Mi16, Mi17 - Oblique impact between two bodies | OCR A-Level Further Mathematics B (MEI)

Solve an oblique collision by applying momentum and restitution along the line of centres, while preserving each body's transverse velocity. Reconstruct the outgoing vectors and calculate energy loss using every component that contributes to speed.

Solve along the line of centres

If two spheres have coefficient of restitution zero, must they stick together? In a direct collision they share a final velocity. In an oblique collision, restitution controls only one component of their relative motion. We will use that distinction to settle the question.

You already know

For smooth impact, the impulse changes velocity only along its own line. Momentum conservation and Newton's experimental law determine the outgoing components along that line; the perpendicular components are unchanged individually.

Model the objects as uniform smooth spheres or discs, with no rotation before or after impact. Their shapes give a line of centres at contact, along which the equal and opposite impulses act. Two particles do not provide that preferred impulse direction and cannot model this oblique two-body collision.

Let $\mathbf{n}$ point from $A$ 's centre towards $B$ 's at contact, and let $\mathbf{t}$ be perpendicular to it. Subscripts $n$ and $t$ mean signed components in those directions. There is approach when $u_{An}>u_{Bn}$ .

Normal equations and transverse invariance

For masses $m_A,m_B>0$ , smooth contact and negligible external impulse,

\begin{aligned} m_Au_{An}+m_Bu_{Bn}&=m_Av_{An}+m_Bv_{Bn} &&\text{system momentum along centres}\\ v_{Bn}-v_{An}&=e(u_{An}-u_{Bn}) &&\text{normal separation and approach}\\ v_{At}&=u_{At},\qquad v_{Bt}=u_{Bt} &&\text{each body's transverse velocity} \end{aligned}

Use $0\le e\le1$ and positive normal approach speed. Afterwards $v_{Bn}\ge v_{An}$ .

Not in the formula booklet: learn it.

The first line conserves momentum for the pair. The last line makes the stronger statement that each body's transverse momentum is separately unchanged, because neither receives a transverse impulse.

[DIAGRAM: asset_name: Normal and transverse components in a two-body impact; asset_slug: oblique_bodies_components; caption: The line of centres controls the impulse, while each transverse component survives unchanged; description: Before and after panels show two equal-radius discs A and B touching on a horizontal line of centres with A left of B. A has mass 2 kilograms and B 1 kilogram. Incoming velocity arrows are 4 right and 3 up for A, and 2 left and 1 up for B. Outgoing arrows are 1 right and 3 up for A, and 4 right and 1 up for B. Common rightwards normal and upwards transverse directions are labelled.]
The line of centres controls the impulse, while each transverse component survives unchanged

Worked example 1: centres on a coordinate axis

Question. Smooth uniform spheres $A$ and $B$ , of masses $2\,\mathrm{kg}$ and $1\,\mathrm{kg}$ , collide on a smooth horizontal table. At contact, $B$ lies due east of $A$ . Their incoming velocities are $(4\mathbf{i}+3\mathbf{j})$ and $(-2\mathbf{i}+\mathbf{j})\,\mathrm{m\,s^{-1}}$ , where $\mathbf{i}$ is east and $\mathbf{j}$ north. The coefficient of restitution is $0.5$ . Find their outgoing velocities.

Plan. Use east as the positive normal direction, solve the two eastward equations and retain each northward component.

\begin{aligned} 2(4)+1(-2)&=2v_{An}+v_{Bn} &&\text{normal momentum}\\ 6&=2v_{An}+v_{Bn} &&\text{simplify}\\ v_{Bn}-v_{An}&=0.5(4-(-2))=3 &&\text{normal restitution}\\ 6&=2v_{An}+(v_{An}+3) &&\text{substitute}\\ v_{An}&=1,\qquad v_{Bn}=4 &&\text{solve}\\ v_{At}&=3,\qquad v_{Bt}=1 &&\text{smooth contact} \end{aligned}

Why this step? Restitution uses the signed normal components $4$ and $-2$ , not the whole incoming speeds $5$ and $\sqrt5$ .

Answer. $\mathbf{v}_A=(\mathbf{i}+3\mathbf{j})\,\mathrm{m\,s^{-1}}$ and $\mathbf{v}_B=(4\mathbf{i}+\mathbf{j})\,\mathrm{m\,s^{-1}}$ .

Check. The total momentum vector is $(6\mathbf{i}+7\mathbf{j})\,\mathrm{kg\,m\,s^{-1}}$ before and after. Normal separation speed $4-1=3$ is half the approach speed $6$ .

Change only $B$ 's transverse incoming component from $1$ to $-2$ . This alters the path of $B$ , but leaves both equations along the line of centres unchanged.

Worked example 2: a different sideways motion

Question. For the same masses, line of centres, $A$ 's incoming velocity and coefficient, let $\mathbf{u}_B=(-2\mathbf{i}-2\mathbf{j})\,\mathrm{m\,s^{-1}}$ . Determine the outgoing velocities.

Plan. Solve the same normal equations, then carry through the new transverse component of $B$ .

\begin{aligned} 2v_{An}+v_{Bn}&=2(4)+1(-2)=6 &&\text{normal momentum}\\ v_{Bn}-v_{An}&=0.5(4-(-2))=3 &&\text{normal restitution}\\ 3v_{An}+3&=6 &&\text{eliminate }v_{Bn}\\ v_{An}&=1,\qquad v_{Bn}=4 &&\text{same normal solution}\\ v_{At}&=3,\qquad v_{Bt}=-2 &&\text{retain each transverse component} \end{aligned}

Why this step? Smoothness removes the coupling between transverse motions; changing one transverse component cannot alter the impulse along the centres.

Answer. $\mathbf{v}_A=(\mathbf{i}+3\mathbf{j})\,\mathrm{m\,s^{-1}}$ and $\mathbf{v}_B=(4\mathbf{i}-2\mathbf{j})\,\mathrm{m\,s^{-1}}$ .

Check. Total transverse momentum is $2(3)+1(-2)=4$ before and after; the normal restitution ratio remains $3/6=0.5$ .

For the opening question, use the first incoming pair but set $e=0$ . Normal restitution gives $v_{An}=v_{Bn}$ , and momentum gives $3v_{An}=6$ . The outgoing velocities are therefore $(2\mathbf{i}+3\mathbf{j})$ and $(2\mathbf{i}+\mathbf{j})\,\mathrm{m\,s^{-1}}$ . Their normal components agree, but their full velocities do not: they need not stick together.

Watch out

Zero restitution means zero relative separation speed along the line of impulse. In an oblique collision, it does not require equal transverse velocities or coalescence.

Change a normal component in the next collision, so the simultaneous equations really do change.

Use an inclined line of centres

The line of centres need not lie along a given coordinate axis. Resolve into its direction first, solve there, then return to the original coordinates. Label each unknown as a component, rather than calling it a whole speed.

For an orthonormal pair $\mathbf{n},\mathbf{t}$ , dot products give the signed components. Equivalently, use ordinary trigonometric resolution against the diagram.

Resolve and reconstruct

For perpendicular unit vectors $\mathbf{n}$ and $\mathbf{t}$ ,

u_n=\mathbf{u}\mathbin{\cdot}\mathbf{n},\qquad u_t=\mathbf{u}\mathbin{\cdot}\mathbf{t},\qquad \mathbf{v}=v_n\mathbf{n}+v_t\mathbf{t}

Apply momentum and restitution to the normal components, and retain each body's transverse component.

Not in the formula booklet: learn it.

For a target initially at rest, its outgoing transverse component is zero. It therefore moves along the line of centres after impact. This gives a useful geometric clue when a question specifies the target's outgoing direction.

[DIAGRAM: asset_name: An inclined line of centres; asset_slug: oblique_inclined_centres; caption: Resolve the incoming horizontal velocity into directions along and perpendicular to the centres before applying the collision laws; description: Two equal smooth spheres touch with the line from A to B in direction n equal to three fifths i plus four fifths j. A horizontal incoming arrow of magnitude 5 is resolved exactly into 3 n and minus 4 t, where t equals minus four fifths i plus three fifths j. A separate after-impact panel for equal masses and e equal to one shows A moving in direction minus t and B in direction n, at right angles.]
Resolve the incoming horizontal velocity into directions along and perpendicular to the centres before applying the collision laws

Worked example 3: reconstruct two velocity vectors

Question. Two smooth uniform spheres of equal mass collide on a smooth horizontal table. At contact, the unit vector from $A$ to $B$ is $\mathbf{n}=(3\mathbf{i}+4\mathbf{j})/5$ . Initially $\mathbf{u}_A=5\mathbf{i}\,\mathrm{m\,s^{-1}}$ and $B$ is at rest. The coefficient of restitution is $0.5$ . Find both outgoing velocities and the deflection of $A$ .

Plan. Choose $\mathbf{t}=(-4\mathbf{i}+3\mathbf{j})/5$ , resolve, solve the normal equations and reconstruct.

\begin{aligned} u_{An}&=5\left(\frac35\right)=3,\qquad u_{At}=5\left(-\frac45\right)=-4 &&\text{signed projections}\\ v_{An}+v_{Bn}&=3 &&\text{equal-mass momentum}\\ v_{Bn}-v_{An}&=0.5(3)=\frac32 &&\text{normal restitution}\\ v_{An}&=\frac34,\qquad v_{Bn}=\frac94 &&\text{solve}\\ \mathbf{v}_A&=\frac34\frac{3\mathbf{i}+4\mathbf{j}}5-4\frac{-4\mathbf{i}+3\mathbf{j}}5 &&\text{retain }u_{At}=-4\\ &=\frac{73}{20}\mathbf{i}-\frac95\mathbf{j} &&\text{collect coordinates}\\ \mathbf{v}_B&=\frac94\frac{3\mathbf{i}+4\mathbf{j}}5=\frac{27}{20}\mathbf{i}+\frac95\mathbf{j} &&\text{target has no transverse motion}\\ \theta&=\tan^{-1}\left(\frac{9/5}{73/20}\right)=26.250\ldots^\circ &&\text{below the initial rightwards direction} \end{aligned}

Why this step? Restitution changes components along an inclined line, so the original horizontal and vertical components both change when the velocity is reconstructed.

Answer. $\mathbf{v}_A=(73\mathbf{i}/20-9\mathbf{j}/5)\,\mathrm{m\,s^{-1}}$ , $\mathbf{v}_B=(27\mathbf{i}/20+9\mathbf{j}/5)\,\mathrm{m\,s^{-1}}$ . Sphere $A$ is deflected $26.3^\circ$ clockwise, to $3$ significant figures.

Check. The outgoing vectors add to $5\mathbf{i}\,\mathrm{m\,s^{-1}}$ , as equal-mass momentum requires. Their normal separation is $9/4-3/4=3/2$ , half the incoming normal approach.

Change only $e$ to $1$ . Equal masses and a stationary target now produce a striking geometric result.

Worked example 4: a right-angle departure

Question. For the same incoming velocities and line of centres, let the collision be perfectly elastic. Find both outgoing velocities and $A$ 's deflection, and show that the outgoing directions are perpendicular.

Plan. Resolve as before, set $e=1$ , then check the outgoing vectors' dot product.

\begin{aligned} v_{An}+v_{Bn}&=3,\qquad v_{Bn}-v_{An}=3 &&\text{momentum and restitution}\\ v_{An}&=0,\qquad v_{Bn}=3 &&\text{normal components exchange}\\ \mathbf{v}_A&=-4\mathbf{t}=\frac{16}5\mathbf{i}-\frac{12}5\mathbf{j} &&\text{only transverse motion remains for }A\\ \mathbf{v}_B&=3\mathbf{n}=\frac95\mathbf{i}+\frac{12}5\mathbf{j} &&\text{only normal motion for }B\\ \mathbf{v}_A\mathbin{\cdot}\mathbf{v}_B&=\frac{16}5\frac95-\frac{12}5\frac{12}5=0 &&\text{perpendicular non-zero vectors}\\ \theta&=\tan^{-1}\left(\frac{12}{16}\right)=36.869\ldots^\circ &&\text{clockwise from the incoming direction} \end{aligned}

Why this step? With equal masses, a stationary target and $e=1$ , the incoming sphere transfers its normal component to the target while retaining its perpendicular component.

Answer. $\mathbf{v}_A=(16\mathbf{i}/5-12\mathbf{j}/5)\,\mathrm{m\,s^{-1}}$ , $\mathbf{v}_B=(9\mathbf{i}/5+12\mathbf{j}/5)\,\mathrm{m\,s^{-1}}$ . The deflection is $36.9^\circ$ clockwise, and the two outgoing directions are at right angles.

Check. Their speeds are $4$ and $3\,\mathrm{m\,s^{-1}}$ , so their squared speeds sum to $4^2+3^2=5^2$ , consistent with no kinetic energy loss.

A vanishing component does not necessarily mean a vanishing velocity.

Watch out

Wrong: $\mathbf v_A=\mathbf0$ because $v_{An}=0$ in the elastic example. Smoothness preserves the non-zero transverse component. Correct: $\mathbf v_A=-4\mathbf t=(16\mathbf i/5-12\mathbf j/5)\,\mathrm{m\,s^{-1}}$ .

This right-angle result holds for this smooth equal-mass, perfectly elastic collision with one body initially at rest, provided both outgoing velocities are non-zero. It is not a rule for every pair of colliding balls.

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Mi16, Mi17 - Oblique impact between two bodies

Solve along the line of centres

Use an inclined line of centres

Continue this lesson.

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