FM1 4.1 - Newton's law of restitution

FM1 4.1 - Newton's law of restitution

Conservation of momentum alone usually leaves two unknown velocities after a collision. Newton's law of restitution supplies the missing relationship through relative speeds. You will combine the two laws, use the permitted coefficient range to interpret directions, and calculate the total kinetic energy lost.

Restitution compares separation with approach

Can a stationary target leave a collision faster than the incoming sphere, without energy being created? A heavier sphere striking a lighter one can produce exactly that result. To explain it, you need a rule about how the spheres move relative to each other, rather than a rule saying their individual speeds are conserved.

You already know

In a direct collision, choose one positive direction and conserve the total signed momentum when the external impulse is negligible. Each sphere's impulse is its own final momentum minus initial momentum; the two contact impulses are equal and opposite.

Suppose AA is behind BB, with both moving right. If AA moves at 5 m s−15\,\mathrm{m\,s^{-1}} and BB at 1 m s−11\,\mathrm{m\,s^{-1}}, their gap closes at 4 m s−14\,\mathrm{m\,s^{-1}}, not 66. If they instead move towards each other at those speeds, the gap closes at 6 m s−16\,\mathrm{m\,s^{-1}}.

Coefficient of restitution

For a direct impact, the coefficient of restitution ee is the relative speed of separation immediately after impact divided by the relative speed of approach immediately before impact. In this model 0≤e≤10\le e\le1. A perfectly elastic impact has e=1e=1; at e=0e=0, the spheres have the same velocity immediately after impact. They need not remain attached permanently.

Smooth elastic spheres in an exam question need not have e=1e=1. “Perfectly elastic” is the condition that identifies this endpoint.

Newton's law of restitution

Let AA be to the left of BB, with right positive. Write their velocities immediately before impact as uA,uBu_A,u_B, and immediately after as vA,vBv_A,v_B. For approaching spheres, uA>uBu_A>u_B, and

vB−vA=e(uA−uB),0≤e≤1v_B-v_A=e(u_A-u_B),\qquad0\le e\le1

Both sides describe non-negative relative speeds. Use signed velocities in the differences. The coefficient belongs to the pair of bodies and the stated model; it is not the ratio of one sphere's final and initial speeds.

Not in the formula booklet: learn it.

The following before-and-after measurements give the coefficient without requiring the masses.

[DIAGRAM: asset_name: Relative speeds in a same-direction collision; asset_slug: 9fm0-fm142-relative-speeds; caption: Subtract the two velocities in the correct order to describe the changing gap; description: Before impact, A is on the left moving right at five metres per second and B on the right moving right at one, with approach speed five minus one equals four labelled between them. After impact, A moves right at one and B right at three, with separation speed three minus one equals two. A fixed positive-right arrow and the ratio e equals two over four equals one half are shown.]
Subtract the two velocities in the correct order to describe the changing gap

Worked example 1: catching a sphere ahead

Question. Immediately before a direct collision, AA is to the left of BB and their velocities are 5 m s−15\,\mathrm{m\,s^{-1}} and 1 m s−11\,\mathrm{m\,s^{-1}}, both rightwards. Immediately afterwards their velocities are 1 m s−11\,\mathrm{m\,s^{-1}} and 3 m s−13\,\mathrm{m\,s^{-1}}, both rightwards. Find the coefficient of restitution.

Plan. Use the rate at which the gap closes before impact and the rate at which it opens afterwards.

Approach speed=5−1=4 m s−1Separation speed=3−1=2 m s−1e=24=12\begin{aligned} \text{Approach speed}&=5-1=4\,\mathrm{m\,s^{-1}} \\ \text{Separation speed}&=3-1=2\,\mathrm{m\,s^{-1}} \\ e&=\frac{2}{4}=\frac12 \end{aligned}

Why this step? Both spheres move in the same direction, so their relative speed is a difference, not the sum of their speeds.

Answer. e=12e=\tfrac12.

Check. The separation speed is half the approach speed, and the coefficient lies in [0,1][0,1].

In a second collision, change only BB's initial direction in the supplied velocity data. The initial relative speed changes even though the two initial speed magnitudes do not.

Worked example 2: opposite initial directions

Question. Immediately before a direct collision, AA is to the left of BB. AA moves right at 5 m s−15\,\mathrm{m\,s^{-1}} and BB left at 1 m s−11\,\mathrm{m\,s^{-1}}. Immediately afterwards AA moves right at 1 m s−11\,\mathrm{m\,s^{-1}} and BB right at 3 m s−13\,\mathrm{m\,s^{-1}}. Find the coefficient of restitution.

Plan. Take right as positive, so uB=−1u_B=-1, and form the two signed differences.

uA−uB=5−(−1)=6 m s−1the gap closes fastervB−vA=3−1=2 m s−1e=26=13\begin{aligned} u_A-u_B&=5-(-1)=6\,\mathrm{m\,s^{-1}} && \text{the gap closes faster} \\ v_B-v_A&=3-1=2\,\mathrm{m\,s^{-1}} \\ e&=\frac{2}{6}=\frac13 \end{aligned}

Why this step? Subtracting the negative initial velocity adds the two approach-speed magnitudes automatically. The signed formula handles both geometries.

Answer. e=13e=\tfrac13.

Check. The spheres approach at 6 m s−16\,\mathrm{m\,s^{-1}} and separate at 2 m s−12\,\mathrm{m\,s^{-1}}, a one-to-three ratio.

The direction convention prevents you having to memorise a separate sign pattern for every picture.

Watch out

Wrong: e=(vA−vB)/(uA−uB)e=(v_A-v_B)/(u_A-u_B) with AA initially on the left and right positive. That numerator is the negative of the separation speed. Also, interchanging approach and separation would give e>1e>1 in these examples, outside the permitted model.

Change only the leftward incoming speed of BB.

Solve momentum and restitution together

For two unknown final velocities, write two equations: one for total momentum and one for relative motion. Keep the same signed unknowns in both. A negative answer then tells you a sphere reverses; it is not a reason to change the equation halfway through.

Two equations for a direct impact

For isolated direct impact of two spheres modelled as particles,

mAuA+mBuB=mAvA+mBvBm_Au_A+m_Bu_B=m_Av_A+m_Bv_B vB−vA=e(uA−uB)v_B-v_A=e(u_A-u_B)

The restitution equation uses the ordering AA left of BB and right positive, with uA>uBu_A>u_B. The horizontal external impulse is negligible on a smooth horizontal plane.

Not in the formula booklet: learn it.

The first example has a moving heavier sphere and a stationary lighter target.

Worked example 3: a partially elastic collision

Question. A sphere AA of mass 2 kg2\,\mathrm{kg} moves right at 6 m s−16\,\mathrm{m\,s^{-1}} and collides directly with a stationary sphere BB of mass 1 kg1\,\mathrm{kg} on a smooth horizontal plane. The coefficient of restitution is 12\tfrac12. Find their velocities immediately afterwards.

Plan. Conserve momentum and use restitution to express one final velocity in terms of the other.

2(6)+1(0)=2vA+vBconservation of momentumvB−vA=12(6−0)=3restitutionvB=vA+312=2vA+(vA+3)substitute into momentumvA=3,vB=6\begin{aligned} 2(6)+1(0)&=2v_A+v_B && \text{conservation of momentum} \\ v_B-v_A&=\tfrac12(6-0)=3 && \text{restitution} \\ v_B&=v_A+3 \\ 12&=2v_A+(v_A+3) && \text{substitute into momentum} \\ v_A&=3,\qquad v_B=6 \end{aligned}

Why this step? Substitution removes one unknown while preserving the two laws. Neither momentum alone nor restitution alone determines both velocities.

Answer. AA moves right at 3 m s−13\,\mathrm{m\,s^{-1}}, and BB right at 6 m s−16\,\mathrm{m\,s^{-1}}.

Check. Final momentum is 2(3)+6=12 kg m s−12(3)+6=12\,\mathrm{kg\,m\,s^{-1}}. The separation speed is 6−3=3 m s−16-3=3\,\mathrm{m\,s^{-1}}, half the approach speed.

Change only the coefficient to 11. The impact is now perfectly elastic.

Worked example 4: a target leaves faster than the striker arrived

Question. A 2 kg2\,\mathrm{kg} sphere AA, moving right at 6 m s−16\,\mathrm{m\,s^{-1}}, collides directly with a stationary 1 kg1\,\mathrm{kg} sphere BB on a smooth horizontal plane. The coefficient of restitution is 11. Find the velocities immediately afterwards.

Plan. Retain the momentum equation and change the relative separation speed to the full approach speed.

12=2vA+vBvB−vA=612=2vA+(vA+6)substitute vB=vA+6vA=2,vB=8\begin{aligned} 12&=2v_A+v_B \\ v_B-v_A&=6 \\ 12&=2v_A+(v_A+6) && \text{substitute }v_B=v_A+6 \\ v_A&=2,\qquad v_B=8 \end{aligned}

Why this step? e=1e=1 preserves the relative-speed magnitude, not either sphere's individual speed. A light target can receive enough energy to exceed the striker's initial speed.

Answer. AA moves right at 2 m s−12\,\mathrm{m\,s^{-1}}, and BB right at 8 m s−18\,\mathrm{m\,s^{-1}}.

Check. Momentum is 2(2)+8=122(2)+8=12. Initial kinetic energy is 12(2)(62)=36 J\tfrac12(2)(6^2)=36\,\mathrm{J}; final kinetic energy is 12(2)(22)+12(1)(82)=4+32=36 J\tfrac12(2)(2^2)+\tfrac12(1)(8^2)=4+32=36\,\mathrm{J}. The faster target has not created energy.

That resolves the opening puzzle. A greater speed for one sphere does not imply a greater total kinetic energy for the pair.

Watch out

Wrong for these two-sphere impacts: vB=6ev_B=6e. This ignores AA's motion after impact; restitution concerns relative speeds. Right: solve vB−vA=6ev_B-v_A=6e together with 12=2vA+vB12=2v_A+v_B.

Show both unsimplified equations before solving. Pair every mass with its own signed velocity, and use the same assumed directions in both equations. Finish by interpreting each sign as a direction.

Change only the restitution coefficient once more.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account