FM1 4.2 - Successive direct impacts

FM1 4.2 - Successive direct impacts

A sequence of impacts is a sequence of short problems, each with its own immediate before-and-after velocities. You will handle a direct impact with a fixed wall, carry velocities into later sphere collisions, and decide whether two spheres meet again. You will also combine constant-speed travel between impacts with restitution at each event.

A fixed wall reverses the velocity

Can a wall reverse a particle's momentum without removing any kinetic energy? Yes: a perfectly elastic direct rebound changes the velocity's sign while preserving its speed. That combination makes the impulse large even though the kinetic energy loss is zero.

You already know

Use signed velocities in I=m(v−u)I=m(v-u). Restitution compares separation with approach, and 0≤e≤10\le e\le1. Kinetic energy depends on the square of speed, so reversing velocity need not change kinetic energy.

For a direct impact, motion is perpendicular to the smooth fixed wall. The wall's velocity is zero, so the relative approach speed is the particle's incoming speed and the separation speed is its rebound speed. Do not conserve the particle's momentum: the wall supplies an external impulse.

Direct impact with a smooth fixed wall

Take the positive direction towards the wall, with incoming velocity u>0u>0, outgoing signed velocity vv and wall coefficient ee. Then

v=−euv=-eu

For mass mm, the impulse on the particle is

I=m(v−u)=−m(1+e)uI=m(v-u)=-m(1+e)u

Its magnitude is m(1+e)um(1+e)u, directed away from the wall. The kinetic energy lost is

ΔK=12mu2−12m(eu)2=12m(1−e2)u2\Delta K=\tfrac12mu^2-\tfrac12m(eu)^2=\tfrac12m(1-e^2)u^2

At e=0e=0 there is no rebound speed; at e=1e=1 the speed is unchanged. These formulae concern this direct impact only.

Not in the formula booklet: learn it.

Use momentum change for impulse and energy difference for loss. They measure different effects of the same collision.

Worked example 1: rebound with a speed loss

Question. A 0.4 kg0.4\,\mathrm{kg} particle moves at 10 m s−110\,\mathrm{m\,s^{-1}} directly towards a smooth fixed wall. The coefficient of restitution is 12\tfrac12. Find its rebound velocity, the magnitude and direction of its impulse, and the kinetic energy lost.

Plan. Take towards the wall as positive; find the signed rebound velocity before calculating momentum and energy changes.

v=−12(10)=−5 m s−1restitution reverses the directionI=0.4(−5−10)=−6 N sfinal minus initial momentumΔK=12(0.4)(102)−12(0.4)(−5)2=20−5=15 J\begin{aligned} v&=-\tfrac12(10)=-5\,\mathrm{m\,s^{-1}} && \text{restitution reverses the direction} \\ I&=0.4(-5-10)=-6\,\mathrm{N\,s} && \text{final minus initial momentum} \\ \Delta K&=\tfrac12(0.4)(10^2)-\tfrac12(0.4)(-5)^2 \\ &=20-5=15\,\mathrm{J} \end{aligned}

Why this step? The velocity reverses, so the momentum change involves −5−10-5-10. Energy uses the squares, so its calculation does not retain that sign reversal.

Answer. The particle rebounds at 5 m s−15\,\mathrm{m\,s^{-1}} away from the wall. Its impulse has magnitude 6 N s6\,\mathrm{N\,s}, directed away from the wall, and its kinetic energy loss is 15 J15\,\mathrm{J}.

Check. The rebound speed is half the incoming speed, so final kinetic energy is a quarter of its initial value: 5 J5\,\mathrm{J} from 20 J20\,\mathrm{J}.

Change only the coefficient to 11, making the impact perfectly elastic.

Worked example 2: a larger impulse with no energy loss

Question. A 0.4 kg0.4\,\mathrm{kg} particle approaches the same smooth fixed wall directly at 10 m s−110\,\mathrm{m\,s^{-1}}. The coefficient of restitution is now 11. Find the rebound velocity, the impulse magnitude and direction, and the kinetic energy lost.

Plan. Reverse the velocity without changing its magnitude, then calculate both changes separately.

v=−10 m s−1I=0.4(−10−10)=−8 N sΔK=12(0.4)(102)−12(0.4)(−10)2=0\begin{aligned} v&=-10\,\mathrm{m\,s^{-1}} \\ I&=0.4(-10-10)=-8\,\mathrm{N\,s} \\ \Delta K&=\tfrac12(0.4)(10^2)-\tfrac12(0.4)(-10)^2=0 \end{aligned}

Why this step? Momentum is a vector and reverses; kinetic energy is a scalar and remains the same. Zero kinetic energy loss does not imply zero impulse.

Answer. The rebound speed is 10 m s−110\,\mathrm{m\,s^{-1}} away from the wall, the impulse magnitude is 8 N s8\,\mathrm{N\,s} away from the wall, and the energy loss is 0 J0\,\mathrm{J}.

Check. Momentum changes from 44 to −4 kg m s−1-4\,\mathrm{kg\,m\,s^{-1}}, a change of −8-8. Kinetic energy is 20 J20\,\mathrm{J} both before and after.

The opening puzzle is therefore resolved. The larger rebound in the second example gives a larger momentum change while losing less energy.

[DIAGRAM: asset_name: Wall impulse and kinetic energy loss are different; asset_slug: 9fm0-fm143-wall-impulses; caption: A perfectly elastic rebound can give a larger impulse while losing no kinetic energy; description: Two direct-wall-impact panels use a mass of 0.4 kilograms and incoming speed ten metres per second towards a fixed wall. The first has coefficient one half and rebound speed five away, impulse six newton seconds away, and energy loss fifteen joules. The second has coefficient one and rebound speed ten away, impulse eight newton seconds away, and zero energy loss. Before and after arrows show reversal clearly.]
A perfectly elastic rebound can give a larger impulse while losing no kinetic energy

Watch out

Wrong: using the positive rebound speed in I=m(v−u)I=m(v-u) while incoming motion is positive. Right: use the negative rebound velocity. Also, the particle's momentum is not conserved when a fixed wall supplies its impulse.

Change only the wall coefficient to a value between the two examples.

Update the velocities after each collision

For a chain of impacts, take the events one at a time. Immediately after one collision becomes immediately before the next for a sphere involved in both. A sphere not involved retains its velocity while moving freely on the smooth horizontal surface.

Model the small spheres as particles. Collisions are instantaneous and separated events; between them the horizontal velocities are constant. Draw the order of the particles and retain one positive direction throughout.

Worked example 3: two successive sphere impacts

Question. Three spheres A,B,CA,B,C lie in that order on a smooth horizontal surface, with positive gaps between them. Their masses are 1,1,2 kg1,1,2\,\mathrm{kg} respectively. AA moves right at 6 m s−16\,\mathrm{m\,s^{-1}}, while B,CB,C are initially at rest. The coefficient of restitution is 12\tfrac12 for each pair that collides. Find all three velocities immediately after BB's first collision with CC.

Plan. Solve the AA-BB impact first, then use BB's new velocity in the BB-CC impact.

For the first collision, with right positive,

6=vA+vBmomentum for A,BvB−vA=12(6)=3restitution for A,BvA=1.5,vB=4.5\begin{aligned} 6&=v_A+v_B && \text{momentum for }A,B \\ v_B-v_A&=\tfrac12(6)=3 && \text{restitution for }A,B \\ v_A&=1.5,\qquad v_B=4.5 \end{aligned}

Now BB moves away from AA and towards the stationary CC, so BB-CC is the next impact. Denote their new velocities by wB,wCw_B,w_C:

4.5=wB+2wCmomentum for B,CwC−wB=12(4.5)=2.254.5=wB+2(wB+2.25)wB=0,wC=2.25\begin{aligned} 4.5&=w_B+2w_C && \text{momentum for }B,C \\ w_C-w_B&=\tfrac12(4.5)=2.25 \\ 4.5&=w_B+2(w_B+2.25) \\ w_B&=0,\qquad w_C=2.25 \end{aligned}

Why this step? The 6 m s−16\,\mathrm{m\,s^{-1}} velocity belonged to AA before the first impact. The second impact uses BB's current 4.5 m s−14.5\,\mathrm{m\,s^{-1}}, with CC still at rest.

Answer. Immediately after BB-CC, AA moves right at 1.5 m s−11.5\,\mathrm{m\,s^{-1}}, BB is at rest, and CC moves right at 2.25 m s−12.25\,\mathrm{m\,s^{-1}}.

Check. Total momentum is now 1(1.5)+1(0)+2(2.25)=6 kg m s−11(1.5)+1(0)+2(2.25)=6\,\mathrm{kg\,m\,s^{-1}}, equal to the initial total. AA's velocity was unchanged by the second impact.

Change only the BB-CC coefficient to 11. The first collision is unchanged.

Worked example 4: a different coefficient for the second pair

Question. The same three spheres have masses 1,1,2 kg1,1,2\,\mathrm{kg}, with AA initially moving right at 6 m s−16\,\mathrm{m\,s^{-1}} and B,CB,C at rest. The AA-BB coefficient is 12\tfrac12, but the BB-CC coefficient is 11. Find all three velocities immediately after BB-CC.

Plan. Retain the first-impact results and replace only the second restitution equation.

6=vA+vB,vB−vA=3vA=1.5,vB=4.5first collision unchanged4.5=wB+2wCsecond momentum equationwC−wB=4.5second coefficient is 14.5=wB+2(wB+4.5)wB=−1.5,wC=3\begin{aligned} 6&=v_A+v_B,\qquad v_B-v_A=3 \\ v_A&=1.5,\qquad v_B=4.5 && \text{first collision unchanged} \\ 4.5&=w_B+2w_C && \text{second momentum equation} \\ w_C-w_B&=4.5 && \text{second coefficient is }1 \\ 4.5&=w_B+2(w_B+4.5) \\ w_B&=-1.5,\qquad w_C=3 \end{aligned}

Why this step? A restitution coefficient belongs to a stated pair. There is no reason to reuse the first pair's coefficient for a different pair.

Answer. AA moves right at 1.5 m s−11.5\,\mathrm{m\,s^{-1}}, BB left at 1.5 m s−11.5\,\mathrm{m\,s^{-1}}, and CC right at 3 m s−13\,\mathrm{m\,s^{-1}}.

Check. Total momentum is 1.5−1.5+2(3)=61.5-1.5+2(3)=6. For the second collision alone, separation speed is 3−(−1.5)=4.53-(-1.5)=4.5, equal to approach speed, as required.

A short state table prevents old velocities from slipping into new equations.

Stage in Worked example 4AABBCC
Before AA-BB660000
After AA-BB1.51.54.54.500
After BB-CC1.51.5−1.5-1.533

All entries are signed velocities in m s−1\mathrm{m\,s^{-1}}, with right positive.

Watch out

Do not restart every impact from the original velocities, and do not assume a third sphere is at rest unless the question says so. Update only the participants in each collision; carry every other velocity forward unchanged.

Change only the second pair's coefficient again. You may use the already established first-impact velocities.

A rebound can bring two spheres together again

After a wall impact, reset the rebounding velocity's sign before considering the next meeting. Prove that the particles approach each other using their positions and directions relative to the wall; comparing speed magnitudes alone can be misleading.

For timing questions below, model the spheres as particles and ignore their radii. Take the first sphere-sphere impact as time t=0t=0 at coordinate x=0x=0, with the wall at x=12 mx=12\,\mathrm{m}. Right, towards the wall, remains positive.

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