3.6 - Oblique impact of smooth spheres

3.6 - Oblique impact of smooth spheres

In an oblique collision, the line joining the sphere centres plays the role of a wall's normal. In this part of 3.6, you will conserve momentum and apply restitution along that line, preserve each sphere's perpendicular component, then reconstruct its final velocity.

The impulse direction

Can a perfectly elastic collision change a sphere's direction without stopping it? For two smooth spheres, the contact impulse acts along their common normal: the line of centres at impact. The perpendicular components survive the collision unchanged.

Choose a unit direction n\mathbf n from AA's centre towards BB's, and a perpendicular unit direction t\mathbf t. These are fixed axes for the instant of impact. Write each velocity as a normal component along n\mathbf n plus a tangential component along t\mathbf t.

[DIAGRAM: asset_name: Smooth spheres exchange a normal impulse; asset_slug: cam9231-sphere-impact-axis; caption: Contact impulses lie along the centres, leaving each tangential component unchanged; description: Two touching equal-radius circles have their centres on an inclined common normal n directed from A to B. Opposite contact impulse arrows on A and B lie on the centre line. A perpendicular t direction is shown separately. A side panel states that each sphere keeps its own tangential velocity, while its normal velocity can change.]
Contact impulses lie along the centres, leaving each tangential component unchanged

Smooth-sphere tangential components

During the short impact of two smooth spheres, neglect other impulses in the plane of motion. Then

vAt=uAt,vBt=uBt.v_{At}=u_{At},\qquad v_{Bt}=u_{Bt}.

Each sphere has zero tangential contact impulse. The two tangential velocities need not be equal to each other.

Not in the formula booklet: learn it.

Along the normal, an approaching pair has uAn>uBnu_{An}>u_{Bn}. It is this relative normal motion that brings the surfaces together, even if either sphere also moves sideways.

Worked example 1: a horizontal line of centres

Question. At impact the line of centres points horizontally right from AA to BB. Their initial velocities are 6i+3j6\mathbf i+3\mathbf j and i−2j m s−1\mathbf i-2\mathbf j\,\mathrm{m\,s^{-1}}, with i\mathbf i rightwards and j\mathbf j upwards. Find their normal approach speed and identify the unchanged tangential components.

Plan. Read horizontal components along the centres and vertical components perpendicular to them.

uAn=6,uBn=1normal approach speed=6−1=5 m s−1vAt=3,vBt=−2 m s−1\begin{aligned} u_{An}&=6,\qquad u_{Bn}=1 \\ \text{normal approach speed}&=6-1=5\,\mathrm{m\,s^{-1}} \\ v_{At}&=3,\qquad v_{Bt}=-2\,\mathrm{m\,s^{-1}} \end{aligned}

Answer. The normal approach speed is 5 m s−15\,\mathrm{m\,s^{-1}}. AA keeps its upward component 33, and BB its downward component 2 m s−12\,\mathrm{m\,s^{-1}}.

Check. Their relative initial velocity is 5i+5j5\mathbf i+5\mathbf j. Its projection onto the centre line is 55, confirming that the full relative speed 525\sqrt2 is not the speed to use in restitution.

Keep the same physical velocities but incline the line of centres. Components must now be resolved along the new impact axes.

Worked example 2: an inclined line of centres

Question. The initial velocities are again 6i+3j6\mathbf i+3\mathbf j and i−2j m s−1\mathbf i-2\mathbf j\,\mathrm{m\,s^{-1}}. Now

n=35i+45j,t=−45i+35j.\mathbf n=\frac35\mathbf i+\frac45\mathbf j, \qquad \mathbf t=-\frac45\mathbf i+\frac35\mathbf j.

Find the normal approach speed and both unchanged tangential components.

Plan. Project each velocity onto the new normal and tangent using their direction cosines.

For example, the normal component of ai+bja\mathbf i+b\mathbf j is 3a/5+4b/53a/5+4b/5; the tangent component is −4a/5+3b/5-4a/5+3b/5. Thus

uAn=35(6)+45(3)=6uBn=35(1)+45(−2)=−1uAt=−45(6)+35(3)=−3uBt=−45(1)+35(−2)=−2\begin{aligned} u_{An}&=\frac35(6)+\frac45(3)=6 \\ u_{Bn}&=\frac35(1)+\frac45(-2)=-1 \\ u_{At}&=-\frac45(6)+\frac35(3)=-3 \\ u_{Bt}&=-\frac45(1)+\frac35(-2)=-2 \end{aligned}

Answer. The normal approach speed is 6−(−1)=7 m s−16-(-1)=7\,\mathrm{m\,s^{-1}}. The unchanged tangential components are −3-3 for AA and −2 m s−1-2\,\mathrm{m\,s^{-1}} for BB.

Check. Reconstructing AA's velocity gives 6n−3t=(18/5+12/5)i+(24/5−9/5)j=6i+3j6\mathbf n-3\mathbf t=(18/5+12/5)\mathbf i+(24/5-9/5)\mathbf j=6\mathbf i+3\mathbf j. For BB, −n−2t=i−2j-\mathbf n-2\mathbf t=\mathbf i-2\mathbf j, so both projections are consistent.

The tangent is defined relative to the line of centres, rather than to either incoming path. Choose it once and retain its sign convention throughout.

Momentum and restitution

The collision along the line of centres is a direct-impact problem. Both spheres can change their normal velocities, so we need momentum and restitution together. Unlike a fixed wall, neither sphere is generally constrained to remain at rest.

For positive normal from AA towards BB, the equations are

mAvAn+mBvBn=mAuAn+mBuBnvBn−vAn=e(uAn−uBn)\begin{aligned} m_Av_{An}+m_Bv_{Bn}&=m_Au_{An}+m_Bu_{Bn} \\ v_{Bn}-v_{An}&=e(u_{An}-u_{Bn}) \end{aligned}

The tangential components do not enter these equations. Once the normal velocities are found, restore each sphere's own unchanged tangent.

Worked example 3: a normal rebound

Question. On a smooth horizontal surface, spheres AA and BB have masses 11 and 3 kg3\,\mathrm{kg}. At impact the line from AA to BB is horizontal, with n\mathbf n rightwards and t\mathbf t upwards. Their initial velocities are 8n+3t8\mathbf n+3\mathbf t and −2t m s−1-2\mathbf t\,\mathrm{m\,s^{-1}}. Given e=1/2e=1/2, find all final components.

Plan. Solve the normal sum and difference equations, then retain the two tangents.

vAn+3vBn=8vBn−vAn=4(vBn−4)+3vBn=8vBn=3,vAn=−1\begin{aligned} v_{An}+3v_{Bn}&=8 \\ v_{Bn}-v_{An}&=4 \\ (v_{Bn}-4)+3v_{Bn}&=8 \\ v_{Bn}&=3,\qquad v_{An}=-1 \end{aligned}

Since the spheres are smooth, vAt=3v_{At}=3 and vBt=−2v_{Bt}=-2.

Answer. vA=−n+3t\mathbf v_A=-\mathbf n+3\mathbf t and vB=3n−2t m s−1\mathbf v_B=3\mathbf n-2\mathbf t\,\mathrm{m\,s^{-1}}. AA's normal component reverses, but its tangent does not.

Check. The normal impulses are 1(−1−8)=−91(-1-8)=-9 and 3(3−0)=9 kg m s−13(3-0)=9\,\mathrm{kg\,m\,s^{-1}}, equal and opposite. The positive separation speed is 3−(−1)=4 m s−13-(-1)=4\,\mathrm{m\,s^{-1}}.

Change only the coefficient to 11. This changes the normal exchange without changing either tangent.

Worked example 4: the elastic normal collision

Question. Repeat Worked example 3 with e=1e=1.

Plan. Keep momentum unchanged and double the required normal separation speed.

vAn+3vBn=8vBn−vAn=8(vBn−8)+3vBn=8vBn=4,vAn=−4\begin{aligned} v_{An}+3v_{Bn}&=8 \\ v_{Bn}-v_{An}&=8 \\ (v_{Bn}-8)+3v_{Bn}&=8 \\ v_{Bn}&=4,\qquad v_{An}=-4 \end{aligned}

The tangential components remain 33 and −2 m s−1-2\,\mathrm{m\,s^{-1}}.

Answer. vA=−4n+3t\mathbf v_A=-4\mathbf n+3\mathbf t and vB=4n−2t m s−1\mathbf v_B=4\mathbf n-2\mathbf t\,\mathrm{m\,s^{-1}}.

Check. Initial normal kinetic energy is 12(1)(82)=32 J\tfrac12(1)(8^2)=32\,\mathrm{J}. Final normal energy is 12(1)(42)+12(3)(42)=32 J\tfrac12(1)(4^2)+\tfrac12(3)(4^2)=32\,\mathrm{J}. The tangential energies are unchanged too, agreeing with e=1e=1.

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