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M2.4 - Collisions

Syllabus
2019
Topic
M2.4
Level
A2

Learning objectives

M2.4.1Momentum as a vectorMomentum as a vector.; The impulse-momentum principle in vector form.; Conservation of linear momentum.M2.4.2Direct impact of elastic particlesUse Newton’s law of restitution for direct impact of elastic particles, with 0 ≤ e ≤ 1, and determine loss of mechanical energy.M2.4.3Successive impacts of up to threeSuccessive impacts of up to three Collision with a plane surface will not involve oblique particles or two particles and a impact. smooth plane surface.

Use vector impulse to change momentum

Linear momentum has the same direction as velocity: for a particle of mass mm, p=mv\mathbf p=m\mathbf v. An impulse is the vector change in momentum, so its components change the corresponding velocity components independently.

J=Δp=m(v−u)\mathbf J=\Delta\mathbf p=m(\mathbf v-\mathbf u)

Impulse is measured in N s=kg m s−1\mathrm{N\,s}=\mathrm{kg\,m\,s^{-1}}. During an interaction, the impulses two particles exert on each other are equal and opposite. If the total external impulse on a chosen system is negligible, those internal impulses cancel and total vector momentum is conserved.

∑mu=∑mv\sum m\mathbf u=\sum m\mathbf v

Step Vector decision
choose axes keep one fixed pair of directions before and after the impulse
write impulse-momentum subtract initial momentum from final momentum
resolve equate i\mathbf i and j\mathbf j components separately
use a magnitude apply Pythagoras only after components are known
use a direction take an inverse tangent and check the quadrant from component signs

A 22 kg particle has velocity (3i−j) m s−1(3\mathbf i-\mathbf j)\,\mathrm{m\,s^{-1}} and receives impulse (−4i+6j) N s(-4\mathbf i+6\mathbf j)\,\mathrm{N\,s}. Thenv=u+Jm=(3i−j)+(−2i+3j)=i+2j.\mathbf v=\mathbf u+\frac{\mathbf J}{m}=(3\mathbf i-\mathbf j)+(-2\mathbf i+3\mathbf j)=\mathbf i+2\mathbf j.Its new speed is 5 m s−1\sqrt5\,\mathrm{m\,s^{-1}} and its direction is tan⁡−12\tan^{-1}2 above the positive i\mathbf i-axis.

Do not replace a velocity vector by its speed inside the impulse equation, add magnitudes instead of components, or conserve the momentum of one particle alone during a collision. Kinetic energy need not be conserved merely because momentum is.

Solve one direct impact with restitution

For a direct impact, choose one positive direction along the line of motion and give every velocity a sign. During the short collision, conserve total signed momentum for the two-particle system and pair it with Newton's law of restitution.

m1u1+m2u2=m1v1+m2v2,v2−v1=e(u1−u2)m_1u_1+m_2u_2=m_1v_1+m_2v_2,\qquad v_2-v_1=e(u_1-u_2)

The restitution equation shown assumes particle 1 approaches particle 2 from behind, so u1−u2u_1-u_2 is the relative speed of approach and v2−v1v_2-v_1 the relative speed of separation. Rebuild the equation from 'separation equals ee times approach' if directions differ; do not memorize signs without the geometry.

0≤e≤10\le e\le1

Value of ee Immediate impact meaning
e=1e=1 separation speed equals approach speed; total kinetic energy is unchanged for the isolated direct impact
0<e<10<e<1 separation speed is reduced and kinetic energy is lost
e=0e=0 separation speed is zero immediately after impact; the two velocities are equal at that instant

A 22 kg particle moving at 5 m s−15\,\mathrm{m\,s^{-1}} catches a 33 kg particle moving in the same direction at 1 m s−11\,\mathrm{m\,s^{-1}}. If e=0.5e=0.5, then13=2v1+3v2,v2−v1=2.13=2v_1+3v_2,\qquad v_2-v_1=2.Thus v1=1.4v_1=1.4 and v2=3.4 m s−1v_2=3.4\,\mathrm{m\,s^{-1}}. Kinetic energy falls from 26.526.5 J to12(2)(1.4)2+12(3)(3.4)2=19.3 J,\tfrac12(2)(1.4)^2+\tfrac12(3)(3.4)^2=19.3\text{ J},so the loss is 7.27.2 J.

Momentum conservation and restitution are independent equations; neither replaces the other. Calculate kinetic-energy loss as Kbefore−KafterK_{\rm before}-K_{\rm after} using speeds squared, and check it is non-negative. Do not assume e=1e=1 from the word 'elastic particle' unless the coefficient or condition establishes it.

Track successive impacts event by event

A successive-impact problem is a sequence of separate direct collisions. After each event, replace the affected velocities with their new values, preserve every unaffected velocity, then decide which objects can meet next from their positions and signed relative velocities.

Event Equation set State update
two particles collide signed momentum conservation + particle restitution update both particle velocities
particle hits a smooth fixed plane normally speed after =ew=e_w times speed before, direction reversed update that particle only
test for another collision compare positions and use closing speed continue only if the gap is decreasing
find time to meet time == current gap / positive closing speed move every particle for that same time

A fixed wall supplies an external impulse, so particle momentum is not conserved across the wall impact. For a normal impact with wall coefficient ewe_w, a velocity +u+u towards the wall becomes −ewu-e_wu after rebound when positive points towards the wall. This syllabus does not include oblique impact with a plane surface.

Immediately after a wall rebound, particle QQ is 66 m to the right of particle PP. Let right be positive. If PP moves right at 1 m s−11\,\mathrm{m\,s^{-1}} and QQ moves left at 3.2 m s−13.2\,\mathrm{m\,s^{-1}}, their closing speed is 1−(−3.2)=4.2 m s−11-(-3.2)=4.2\,\mathrm{m\,s^{-1}}. They therefore collide after 6/4.2=1.436/4.2=1.43 s. If QQ had instead moved right faster than PP, the gap would grow and no second collision would occur.

When a coefficient is unknown, conditions such as 'a second collision occurs' become strict inequalities on signed speeds after the preceding event. Intersect the resulting condition with 0≤e≤10\le e\le1. At a boundary where closing speed is zero, the particles never close a positive gap, so equality usually does not produce another collision.

Do not reuse pre-collision velocities in a later event, conserve particle momentum through a wall impact, or decide collision order from speed alone without position and direction. The scope is at most three particles, or two particles with a smooth plane, and plane impacts are normal rather than oblique.