pro tip. A common wrong answer multiplies the whole speed by e. Restitution never touches the tangential component in a smooth impact.
In a smooth oblique impact with a wall, the component unchanged by the collision is
- Normal component
- Tangential component
- Both components
With no tangential impulse, the tangential component is unchanged.
3One-dimensional collision equations
In one dimension, momentum conservation always gives one equation. A second relation comes from restitution, or from kinetic energy if the collision is perfectly elastic. For e=1, relative speed of separation equals relative speed of approach; equal masses exchange velocities when the collision is head-on and elastic.
Figure. Equal masses, elastic head-on: the velocities swap. Momentum conservation plus e = 1 is enough — no picture of the blocks is required beyond the signed speeds.
How it works
- Momentum firstWrite m_1u_1+m_2u_2=m_1v_1+m_2v_2.
- Second relationUse restitution, sticking condition or kinetic-energy conservation.
- Check limiting casesEqual masses elastic should exchange velocities.
Equal masses elastic
A ball moving at 5\text{ m s}^{-1} hits an identical ball at rest elastically in a line.
- Equal masses, elastic, target at restvelocities exchange
- First ball final speed0
- Second ball final speed5\text{ m s}^{-1}
Pro tip. This shortcut is safe only for equal masses in a head-on perfectly elastic collision.
In a perfectly inelastic 1D collision, the two bodies after collision have
- Equal speeds in opposite directions
- A common velocity
- Unchanged kinetic energy
Perfectly inelastic means they stick or move together, so there is one final velocity.
4Two-dimensional and oblique collisions
In two dimensions, momentum conservation becomes two component equations. The collision law still acts along the line of impact. For two smooth spheres, the impulse is along the common normal at contact; components perpendicular to that normal pass through unchanged. In an elastic glancing collision, the outgoing directions are constrained by both vector momentum and kinetic energy.
Figure. For smooth spheres the impulse lies along the line of centres. Momentum splits into normal and tangential books; restitution acts only on the normal pair.
How it works
- Choose normal-tangent axesUse axes along and perpendicular to the line of impact when possible.
- Conserve both componentsTotal x-momentum and y-momentum are separately conserved.
- Add collision lawApply restitution along the normal component only.
Identical spheres, glancing and elastic
A smooth ball travelling at 10\text{ m s}^{-1} strikes an identical ball at rest. At contact the line of centres makes 60^\circ with the incoming velocity and the impact is perfectly elastic. Find how each ball leaves.
- Along the normal: 10\cos60^\circ5.0\text{ m s}^{-1}
- Along the tangent: 10\sin60^\circ8.66\text{ m s}^{-1}
- Equal masses, elastic, target at rest, so the normal components exchangestruck ball leaves at 5.0\text{ m s}^{-1} along the line of centres
- Smooth contact, so no impulse acts along the tangentstriker leaves at 8.66\text{ m s}^{-1} along the tangent
Pro tip. The two outgoing velocities come out perpendicular. That is not an accident of these numbers — equal masses in a smooth elastic impact always separate at a right angle.
For a smooth sphere-sphere oblique collision, impulse acts along
- The tangent at contact
- The common normal at contact
- The initial velocity of the faster sphere always
Smooth contact gives normal impulse along the line joining centres at impact.
5COM frame makes collision symmetry visible
In the center-of-mass frame the total momentum is zero. Before and after a two-body collision, the two momenta are equal and opposite. For an elastic collision, the magnitudes of those relative momenta stay the same; only directions change. This is often the cleanest way to understand scattering and oblique elastic collisions.
Figure. Subtract V_cm from every velocity and the momenta become equal-and-opposite. Elastic collisions then keep those magnitudes; only the signs flip.
How it works
- Find COM velocityV_{cm}=P/M.
- Subtract itTransform each particle velocity to the COM frame.
- Solve symmetryUse equal-and-opposite momenta, then transform back if needed.
COM speed for two bodies
Masses 1 kg and 3 kg move in a line at 6 m/s and 2 m/s in the same direction.
- V_{cm}=(1\times6+3\times2)/43\text{ m s}^{-1}
- COM-frame velocities+3\text{ m s}^{-1} and -1\text{ m s}^{-1}
- COM-frame momenta+3 and -3\text{ kg m s}^{-1}
Pro tip. Equal and opposite momenta in the COM frame are guaranteed; equal and opposite velocities are not unless masses are equal.
In the COM frame of a two-particle system, total momentum is
- Zero
- Mv
- Always positive
The COM frame moves with V_{cm}=P/M, so total momentum measured there is zero.
Notes
- During the short contact time of a collision, external impulse is often negligible, so total momentum is conserved. Kinetic energy is conserved only in a perfectly elastic collision.
- Impulse is change in momentum: \vec J=\int \vec F\,dt=\Delta\vec p. The average impact force is F_{avg}=\Delta p/\Delta t along the impulse direction.
- The coefficient of restitution is defined along the line of impact: e=\dfrac{\text{relative speed of separation}}{\text{relative speed of approach}}. Components perpendicular to a smooth line of impact are unchanged by the normal impulse.
- For a 1D perfectly elastic collision, use momentum plus kinetic energy, or the equivalent relative-speed reversal. Equal masses exchange velocities in a head-on elastic collision.
- For 2D/oblique collisions, conserve momentum component-wise and apply restitution only along the common normal/line of impact. The COM frame often turns the before-and-after momenta into equal and opposite vectors.
Formulas
- \vec J=\Delta\vec p
- F_{avg}=\dfrac{\Delta p}{\Delta t}
- m_1u_1+m_2u_2=m_1v_1+m_2v_2
- e=\dfrac{v_2-v_1}{u_1-u_2} for 1D approach with u_1>u_2
- Perfectly inelastic common speed: v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}
- Energy loss in perfectly inelastic 1D collision: \Delta K=\dfrac12\mu(u_1-u_2)^2, \mu=\dfrac{m_1m_2}{m_1+m_2}
Exam traps & shortcuts
- Momentum is vector conservation; kinetic energy conservation is an extra condition only for elastic collisions.
- Restitution belongs only along the line of impact, not automatically to every component in a 2D problem.
- Equal masses in a 1D elastic collision exchange velocities; equal masses in a 2D elastic collision with one initially at rest leave at right angles if both continue moving.
Reference tables
| Collision type | Momentum | Kinetic energy | Extra relation |
|---|---|---|---|
| Any isolated collision | Conserved | May change | external impulse negligible |
| Perfectly elastic | Conserved | Conserved | e=1 |
| Perfectly inelastic | Conserved | Maximum loss allowed | common final velocity |
| Oblique smooth | Conserved in x and y | depends on e | restitution only along normal |
Recap
Read only this before a collision problem.
- Impulse
- J=\Delta p; longer contact time lowers average force for the same momentum change.
- Momentum
- Conserve vector momentum when external impulse during impact is negligible.
- Energy
- Use kinetic-energy conservation only for perfectly elastic collisions.
- Restitution
- e is along the line of impact; do not apply it blindly to tangential components.
- COM frame
- In the COM frame, total momentum is zero and two-body momenta are equal and opposite.
Practise Collisions in One and Two Dimensions
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