3.6 - Direct impact of two spheres

3.6 - Direct impact of two spheres

Restitution tells us how two spheres separate, but it does not determine their individual velocities. In this part of 3.6, you will combine it with conservation of momentum, then use the final velocities to check directions and kinetic-energy loss.

Conserving linear momentum

Can a light sphere rebound while a heavier sphere moves forwards? To predict both velocities, we need a second equation linking the spheres. During their short impact, they exert equal and opposite impulses on each other, so these internal effects cancel for the pair.

Linear momentum

A particle of mass mm and signed velocity vv has linear momentum mvmv, measured in kg m s−1\mathrm{kg\,m\,s^{-1}}. Momentum has direction; kinetic energy, 12mv2\tfrac12mv^2, does not.

On a smooth horizontal surface, the vertical weight and support forces balance. Their horizontal impulse is zero, and any other external impulse during the short impact is neglected. Hence the pair's total horizontal momentum is conserved.

Momentum in a direct impact

For masses mA,mBm_A,m_B, using the same positive direction for both spheres before and after impact,

mAuA+mBuB=mAvA+mBvB.m_Au_A+m_Bu_B=m_Av_A+m_Bv_B.

This requires negligible resultant external impulse along the impact line. The internal contact impulses cancel.

Not in the formula booklet: learn it.

Write the masses explicitly. Adding velocities alone conserves momentum only in the special case of equal masses, when the common mass cancels.

Worked example 1: finding one final velocity

Question. Smooth spheres AA and BB, of masses 22 and 3 kg3\,\mathrm{kg}, move along the same horizontal line. AA is behind BB, and their rightward velocities are 66 and 1 m s−11\,\mathrm{m\,s^{-1}}. After impact AA is at rest. Find BB's velocity.

Plan. Conserve the total signed momentum of the two spheres, taking rightwards as positive.

2(6)+3(1)=2(0)+3vB15=3vBvB=5 m s−1\begin{aligned} 2(6)+3(1)&=2(0)+3v_B \\ 15&=3v_B \\ v_B&=5\,\mathrm{m\,s^{-1}} \end{aligned}

Answer. BB moves right at 5 m s−15\,\mathrm{m\,s^{-1}}.

Check. AA's momentum changes by 2(0−6)=−122(0-6)=-12, while BB's changes by 3(5−1)=12 kg m s−13(5-1)=12\,\mathrm{kg\,m\,s^{-1}}. The two changes cancel.

Change only BB's initial direction. It now contributes negative initial momentum, even though its speed remains 1 m s−11\,\mathrm{m\,s^{-1}}.

Worked example 2: an initially opposing sphere

Question. Use the same masses and AA's initial velocity, but let BB initially move left at 1 m s−11\,\mathrm{m\,s^{-1}}. Again AA is at rest after impact. Find BB's final velocity.

Plan. Put uB=−1u_B=-1 in the common rightward coordinate system.

2(6)+3(−1)=3vB9=3vBvB=3 m s−1\begin{aligned} 2(6)+3(-1)&=3v_B \\ 9&=3v_B \\ v_B&=3\,\mathrm{m\,s^{-1}} \end{aligned}

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

Check. AA again loses 12 kg m s−112\,\mathrm{kg\,m\,s^{-1}}, while BB gains 3(3−(−1))=12 kg m s−13(3-(-1))=12\,\mathrm{kg\,m\,s^{-1}}. The impulse exchange balances.

The mass-weighted equation handles a reversal automatically. A negative answer means motion opposite to the chosen positive direction; it is not a negative speed.

Using both impact equations

Momentum gives a mass-weighted sum of final velocities. Restitution gives their difference. Together, these two equations determine both unknown final velocities.

With AA initially left of BB and positive towards BB, use

mAvA+mBvB=mAuA+mBuBvB−vA=e(uA−uB)\begin{aligned} m_Av_A+m_Bv_B&=m_Au_A+m_Bu_B \\ v_B-v_A&=e(u_A-u_B) \end{aligned}

The first equation contains masses; the second does not. Restitution relates velocities rather than momenta. Neither equation assumes that AA actually rebounds.

Worked example 3: solving for both velocities

Question. On a smooth horizontal line, AA of mass 2 kg2\,\mathrm{kg} moves at 5 m s−15\,\mathrm{m\,s^{-1}} towards stationary BB of mass 3 kg3\,\mathrm{kg}. Their coefficient of restitution is 1/21/2. Find their velocities after direct impact.

Plan. Express vBv_B in terms of vAv_A using restitution, then substitute into momentum.

Taking AA's initial direction as positive gives

2vA+3vB=10momentumvB−vA=12(5)=52restitution2vA+3(vA+52)=105vA=52vA=12,vB=3\begin{aligned} 2v_A+3v_B&=10 &&\text{momentum} \\ v_B-v_A&=\frac12(5)=\frac52 &&\text{restitution} \\ 2v_A+3\left(v_A+\frac52\right)&=10 \\ 5v_A&=\frac52 \\ v_A&=\frac12,\qquad v_B=3 \end{aligned}

Answer. Both move forwards: AA at 0.5 m s−10.5\,\mathrm{m\,s^{-1}}, BB at 3 m s−13\,\mathrm{m\,s^{-1}}.

Check. The exchanged impulses are 2(0.5−5)=−92(0.5-5)=-9 and 3(3−0)=9 kg m s−13(3-0)=9\,\mathrm{kg\,m\,s^{-1}}. Also BB is faster afterwards, so the spheres separate.

Keep the initial collision, but now a final velocity is measured and ee is unknown. The same equations can recover the material coefficient.

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