Oblique Collisions

In a direct collision everything happens along one line. In an oblique collision the velocity is not along the line of impact, so we resolve it into two components — one along the line of impact and one perpendicular to it — and treat them separately. The perpendicular component (along a smooth surface) is untouched; the component along the line of impact obeys restitution. This "resolve, treat independently, recombine" strategy is the single most important technique in this lesson.

1 Where this sits in AQA 7367

Oblique impact is the most demanding collision topic in the Mechanics option, Paper 3 (7367/3M), drawing together momentum, restitution, vectors and trigonometry. It is heavily AO3 (Paper 3 is AO1 40% / AO2 25% / AO3 35%): a typical question asks you to set up components, apply two conservation laws along one direction, and recombine to a speed and a direction. It builds on direct restitution (previous lesson) and on the resolution of vectors from A-Level Maths.

2 Core theory — impact with a smooth wall

The key fact: a smooth surface exerts a force only along its normal (perpendicular), never along itself. Therefore when a particle strikes a smooth wall:

the component of velocity parallel to the wall is unchanged (no force acts along the wall);
the component perpendicular to the wall obeys Newton's law of restitution — its magnitude is multiplied by $e$ and its direction reverses.

If the particle approaches at angle $\alpha$ to the wall with speed $c$ , then before impact the components are

$\text{perpendicular} = c\sin\alpha, \qquad \text{parallel} = c\cos\alpha,$

and after impact

$\text{perpendicular} = e\,c\sin\alpha \ (\text{reversed}), \qquad \text{parallel} = c\cos\alpha \ (\text{unchanged}).$

The $\tan\beta = e\tan\alpha$ rule

If the particle rebounds at angle $\beta$ to the wall, then $\tan\alpha = \dfrac{c\sin\alpha}{c\cos\alpha}$ (perpendicular over parallel before) and $\tan\beta = \dfrac{e\,c\sin\alpha}{c\cos\alpha}$ after, so

$\boxed{\;\tan\beta = e\tan\alpha\;}.$

Since $e \le 1$ , we get $\beta \le \alpha$ : the rebound is shallower to the wall (the path flattens toward the wall). When $e = 1$ , $\beta = \alpha$ — angle of incidence equals angle of reflection, exactly like light off a mirror.

Watch the reference angle. The relation $\tan\beta = e\tan\alpha$ holds when $\alpha,\beta$ are measured from the wall (the surface). If instead you measure from the normal (call these $\alpha',\beta'$ ), then $\alpha' = 90^\circ - \alpha$ and the relation inverts to $\tan\beta' = \dfrac{1}{e}\tan\alpha'$ — or more safely, just resolve from scratch each time. State your reference angle explicitly to avoid this trap.

Two smooth spheres — the line of centres

For an oblique collision between two smooth spheres, the contact force acts along the line of centres (LOC) — the line joining the two centres at the instant of impact. Smoothness means there is no force perpendicular to the LOC, so:

the velocity component of each sphere perpendicular to the LOC is unchanged;
along the LOC, the spheres obey conservation of momentum and Newton's law of restitution — exactly a one-dimensional (direct) collision.

The full method is therefore:

Resolve each sphere's velocity into components along and perpendicular to the LOC.
The perpendicular-to-LOC components carry through unchanged.
Apply momentum and restitution to the along-LOC components (a 1-D collision).
Recombine each sphere's two components into a resultant speed and direction.

The beauty of this is that oblique sphere–sphere impact is not a new problem — it is the direct collision of the previous lesson, applied to one axis, with two spectator components riding along untouched.

3 Worked examples with mark scheme

Worked Example 1 — speed and direction after a wall bounce

A ball hits a smooth wall at $30^\circ$ to the wall with speed $12\ \mathrm{m\,s^{-1}}$ ; $e = 0.8$ . Find the rebound speed and the angle to the wall.

Resolve (perpendicular = toward wall, parallel = along wall):

\begin{aligned} \text{perp.} &= 12\sin 30^\circ = 6, & \text{parallel} &= 12\cos 30^\circ = 6\sqrt{3}\\ \text{perp. after} &= 0.8 \times 6 = 4.8, & \text{parallel after} &= 6\sqrt{3}\ (\text{unchanged}). \end{aligned}

Rebound speed $= \sqrt{4.8^2 + (6\sqrt{3})^2} = \sqrt{23.04 + 108} = \sqrt{131.04} \approx 11.4\ \mathrm{m\,s^{-1}}$ .

Angle to the wall: $\tan\beta = \dfrac{4.8}{6\sqrt{3}} = \dfrac{4.8}{10.39} \approx 0.462$ , so $\beta \approx 24.8^\circ$ .

Mark scheme: M1 resolve into perpendicular/parallel; A1 components $6$ and $6\sqrt3$ ; M1 apply $e$ to perpendicular only; M1 recombine by Pythagoras; A1 $11.4\ \mathrm{m\,s^{-1}}$ ; A1 $\beta \approx 24.8^\circ$ . (6 marks.)

Worked Example 2 — the angle rule directly

A ball strikes a smooth wall at $45^\circ$ to the wall; $e = 0.5$ . Find the rebound angle to the wall.

$\tan\beta = e\tan\alpha = 0.5\tan 45^\circ = 0.5 \Rightarrow \beta = \arctan 0.5 \approx 26.6^\circ.$

Mark scheme: M1 $\tan\beta = e\tan\alpha$ ; A1 $26.6^\circ$ . As expected $\beta < \alpha$ , the path flattens toward the wall.

Worked Example 3 — two smooth spheres, oblique

Two equal smooth spheres collide. Sphere A moves at speed $u$ at $30^\circ$ to the line of centres; B is at rest; $e = 1$ . Find both velocities afterwards.

Resolve A's velocity relative to the line of centres (LOC):

$\text{along LOC} = u\cos 30^\circ = \tfrac{\sqrt3}{2}u, \qquad \text{perp. to LOC} = u\sin 30^\circ = \tfrac12 u.$

The perpendicular-to-LOC component of each smooth sphere is unchanged (no force perpendicular to the LOC). Along the LOC, apply momentum and restitution (equal masses $m$ ):

\begin{aligned} m\big(\tfrac{\sqrt3}{2}u\big) &= m v_A + m v_B &&\Rightarrow\; v_A + v_B = \tfrac{\sqrt3}{2}u\\ v_B - v_A &= 1\cdot\tfrac{\sqrt3}{2}u &&(e=1) \end{aligned}

Adding: $2v_B = \sqrt3\,u \Rightarrow v_B = \tfrac{\sqrt3}{2}u,\ v_A = 0$ (along the LOC).

B has only the LOC component: speed $\tfrac{\sqrt3}{2}u$ along the line of centres.
A has zero LOC component but keeps its perpendicular component $\tfrac12 u$ : speed $\tfrac12 u$ perpendicular to the LOC.

Mark scheme: M1 resolve relative to the LOC; B1 perpendicular components unchanged; M1 momentum along LOC; M1 restitution along LOC; A1 $v_B = \tfrac{\sqrt3}{2}u$ ; A1 A moves at $\tfrac12 u$ perpendicular to the LOC. With equal masses and $e=1$ the LOC components exchange, exactly as in the direct case — a neat consistency check.

Worked Example 4 — wall bounce in $\vec i, \vec j$ form

A particle moving with velocity $(6\vec i - 4\vec j)\ \mathrm{m\,s^{-1}}$ strikes a smooth wall lying along the $\vec i$ -axis (so the wall's normal is $\vec j$ ); $e = 0.75$ . Find the velocity after impact.

The wall is along $\vec i$ , so the $\vec i$ -component is parallel (unchanged) and the $\vec j$ -component is perpendicular (obeys restitution: reverse and scale by $e$ ):

\begin{aligned} \vec i\text{-component} &= 6 \quad(\text{unchanged}),\\ \vec j\text{-component} &= -0.75 \times (-4) = +3 \quad(\text{reversed and scaled}). \end{aligned}

So the velocity after impact is $(6\vec i + 3\vec j)\ \mathrm{m\,s^{-1}}$ . Mark scheme: M1 identify which component is normal to the wall; M1 apply $\times(-e)$ to the normal component, leave the parallel unchanged; A1 $(6\vec i + 3\vec j)$ . When the wall is conveniently aligned with an axis, the resolution is automatic — you simply scale the normal component by $-e$ and leave the other alone.

Worked Example 5 — direct collision as a special case

Sphere A (mass $2\ \mathrm{kg}$ ) moves at $6\ \mathrm{m\,s^{-1}}$ straight along the line of centres toward sphere B (mass $3\ \mathrm{kg}$ ) at rest; $e = 0.4$ . Find the velocities afterwards.

Here A's velocity is entirely along the LOC (no perpendicular component), so this is just a direct collision:

\begin{aligned} 2(6) &= 2v_A + 3v_B &&\Rightarrow\; 12 = 2v_A + 3v_B\\ v_B - v_A &= 0.4(6) = 2.4 &&\Rightarrow\; v_B = v_A + 2.4 \end{aligned}

So $12 = 2v_A + 3(v_A + 2.4) = 5v_A + 7.2 \Rightarrow v_A = 0.96,\ v_B = 3.36\ \mathrm{m\,s^{-1}}$ . Mark scheme: M1 momentum; M1 restitution; A1 $v_A = 0.96$ ; A1 $v_B = 3.36$ . This confirms that oblique sphere–sphere impact contains the direct collision as the zero-perpendicular-component limit — the methods are one and the same.

Worked Example 6 — symbolic wall-rebound speed

A particle hits a smooth wall at angle $\alpha$ to the wall with speed $c$ ; the coefficient of restitution is $e$ . Show that the rebound speed is $c\sqrt{\cos^2\alpha + e^2\sin^2\alpha}$ , and deduce the rebound speed when $\alpha = 30^\circ,\ c = 10,\ e = 0.8$ .

Resolve at speed $c$ : parallel component $c\cos\alpha$ (unchanged), perpendicular component $c\sin\alpha$ (becomes $ec\sin\alpha$ on rebound). Recombine by Pythagoras:

$v = \sqrt{(c\cos\alpha)^2 + (ec\sin\alpha)^2} = c\sqrt{\cos^2\alpha + e^2\sin^2\alpha}.$

For $\alpha = 30^\circ,\ c = 10,\ e = 0.8$ :

$v = 10\sqrt{\cos^2 30^\circ + 0.64\sin^2 30^\circ} = 10\sqrt{0.75 + 0.64(0.25)} = 10\sqrt{0.91} \approx 9.54\ \mathrm{m\,s^{-1}}.$

Mark scheme: M1 resolve into parallel and perpendicular; M1 apply $e$ to the perpendicular only; M1 recombine by Pythagoras; A1 $c\sqrt{\cos^2\alpha + e^2\sin^2\alpha}$ ; A1 $\approx 9.54\ \mathrm{m\,s^{-1}}$ . The symbolic form makes the physics transparent: when $e=1$ the surd collapses to $\sqrt{\cos^2\alpha + \sin^2\alpha} = 1$ , so $v = c$ — a perfectly elastic bounce keeps the speed, as it must.

Worked Example 7 — unequal smooth spheres, oblique

A smooth sphere A of mass $3\ \mathrm{kg}$ moving at $4\ \mathrm{m\,s^{-1}}$ at $60^\circ$ to the line of centres strikes a smooth sphere B of mass $1\ \mathrm{kg}$ at rest; $e = 0.5$ . Find the speed of each sphere afterwards.

Resolve A relative to the line of centres (LOC):

$\text{along LOC} = 4\cos 60^\circ = 2, \qquad \text{perp. to LOC} = 4\sin 60^\circ = 2\sqrt3 \approx 3.46.$

Perpendicular-to-LOC components are unchanged: A keeps $2\sqrt3$ ; B has none.

Along the LOC (momentum and restitution, with B at rest):

\begin{aligned} 3(2) + 1(0) &= 3v_A + 1v_B &&\Rightarrow\; 6 = 3v_A + v_B\\ v_B - v_A &= 0.5(2 - 0) = 1 &&\Rightarrow\; v_B = v_A + 1 \end{aligned}

Substitute: $6 = 3v_A + (v_A + 1) = 4v_A + 1 \Rightarrow v_A = 1.25,\ v_B = 2.25$ (along LOC).

B has only the LOC component: speed $2.25\ \mathrm{m\,s^{-1}}$ along the line of centres.
A has LOC component $1.25$ and perpendicular component $2\sqrt3 \approx 3.46$ : speed $\sqrt{1.25^2 + (2\sqrt3)^2} = \sqrt{1.5625 + 12} = \sqrt{13.5625} \approx 3.68\ \mathrm{m\,s^{-1}}$ .

Mark scheme: M1 resolve A relative to LOC; B1 perpendicular component of A unchanged; M1 momentum along LOC; M1 restitution along LOC; A1 $v_A = 1.25,\ v_B = 2.25$ along LOC; A1 A's resultant speed $\approx 3.68\ \mathrm{m\,s^{-1}}$ . With unequal masses the LOC components do not simply exchange — you must solve the simultaneous pair, but the perpendicular component still rides along untouched.

4 Specimen-style exam question

(Specimen-style.) A smooth sphere A of mass $2\ \mathrm{kg}$ moving at $5\ \mathrm{m\,s^{-1}}$ at $60^\circ$ to the line of centres strikes an identical-radius smooth sphere B of mass $3\ \mathrm{kg}$ at rest; $e = \tfrac12$ . Find the velocity of each sphere after impact.

Resolve A relative to the line of centres (LOC):

$\text{along LOC} = 5\cos 60^\circ = 2.5, \qquad \text{perp. to LOC} = 5\sin 60^\circ = 2.5\sqrt3 \approx 4.33.$

Perpendicular-to-LOC components are unchanged: A keeps $2.5\sqrt3$ ; B has none.

Along the LOC (momentum and restitution):

\begin{aligned} 2(2.5) + 3(0) &= 2v_A + 3v_B &&\Rightarrow\; 5 = 2v_A + 3v_B\\ v_B - v_A &= \tfrac12(2.5 - 0) = 1.25 &&\Rightarrow\; v_B = v_A + 1.25 \end{aligned}

Substitute: $5 = 2v_A + 3(v_A + 1.25) = 5v_A + 3.75 \Rightarrow v_A = 0.25,\ v_B = 1.5$ (along LOC).

Oblique Collisions

Oblique Collisions

1 Where this sits in AQA 7367

2 Core theory — impact with a smooth wall

The tan⁡β=etan⁡α\tan\beta = e\tan\alphatanβ=etanα rule

Two smooth spheres — the line of centres

3 Worked examples with mark scheme

Worked Example 1 — speed and direction after a wall bounce

Worked Example 2 — the angle rule directly

Worked Example 3 — two smooth spheres, oblique

Worked Example 4 — wall bounce in i⃗,j⃗\vec i, \vec ji,j​ form

Worked Example 5 — direct collision as a special case

Worked Example 6 — symbolic wall-rebound speed

Worked Example 7 — unequal smooth spheres, oblique

4 Specimen-style exam question

More in Mathematics

The $\tan\beta = e\tan\alpha$ rule

Worked Example 4 — wall bounce in $\vec i, \vec j$ form