Collisions: 2 Spheres Oblique

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Edexcel

Edexcel · Old spec

A2 June 2025 Q7

EdexcelCurrent spec13 marksCollisions: 2 Spheres Oblique

Figure 4: plan view of spheres P (0.3 kg) and Q (0.4 kg) touching, with the line of centres dashed; P approaches at 5 m s⁻¹ at 30° and leaves at v m s⁻¹ at 60°; Q approaches at 3 m s⁻¹ at 60° and leaves at w m s⁻¹ at θ° — Figure 4

Two uniform spheres \(P\) and \(Q\) have equal radii. The mass of \(P\) is 0.3 kg and the mass of \(Q\) is 0.4 kg. The spheres are moving on a smooth horizontal plane when they collide obliquely.

Immediately before the collision,

\(P\) is moving with speed \(5\ \text{m s}^{-1}\) at \(30^\circ\) to the line of centres of the spheres
\(Q\) is moving with speed \(3\ \text{m s}^{-1}\) at \(60^\circ\) to the line of centres of the spheres

Immediately after the collision,

\(P\) is moving with speed \(v\ \text{m s}^{-1}\) at \(60^\circ\) to the line of centres of the spheres
\(Q\) is moving with speed \(w\ \text{m s}^{-1}\) at \(\theta^\circ\) to the line of centres of the spheres

as in the plan view shown in Figure 4.

(a) Show that \(v = \dfrac{5\sqrt{3}}{3}\) (3)

(b) Find

(i) the value of \(w\)

(ii) the value of \(\theta\)

(7)

Figure 4 with velocity components marked: 5cos30 and 5sin30 for P before, vcos60 (or x) and vsin60 (or y) after; 3cos60 and 3sin60 for Q before, wcosθ (or b) and wsinθ (or a) after — Mark scheme (a)

Scheme	Marks	AO

Velocity of \(P\) perpendicular to the line of centres unchanged	M1	3.4
\(5\sin 30^\circ = v\sin 60^\circ\)	A1	1.1b
\(\dfrac{5}{2} = \dfrac{v\sqrt{3}}{2} \quad\Rightarrow v = \dfrac{5\sqrt{3}}{3}\) *	A1*	2.2a
	(3)

Scheme	Marks	AO

Velocity of \(Q\) perpendicular to the line of centres unchanged	M1	3.4
Either \(w\sin\theta^\circ = 3\sin 60^\circ \left(= \dfrac{3\sqrt{3}}{2}\right)\) or \(a = 3\sin 60^\circ \left(= \dfrac{3\sqrt{3}}{2}\right)\)	A1	1.1b
CLM parallel to the line of centres	M1	3.4
Either \(0.3 \times 5\cos 30^\circ - 0.4 \times 3\cos 60^\circ = 0.4w\cos\theta^\circ - 0.3v\cos 60^\circ\) Or \(0.3 \times 5\cos 30^\circ - 0.4 \times 3\cos 60^\circ = 0.4b - 0.3x\)	A1	1.1b
Solve for \(w\) or \(\theta\) using \(\begin{cases} w\sin\theta^\circ = \dfrac{3\sqrt{3}}{2} \\[4pt] w\cos\theta^\circ = \dfrac{5\sqrt{3} - 3}{2} \end{cases}\) or \(\begin{cases} a = \dfrac{3\sqrt{3}}{2} \\[4pt] b = \dfrac{5\sqrt{3} - 3}{2} \end{cases}\)	dM1	3.1a
\(\theta = 43 \;\; (42.552\ldots)\) or \(w = 3.8 \;\; (3.84182\ldots)\)	A1	2.2a
\(\theta = 43 \;\; (42.552\ldots)\) and \(w = 3.8 \;\; (3.84182\ldots)\)	A1	2.2a
	(7)

Scheme	Marks	AO

NEL parallel to the line of centres	M1	3.1b
\(v\cos 60^\circ + w\cos\theta^\circ = e(5\cos 30^\circ + 3\cos 60^\circ)\) \(x + b = e(5\cos 30^\circ + 3\cos 60^\circ)\)	A1ft	1.1b
\(e = 0.73 \quad (0.73300\ldots)\)	A1	1.1b
	(3)
	(13 marks)

A2 June 2024 Q7

EdexcelCurrent spec15 marksCollisions: 2 Spheres Oblique

Figure 1: sphere A (m) approaching sphere B (3m) at speed U at angle α to the line of centres, then moving off at right angles to its original direction — Figure 1

A smooth uniform sphere \(A\) of mass \(m\) is moving with speed \(U\) on a smooth horizontal plane. The sphere \(A\) collides obliquely with a smooth uniform sphere \(B\) of mass \(3m\) which is at rest on the plane. The two spheres have the same radius.

Immediately before the collision, the direction of motion of \(A\) makes an angle \(\alpha\), where \(0^\circ \lt \alpha \lt 90^\circ\), with the line joining the centres of the spheres.

Immediately after the collision, the direction of motion of \(A\) is perpendicular to its original direction, as shown in Figure 1.

The coefficient of restitution between the spheres is \(e\).

(a) Show that the speed of \(B\) immediately after the collision is \[\frac{1}{4}(1 + e)U\cos\alpha\] (6)

(b) Show that \(e \gt \dfrac{1}{3}\) (4)

Figure 1 twice with velocity components: before, U cos α along the line of centres and U sin α perpendicular, B at rest (0); after, A has v_A along the line of centres (left), or speed V with components V sin α and V cos α (right), and B has v_B — Mark scheme (a)

Scheme	Marks	AO

CLM along line of centres	M1	3.1b
\(mU\cos\alpha = -mv_A + 3mv_B\) OR \(mU\cos\alpha = -mV\sin\alpha + 3mv_B\)	A1	1.1b
Impact Law used along line of centres	M1	3.3
\(eU\cos\alpha = v_A + v_B\) OR \(eU\cos\alpha = V\sin\alpha + v_B\)	A1	1.1b
Solve for \(v_B\)	DM1	2.1
\(\dfrac{1}{4}(1 + e)U\cos\alpha\) *	A1*	2.2a
	(6)

Scheme	Marks	AO

Unsimplified expression for \(v_A\) seen OR Unsimplified expression for \(V\sin\alpha\) seen	M1	2.1
\(= \dfrac{1}{4}(3e - 1)U\cos\alpha\) oe seen	A1	1.1b
Solve \(v_A \gt 0\) to find an inequality for \(e\) OR Solve \(V\sin\alpha \gt 0\) to find an inequality for \(e\)	M1	3.1b
Correct and complete reasoning leading to \(e \gt \dfrac{1}{3}\) * Must include: correct inequality for \(v_A\), \(0^\circ \lt \alpha \lt 90^\circ\) or \(\alpha\) is acute, \(\cos\alpha \gt 0\).	A1*	2.2a
	(4)

Scheme	Marks	AO

Velocity component of \(A\), perpendicular to the line of centres, after collision \(= U\sin\alpha\)	B1	3.3
Use of \(90^\circ\) deflection: \(\tan\alpha = \dfrac{\frac{1}{4}(3e - 1)U\cos\alpha}{U\sin\alpha}\)	M1	2.1
Correct equation in \(e\) and \(\tan\alpha\) \(\tan^2\alpha = \dfrac{1}{4}(3e - 1)\) oe	A1	1.1b
\(e \leqslant 1 \;\Rightarrow\; \tan^2\alpha \leqslant \dfrac{1}{2}\)	DM1	3.1b
\(0 \lt \tan\alpha \leqslant \dfrac{1}{\sqrt{2}}\) since \(0^\circ \lt \alpha \leqslant 90^\circ\) *	A1*	2.2a
	(5)
	(15 marks)

A2 June 2023 Q5

EdexcelCurrent spec10 marksCollisions: 2 Spheres Oblique

Figure 1: two equal spheres touching, S (m) on the left and a sphere of mass M on the right, with the line of centres dashed; S approaches with speed U at angle α to the line of centres and leaves at angle β to it — Figure 1

A smooth uniform sphere \(S\) of mass \(m\) is moving with speed \(U\) on a smooth horizontal plane. The sphere \(S\) collides obliquely with another uniform sphere of mass \(M\) which is at rest on the plane. The two spheres have the same radius.

Immediately before the collision the direction of motion of \(S\) makes an angle \(\alpha\), where \(0 < \alpha < 90^\circ\), with the line joining the centres of the spheres.

Immediately after the collision the direction of motion of \(S\) makes an angle \(\beta\) with the line joining the centres of the spheres, as shown in Figure 1.

The coefficient of restitution between the spheres is \(e\).

(a) Show that \(\tan\beta = \dfrac{(m + M)\tan\alpha}{(m - eM)}\) (8)

Given that \(m = eM\),

(b) show that the directions of motion of the two spheres immediately after the collision are perpendicular. (2)

Diagram: S (m) approaching at U at angle α to the line of centres and leaving at angle β; the sphere (M) leaves along the line of centres; components after impact: v₁ along the line of centres and U sin α perpendicular for S, or V cos β and V sin β; v₂ for the other sphere — Mark scheme (a)

Scheme	Marks	AO
If it helps the candidate, ignore their diagram.
\(U\sin\alpha\) seen as velocity component of \(S\), perpendicular to line of centres after impact.	B1	3.4
CLM along line of centres	M1	3.1b
\(mU\cos\alpha = mv_1 + Mv_2\)	A1	1.1b
NEL used along line of centres	M1	3.3
\(eU\cos\alpha = -v_1 + v_2\)	A1	1.1b
\(\tan\beta = \dfrac{U\sin\alpha}{v_1}\)	dM1	2.1
Solve to produce an equation for \(\tan\beta\) in \(m\), \(M\), \(e\) and \(\alpha\)	dM1	1.1b
\(\tan\beta = \dfrac{(m + M)\tan\alpha}{(m - eM)}\) *	A1*	1.1b
	(8)

Scheme	Marks	AO
\(U\sin\alpha\) seen as velocity cpt of \(S\), perpendicular to line of centres after impact.	B1	3.4
CLM along line of centres	M1	3.1b
\(mU\cos\alpha = mV\cos\beta + Mv_2\)	A1	1.1b
NEL used along line of centres	M1	3.3
\(eU\cos\alpha = -V\cos\beta + v_2\)	A1	1.1b
\(\tan\beta = \dfrac{U\sin\alpha}{V\cos\beta}\) or \(V\sin\beta = U\sin\alpha\)	dM1	2.1
Solve to produce an equation for \(\tan\beta\) in \(m\), \(M\), \(e\) and \(\alpha\)	dM1	1.1b
\(\tan\beta = \dfrac{(m + M)\tan\alpha}{(m - eM)}\) *	A1*	1.1b
	(8)

Mark scheme (b)

Scheme	Marks	AO
Use the given condition to find the direction of \(S\) after impact. Eg \(m = eM \;\Rightarrow\; \tan\beta = \infty\) or \(\tan\beta\) is undefined so \(\beta = 90^\circ\) oe \(m = eM \;\Rightarrow\; v_1 = 0\) so velocity component of \(S\) parallel to line of centres is zero.	M1	3.1b
Conclusion: After the collision, \(S\) moves perpendicular to the line of centres and the other sphere moves parallel to the line of centres i.e. they move at right angles oe *	A1*	2.4
	(2)
	(10 marks)

Notes

M1: Use of given condition to deduce that \(\beta = 90^\circ\) or that velocity component parallel to line of centres is zero.

A1*: Correct explanation using given information. Must refer correctly to the direction of both particles, eg perpendicular, at right angles, parallel and perpendicular to the line of centres,
Do not accept horizontally and vertically since the surface is defined as horizontal.

A2 June 2022 Q4

EdexcelCurrent spec9 marksCollisions: 2 Spheres Oblique

Figure 3: spheres A (3m) and B (4m) colliding; A approaches with speed 3u at 30° to the line of centres, B approaches with speed 2u at 30° to the line of centres, and B leaves at 60° to the line of centres — Figure 3

Two smooth uniform spheres, \(A\) and \(B\), have equal radii. The mass of \(A\) is \(3m\) and the mass of \(B\) is \(4m\). The spheres are moving on a smooth horizontal plane when they collide obliquely. Immediately before they collide, \(A\) is moving with speed \(3u\) at \(30^\circ\) to the line of centres of the spheres and \(B\) is moving with speed \(2u\) at \(30^\circ\) to the line of centres of the spheres. The direction of motion of \(B\) is turned through an angle of \(90^\circ\) by the collision, as shown in Figure 3.

(i) Find the size of the angle through which the direction of motion of \(A\) is turned as a result of the collision.

(ii) Find, in terms of \(m\) and \(u\), the magnitude of the impulse received by \(B\) in the collision. (9)

Figure 3 with A leaving with speed w at angle θ to the line of centres and B leaving with speed v at 60° to the line of centres — Mark scheme (i)

Scheme	Marks	AO

Parallel to line of centres:	M1	3.1b
\(9mu\cos 30^\circ - 8mu\cos 30^\circ = 4mv\cos 60^\circ - 3mw\cos\theta\) \((u\cos 30^\circ = 2v - 3w\cos\theta)\) \((u\cos 30^\circ = 2v - 3w_x)\)	A1	1.1b
\(\updownarrow A:\ (w_y =)\, w\sin\theta = 3u\sin 30^\circ \left(= \dfrac{3u}{2}\right)\) \(\updownarrow B:\ v\sin 60^\circ = 2u\sin 30^\circ\ (= u)\) \(\left(v = \dfrac{2u}{\sqrt{3}}\right)\)	B1 B1	3.4 3.4
\((w_x =)\, w\cos\theta = \dfrac{1}{3}(2v - u\cos 30^\circ) = \dfrac{5u\sqrt{3}}{18}\) \((w_y =)\, w\sin\theta = \dfrac{3u}{2}\) \(\left(\Rightarrow \tan\theta = \dfrac{9\sqrt{3}}{5},\ \ \theta = 72.2^\circ\right)\)	M1	1.1b
Direction deflected by \(77.8^\circ\) (\(78^\circ\) or better)	A1	2.2a

Mark scheme (ii)

Scheme	Marks	AO
Magnitude of impulse	M1	3.1b
\(= 4m\big(v\cos 60^\circ - (-2u\cos 30^\circ)\big)\)	A1	1.1b
\(= 4m\left(\dfrac{1}{2} \times \dfrac{u}{\sin 60^\circ} - \left(-2u\dfrac{\sqrt{3}}{2}\right)\right) = \dfrac{16\sqrt{3}}{3}mu\)	A1	2.2a
OR: magnitude \(= 3m(3u\cos 30^\circ + w\cos\theta)\) \(= 3m\left(+\dfrac{5u\sqrt{3}}{18} - \left(-\dfrac{3u\sqrt{3}}{2}\right)\right) = \dfrac{16\sqrt{3}}{3}mu\)
	(9)
	(9 marks)

Notes

M1: Use of \(I = mv - mu\) in direction of line of centres. Condone subtraction in either order
Allow M1 if they think that they have subtracted but they have not actually taken account of the change of direction.
Allow M1 if they go direct to the correct expression with a + without telling you that they have taken account of the change in direction
Allow M1 if they go straight to an unsimplified expression in surds using values already found earlier.

A1: Correct unsimplified expression.
Allow the negative of this

A1: Any equivalent simplified form. Must be positive. Condone if they change sign at the very end without explaining why. Accept \(9.2(376\ldots)mu\) (2 sf or better)

NB You might see candidates using the right angle and matrix multiplication to rotate the initial velocity of \(B\) to find the correct components of the velocity of \(B\) after impact.

A2 October 2021 Q3

EdexcelCurrent spec14 marksCollisions: 2 Spheres Oblique

3. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.]

A smooth uniform sphere \(P\) has mass 0.3 kg. Another smooth uniform sphere \(Q\), with the same radius as \(P\), has mass 0.5 kg.

The spheres are moving on a smooth horizontal surface when they collide obliquely. Immediately before the collision the velocity of \(P\) is \((u\mathbf{i} + 2\mathbf{j})\ \text{m s}^{-1}\), where \(u\) is a positive constant, and the velocity of \(Q\) is \((-4\mathbf{i} + 3\mathbf{j})\ \text{m s}^{-1}\)

At the instant when the spheres collide, the line joining their centres is parallel to \(\mathbf{i}\).

The coefficient of restitution between \(P\) and \(Q\) is \(\dfrac{3}{5}\)

As a result of the collision, the direction of motion of \(P\) is deflected through an angle of \(90^\circ\) and the direction of motion of \(Q\) is deflected through an angle of \(\alpha^\circ\)

(a) Find the value of \(u\) (8)

(b) Find the value of \(\alpha\) (5)

Diagram: P (0.3 kg) arrives with (ui + 2j) m/s and leaves with (−(4/u)i + 2j) m/s; Q (0.5 kg) arrives with (−4i + 3j) m/s and leaves with (wi + 3j) m/s — Mark scheme (a)

Scheme	Marks	AO

For \(P\) after: Component in \(\mathbf{j}\) direction \(= 2\)	B1	3.4
Deflected through \(90^\circ\) so velocity after \(= \left(-\dfrac{4}{u}\mathbf{i} + 2\mathbf{j}\right)\ (\text{m s}^{-1})\)	B1	3.4
CLM parallel to line of centres:	M1	3.1a
\(0.3\left(u + \dfrac{4}{u}\right) = 0.5(4 + w)\)	A1ft	1.1b
Impact law parallel to line of centres:	M1	3.1a
\(w + \dfrac{4}{u} = \dfrac{3}{5}(u + 4)\)	A1ft	1.1b
\(\left\{\begin{aligned} 3u + \dfrac{12}{u} &= 20 + 5w \\ \dfrac{20}{u} + 5w &= 3u + 12 \end{aligned}\right.\)	M1	1.1b
\(\left(\Rightarrow \dfrac{32}{u} = 32,\right)\ \ u = 1\)	A1	2.2a
	(8)

Mark scheme (b)

Scheme	Marks	AO
For \(Q\) after: \(w = -1\)	B1	1.1b
\(\mathbf{v} = -\mathbf{i} + 3\mathbf{j}\)	B1ft	1.1b
Find relevant angle between directions	M1	3.1a
\(\alpha^\circ = \tan^{-1}3 - \tan^{-1}\dfrac{3}{4}\) or \(\alpha^\circ = \cos^{-1}\left(\dfrac{4 + 9}{5 \times \sqrt{10}}\right)\)	A1ft	1.1b
\(\alpha = 34.7\ \ (35)\)	A1	1.1b
	(5)

Notes

B1: Correct only

B1ft: Follow their \(w\)

M1: Correct method to find a relevant angle between the directions

A1ft: Correct unsimplified expression. Follow their \(\mathbf{v}\)

A1: 35 or better (34.695…) 0.61 radians

Mark scheme (c)

Scheme	Marks	AO
The line of centres is parallel to the surface the spheres are moving on, so the impulse acts parallel to the surface.	B1	3.5a
	(1)
	(14 marks)

Notes

B1: Or equivalent that explains that the line of centres is parallel to the surface

A2 October 2020 Q5

EdexcelCurrent spec14 marksCollisions: 2 Spheres Oblique

5. A smooth uniform sphere \(P\) has mass 0.3 kg. Another smooth uniform sphere \(Q\), with the same radius as \(P\), has mass 0.2 kg.

The spheres are moving on a smooth horizontal surface when they collide obliquely. Immediately before the collision the velocity of \(P\) is \((4\mathbf{i} + 2\mathbf{j})\ \text{m s}^{-1}\) and the velocity of \(Q\) is \((-3\mathbf{i} + \mathbf{j})\ \text{m s}^{-1}\).

At the instant of collision, the line joining the centres of the spheres is parallel to \(\mathbf{i}\).

The kinetic energy of \(Q\) immediately after the collision is half the kinetic energy of \(Q\) immediately before the collision.

(a) Find

(i) the velocity of \(P\) immediately after the collision,

(ii) the velocity of \(Q\) immediately after the collision,

(iii) the coefficient of restitution between \(P\) and \(Q\),

carefully justifying your answers.

(11)

(b) Find the size of the angle through which the direction of motion of \(P\) is deflected by the collision. (3)

Diagram: P (0.3 kg) arrives with (4i + 2j) m/s and leaves with components −ui m/s and 2j m/s; Q (0.2 kg) arrives with (−3i + j) m/s and leaves with components vi m/s and j m/s — Mark scheme (a)

Scheme	Marks	AO

Components perpendicular to the line of centres after the collision: \(\mathbf{v}_{P\mathbf{j}} = 2\mathbf{j}\ (\text{m s}^{-1})\), \(\mathbf{v}_{Q\mathbf{j}} = \mathbf{j}\ (\text{m s}^{-1})\)	B1	3.4
Kinetic energy:	M1	3.1a
\(\dfrac{1}{2} \times 0.2 \times (v^2 + 1) = \dfrac{1}{2} \times \dfrac{1}{2} \times 0.2 \times (9 + 1)\)	A1	1.1b
CLM parallel to line of centres:	M1	3.1a
\(0.3 \times 4 - 0.2 \times 3 = 0.2v - 0.3u\) \((6 = 2v - 3u)\)	A1	1.1b
Impact law parallel to line of centres	M1	3.1a
\(v + u = e(4 + 3)\)	A1	1.1b
Solve for \(\mathbf{v}_P\), \(\mathbf{v}_Q\) or \(e\)	M1	1.1b
\(\mathbf{v}_P = \dfrac{2}{3}\mathbf{i} + 2\mathbf{j}\ (\text{m s}^{-1})\) and \(\mathbf{v}_Q = 2\mathbf{i} + \mathbf{j}\ (\text{m s}^{-1})\)	A1	1.1b
\(e = \dfrac{4}{21}\)	A1	1.1b
\(v = -2\ \Rightarrow\ u = -\dfrac{10}{3}\ \Rightarrow\) \(P\) and \(Q\) have passed through each other: impossible, so solution is unique *	A1*	2.4
	(11)

Mark scheme (b)

Scheme	Marks	AO
Use trig to find angle between velocities	M1	3.1a
\(\cos\theta = \left(\dfrac{\frac{8}{3} + 4}{\sqrt{20}\sqrt{4\frac{4}{9}}}\right)\) or \(\theta = \tan^{-1}\dfrac{2}{2/3} - \tan^{-1}\dfrac{1}{2}\)	A1ft	1.1b
\(\theta = 45^\circ\ \ \left(\dfrac{\pi}{4}\ \text{rads}\right)\)	A1	1.1b
	(3)
	(14 marks)

Notes

M1: Use of trig or equivalent to find a relevant angle between two velocities
e.g by scalar product or difference between angles.

A1ft: Correct unsimplified equation in \(\theta\). Follow their \(\mathbf{v}_P\)

A1: Correct only. (0.785… radians) Do not ISW

A2 June 2019 Q6

EdexcelCurrent spec12 marksCollisions: 2 Spheres Oblique Impulse & Momentum

6. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.]

A smooth uniform sphere \(A\) has mass 0.2 kg and another smooth uniform sphere \(B\), with the same radius as \(A\), has mass 0.4 kg.

The spheres are moving on a smooth horizontal surface when they collide obliquely. Immediately before the collision, the velocity of \(A\) is \((3\mathbf{i} + 2\mathbf{j})\ \text{m s}^{-1}\) and the velocity of \(B\) is \((-4\mathbf{i} - \mathbf{j})\ \text{m s}^{-1}\)

At the instant of collision, the line joining the centres of the spheres is parallel to \(\mathbf{i}\)

The coefficient of restitution between the spheres is \(\dfrac{3}{7}\)

(a) Find the velocity of \(A\) immediately after the collision. (7)

(b) Find the magnitude of the impulse received by \(A\) in the collision. (2)

(c) Find, to the nearest degree, the size of the angle through which the direction of motion of \(A\) is deflected as a result of the collision. (3)

Diagram: A (0.2 kg) arrives with (3i + 2j) m/s and leaves with components vi to the left and 2j; B (0.4 kg) arrives with (−4i − j) m/s and leaves with components wi and (−j) — Mark scheme (a)

Scheme	Marks	AO

Perpendicular to line of centres: \(2\mathbf{j}\)	B1	3.4
CLM parallel to the line of centres	M1	3.1b
\(0.2 \times 3 - 0.4 \times 4 = 0.4w - 0.2v\) \((-5 = 2w - v)\)	A1	1.1b
Impact law parallel to the line of centres	M1	3.4
\(7e = v + w\ \ \Rightarrow 3 = v + w\)	A1	1.1b
Complete strategy to find \(\mathbf{v}_A\)	M1	3.1b
\(\mathbf{v}_A = -\dfrac{11}{3}\mathbf{i} + 2\mathbf{j}\ (\text{m s}^{-1})\) follow their \(2\mathbf{j}\)	A1ft	1.1b
	(7)

Mark scheme (b)

Scheme	Marks	AO
Magnitude of impulse on \(A\): \(0.2\left(\dfrac{11}{3} - (-3)\right)\)	M1	3.1b
\(= 0.2\left(\dfrac{11}{3} + 3\right) = \dfrac{4}{3}\) (N s)	A1	1.1b
	(2)

Notes

M1: Evidence of use of \(m(v - u)\) parallel to the line of centres

A1: 1.3 (N s) or better

Mark scheme (c)

Scheme	Marks	AO
Use of scalar product to find the angle	M1	3.1a
\(\cos\theta = \dfrac{(3\mathbf{i} + 2\mathbf{j}).\left(-\frac{11}{3}\mathbf{i} + 2\mathbf{j}\right)}{\sqrt{13} \times \sqrt{\frac{157}{9}}}\)	A1ft	1.1b
\(\theta = 118^\circ\)	A1	1.1b
Alternative method: \(180^\circ - \tan^{-1}\dfrac{2}{3} - \tan^{-1}\dfrac{6}{11}\) Or \(\tan^{-1}\dfrac{3}{2} + \tan^{-1}\dfrac{11}{6}\)
	(3)
	(12 marks)

Notes

M1: Complete method for finding the required angle.
Allow for \(\tan^{-1}\dfrac{3}{2}\) or \(\tan^{-1}\dfrac{2}{3}\) and \(\tan^{-1}\dfrac{6}{11}\) or \(\tan^{-1}\dfrac{11}{6}\)

A1ft: A correct unsimplified expression Follow their \(\mathbf{v}_A\). Do not ISW

A1: Correct answer only. (Q asks for the nearest degree) Do not ISW

\(62^\circ\) probably scores M1A0A0

M4 June 2018 Q7

EdexcelOld spec13 marksCollisions: 2 Spheres Oblique Impulse & Momentum

7. Two smooth uniform spheres \(A\) and \(B\), of mass 2 kg and 3 kg respectively, and of equal radius, are moving on a smooth horizontal plane when they collide.

Immediately before the collision the velocity of \(A\) is \((3\mathbf{i} + \mathbf{j})\) m s\(^{-1}\) and the velocity of \(B\) is \((-\mathbf{i} + 2\mathbf{j})\) m s\(^{-1}\). Immediately after the collision the velocity of \(A\) is \((\mathbf{i} + 3\mathbf{j})\) m s\(^{-1}\).

(a) Show that, at the instant when \(A\) and \(B\) collide, their line of centres is parallel to \(-\mathbf{i} + \mathbf{j}\). (4)

(b) Find the velocity of \(B\) immediately after the collision. (3)

Mark scheme (a)

Scheme	Marks
Impulse on \(A\): \(I = 2(\mathbf{i} + 3\mathbf{j} - 3\mathbf{i} - \mathbf{j})\)	M1A1
\(= -4\mathbf{i} + 4\mathbf{j} = 4(-\mathbf{i} + \mathbf{j})\)	A1
Impulse parallel to l.o.c., hence l.o.c. parallel to \(-\mathbf{i} + \mathbf{j}\) (Given answer)	A1
	(4)

Mark scheme (b)

Scheme	Marks
Impulse equal and opposite: \(4\mathbf{i} - 4\mathbf{j} = 3(\mathbf{v} + \mathbf{i} - 2\mathbf{j})\)	M1A1
\(3\mathbf{v} = \mathbf{i} + 2\mathbf{j},\quad \mathbf{v} = \dfrac{1}{3}(\mathbf{i} + 2\mathbf{j})\)	A1
	(3)

Alt using CLM

\(2(3\mathbf{i} + \mathbf{j}) + 3(-\mathbf{i} + 2\mathbf{j}) = 2(\mathbf{i} + 3\mathbf{j}) + 3\mathbf{v}\)	M1A1
\(3\mathbf{v} = \mathbf{i} + 2\mathbf{j},\quad \mathbf{v} = \dfrac{1}{3}(\mathbf{i} + 2\mathbf{j})\)	A1

Diagram: line of centres with velocity components of A (speed root 10, angles beta and 135 minus beta) and B (speeds root 5 and one third root 5, angles alpha and 45) — Mark scheme (c)

Scheme	Marks
Components of velocities parallel to \(-\mathbf{i} + \mathbf{j}\): \(A\) before : \((3\mathbf{i} + \mathbf{j})\cdot\dfrac{1}{\sqrt{2}}(-\mathbf{i} + \mathbf{j}) = \dfrac{-2}{\sqrt{2}}\) \(A\) after : \((\mathbf{i} + 3\mathbf{j})\cdot\dfrac{1}{\sqrt{2}}(-\mathbf{i} + \mathbf{j}) = \dfrac{2}{\sqrt{2}}\) \(B\) before : \((-\mathbf{i} + 2\mathbf{j})\cdot\dfrac{1}{\sqrt{2}}(-\mathbf{i} + \mathbf{j}) = \dfrac{3}{\sqrt{2}}\) \(B\) after : \(\dfrac{1}{3}(\mathbf{i} + 2\mathbf{j})\cdot\dfrac{1}{\sqrt{2}}(-\mathbf{i} + \mathbf{j}) = \dfrac{1}{3\sqrt{2}}\) follow through from 7(b) NB: the marks are all available if the unit vector \(\left(\sqrt{2}\right)\) is not used.	M1A3
Coefficient of restitution: \(\dfrac{1}{\sqrt{2}}\left(2 - \dfrac{1}{3}\right) = \dfrac{e}{\sqrt{2}}(3 + 2)\)	M1
\(e = \dfrac{1}{3}\)	A1
	(6)
	(13 marks)

M4 June 2017 Q2

EdexcelOld spec12 marksCollisions: 2 Spheres Oblique

Figure 1: sphere A (3m kg) with velocity 5i minus 2j and sphere B (m kg) with velocity 3i plus 4j colliding with line of centres parallel to j — Figure 1

Two smooth uniform spheres \(A\) and \(B\) have masses \(3m\) kg and \(m\) kg respectively and equal radii. The spheres are moving on a smooth horizontal surface. Initially, sphere \(A\) has velocity \((5\mathbf{i} - 2\mathbf{j})\) m s\(^{-1}\) and sphere \(B\) has velocity \((3\mathbf{i} + 4\mathbf{j})\) m s\(^{-1}\). When the spheres collide, the line joining their centres is parallel to \(\mathbf{j}\), as shown in Figure 1.

The coefficient of restitution between the two spheres is \(e\).

The kinetic energy of sphere \(B\) immediately after the collision is 85% of its kinetic energy immediately before the collision.

Find

(a) the velocity of each sphere immediately after the collision, (9)

(b) the value of \(e\). (3)

Diagram: after impact A has components 5i and w along j; B has components 3i and v in the negative j direction — Mark scheme (a)

Scheme	Marks
For \(A\), component perpendicular to loc = 5	B1
For \(B\), component perpendicular to loc = 3	B1
\(\dfrac{1}{2}m\times 25\times\dfrac{85}{100} = \dfrac{1}{2}m\left(3^2 + v^2\right)\)	M1
\(\dfrac{85}{4} = 9 + v^2,\quad v^2 = \dfrac{49}{4}\)	A1
\(-6m + 4m = 3mw - mv\) \((= 3mw - 3.5m)\)	M1
	A1ft
\(w = 0.5\)
Select correct root and state velocities:	DM1
\(\mathbf{v}_B = (3\mathbf{i} - 3.5\mathbf{j})\) (m s\(^{-1}\))	A1
\(\mathbf{v}_A = (5\mathbf{i} + 0.5\mathbf{j})\) (m s\(^{-1}\))	A1
	(9)

Mark scheme (b)

Scheme	Marks
\(v + w = e(2 + 4)\)	M1
\(0.5 + 3.5 = 6e\)	A1ft
\(e = \dfrac{2}{3}\)	A1
	(3)
	(12 marks)

Notes

M1 Impact law parallel to loc. Used the right way round. Condone sign error(s)

A1ft Correct unsimplified or with their values

M4 June 2016 Q1

EdexcelOld spec8 marksCollisions: 2 Spheres Oblique

Figure 1: sphere A of mass m moving at angle alpha to the line of centres towards sphere B of mass 4m — Figure 1

A smooth uniform sphere \(A\) of mass \(m\) is moving on a smooth horizontal plane when it collides with a second smooth uniform sphere \(B\), which is at rest on the plane. The sphere \(B\) has mass \(4m\) and the same radius as \(A\). Immediately before the collision the direction of motion of \(A\) makes an angle \(\alpha\) with the line of centres of the spheres, as shown in Figure 1. The direction of motion of \(A\) is turned through an angle of \(90^\circ\) by the collision and the coefficient of restitution between the spheres is \(\dfrac{1}{2}\)

Find the value of \(\tan\alpha\). (8)

Diagram: A (m) with speed u at angle alpha to the line of centres; after impact A has components v backwards along the line of centres and w perpendicular to it; B (4m) moves with speed x along the line of centres — Mark scheme

Diagram: A (m) with speed u at angle alpha to the line of centres; after impact A moves with speed v at angle 90 minus alpha to the line of centres; B (4m) moves with speed x — Mark scheme

Scheme	Marks
Along line of centres:
Con of mom: \(mu\cos\alpha = 4mx - mv\)	M1
\((u\cos\alpha = 4x - v)\)	A1
NLR: \(\dfrac{1}{2}u\cos\alpha = x + v\)	M1
\((2u\cos\alpha = 4x + 4v)\)	A1
\((5v = u\cos\alpha)\)
Perp to line of centres: no change to velocity so vel \(= w = u\sin\alpha\)	B1 (A1)
Deflected through \(90^\circ\) \(\left(\tan\alpha = \dfrac{v}{w}\right)\)	B1
\(\tan\alpha = \dfrac{\frac{1}{5}u\cos\alpha}{u\sin\alpha}\)	M1
\(\tan^2\alpha = \dfrac{1}{5}\qquad \tan\alpha = \sqrt{\dfrac{1}{5}}\) or 0.4472....	A1
	(8 marks)

M4 June 2015 Q3

EdexcelOld spec12 marksCollisions: 2 Spheres Oblique

Figure 1: spheres A (m) and B (2m) in contact; A moving with speed 3u and B with speed u, each at angle theta to the line of centres, in opposite directions — Figure 1

Two smooth uniform spheres \(A\) and \(B\) with equal radii have masses \(m\) and \(2m\) respectively. The spheres are moving in opposite directions on a smooth horizontal surface and collide obliquely. Immediately before the collision, \(A\) has speed \(3u\) with its direction of motion at an angle \(\theta\) to the line of centres, and \(B\) has speed \(u\) with its direction of motion at an angle \(\theta\) to the line of centres, as shown in Figure 1. The coefficient of restitution between the spheres is \(\dfrac{1}{8}\)

Immediately after the collision, the speed of \(A\) is twice the speed of \(B\).

Find the size of the angle \(\theta\). (12)

Velocity components before: A 3u cos theta and 3u sin theta, B u cos theta and u sin theta; after: A r and 3u sin theta, B s and u sin theta — Mark scheme

Scheme	Marks
After collision \(u\sin\theta\) and \(3u\sin\theta\) perpendicular to \(l\) of \(c\)	B1
CLM : \(r + 2s = 3u\cos\theta - 2u\cos\theta\ (= u\cos\theta)\)	M1 A1
Impact: \(s - r = e \times 4u\cos\theta\left(= \dfrac{u\cos\theta}{2}\right)\)	M1 A1
\(\Rightarrow r = 0,\ s = \dfrac{u\cos\theta}{2}\)	DM1 A1
After the collision: \((3u\sin\theta)^2 + r^2 = 4\left((u\sin\theta)^2 + s^2\right)\)	M1 A1
\(9u^2\sin^2\theta = 4u^2\sin^2\theta + 4.\dfrac{u^2}{4}\cos^2\theta\)	A1
\(\tan^2\theta = \dfrac{1}{5},\quad \theta = 24.1(^\circ)\ \ (0.421\text{ radians})\)	DM1 A1
	(12)
	(12 marks)

M4 June 2014 (R) Q4

EdexcelOld spec15 marksCollisions: 2 Spheres Oblique

4. A smooth uniform sphere \(S\) is moving on a smooth horizontal plane when it collides obliquely with an identical sphere \(T\) which is at rest on the plane. Immediately before the collision \(S\) is moving with speed \(U\) in a direction which makes an angle of 60\(^\circ\) with the line joining the centres of the spheres. The coefficient of restitution between the spheres is \(e\).

(a) Find, in terms of \(e\) and \(U\) where necessary,

(i) the speed and direction of motion of \(S\) immediately after the collision,

(ii) the speed and direction of motion of \(T\) immediately after the collision. (12)

The angle through which the direction of motion of \(S\) is deflected is \(\delta^\circ\).

(b) Find

(i) the value of \(e\) for which \(\delta\) takes the largest possible value,

(ii) the value of \(\delta\) in this case. (3)

Mark scheme (a)

Scheme	Marks
\(mv_1 + mv_2 = mu\cos 60^\circ\) Momentum	M1 A1
\(-v_1 + v_2 = eu\cos 60^\circ\) Impact law	M1 A1
\(v_1 = \dfrac{u(1 - e)}{4}\)
Speed of \(S = \sqrt{\dfrac{u^2(1 - e)^2}{16} + \dfrac{3u^2}{4}}\) Solve for \(v_1\) and find speed	M1 A1
\(= \dfrac{u}{4}\sqrt{e^2 - 2e + 13}\)	A1
\(\tan\theta = \dfrac{u\sqrt{3}}{2v_1} = \dfrac{2\sqrt{3}}{(1 - e)}\) Use components to find dirn	M1 A1
\(S\) moves at \(\arctan\dfrac{2\sqrt{3}}{(1 - e)}\) to the line of centres	M1 A1
\(v_2 = \dfrac{u(1 + e)}{4}\) \(v_2\) in terms of \(u,\ e\)	B1
\(T\) has speed \(\dfrac{u(1 + e)}{4}\) along the line of centres Conclusion
	(12)

Scheme	Marks
\(\theta\) is a max when \(e = 1\) then \(\theta = 90^\circ\)	M1 A1
then deflection angle is \(90^\circ - 60^\circ = 30^\circ\) \(\delta = 30\)	A1
	(3)
	(15 marks)

M4 June 2014 Q5

EdexcelOld spec12 marksCollisions: 2 Spheres Oblique Impulse & Momentum

Figure 1: spheres A(m) and B(3m) in contact; A moving with speed 3u at angle alpha to the line of centres, B moving with speed u at angle beta to the line of centres — Figure 1

Two smooth uniform spheres \(A\) and \(B\) have equal radii. The mass of \(A\) is \(m\) and the mass of \(B\) is \(3m\). The spheres are moving on a smooth horizontal plane when they collide obliquely. Immediately before the collision, \(A\) is moving with speed \(3u\) at angle \(\alpha\) to the line of centres and \(B\) is moving with speed \(u\) at angle \(\beta\) to the line of centres, as shown in Figure 1. The coefficient of restitution between the two spheres is \(\dfrac{1}{5}\). It is given that \(\cos\alpha = \dfrac{1}{3}\) and \(\cos\beta = \dfrac{2}{3}\) and that \(\alpha\) and \(\beta\) are both acute angles.

(a) Find the magnitude of the impulse on \(A\) due to the collision in terms of \(m\) and \(u\). (8)

(b) Express the kinetic energy lost by \(A\) in the collision as a fraction of its initial kinetic energy. (4)

Before: A components u along and 3u sin alpha perpendicular; B components 2u/3 along and u sin beta perpendicular. After: A components x and 3u sin alpha; B components y and u sin beta — Mark scheme (a)

Scheme	Marks
CLM: \(mx + 3my = 3m \times u\cos\beta - m \times 3u\cos\alpha = mu \quad (x + 3y = u)\)	M1 A1
NEL: \(x - y = \dfrac{1}{5}(3u\cos\alpha + u\cos\beta)\left(= \dfrac{1}{5}\left(u + \dfrac{2}{3}u\right) = \dfrac{1}{3}u\right)\)	M1 A1
\(x = \dfrac{u}{2}\), or \(\ y = \dfrac{u}{6}\)	DM1 A1
Magnitude of the impulse on \(A\) \(= mu - \left(m \times -\dfrac{u}{2}\right) = \dfrac{3mu}{2}\)	M1 A1
	(8)

Mark scheme (b)

Scheme	Marks
Component of velocity perpendicular to the line of centres before = component after \(= 3u\sin\alpha = 3u \times \dfrac{\sqrt{8}}{3} = \sqrt{8}u\)	B1
KE lost \(= \dfrac{m}{2}\left(9u^2 - \left(8u^2 + \dfrac{1}{4}u^2\right)\right)\left[= \dfrac{3}{8}mu^2\right]\)	M1 A1
Fraction lost \(= \dfrac{3/8}{9/2} = \dfrac{3}{8} \times \dfrac{2}{9} = \dfrac{1}{12}\)	A1
	(4)
	(12 marks)

Notes

M1 Change in KE. Does not need to be a fraction at this stage. Does not need to include the (cancelling) component perpendicular to the line of centre.

A1 Correct unsimplified

M4 June 2013 (R) Q3

EdexcelOld spec9 marksCollisions: 2 Spheres Oblique

3. A smooth uniform sphere \(A\), of mass \(5m\) and radius \(r\), is at rest on a smooth horizontal plane. A second smooth uniform sphere \(B\), of mass \(3m\) and radius \(r\), is moving in a straight line on the plane with speed \(u\) m s\(^{-1}\) and strikes \(A\). Immediately before the impact the direction of motion of \(B\) makes an angle of 60\(^\circ\) with the line of centres of the spheres. The direction of motion of \(B\) is turned through an angle of 30\(^\circ\) by the impact.

Find

(a) the speed of \(B\) immediately after the impact, (3)

(b) the coefficient of restitution between the spheres. (6)

Before: B (3m) moving with speed u towards A (5m) at rest. After: B moves with speed v perpendicular to the line of centres, A moves with speed w along it — Mark scheme (a)

Scheme	Marks
After impact \(B\) moves perpendicular to the line of centres	B1
Perp. to line of centres: \(\quad v = u\sin 60 = u\dfrac{\sqrt{3}}{2}\)	M1A1
	(3)

Mark scheme (b)

Scheme	Marks
Parallel to line of centres:
Con of Mom \(\quad 3mu\cos 60 + 5m \times 0 = 3m \times 0 + 5mw\)	M1A1
N.L.R. \(\quad eu\cos 60 = w\)	M1A1
\(\dfrac{1}{2}eu = w\ \ \&\ \ \dfrac{3}{2}u = 5w\)
\(\to\ \dfrac{1}{2}eu = \dfrac{3}{10}u\)	DM1
\(e = \dfrac{3}{5}\)	A1
	(6)
	(9 marks)

Notes

DM1 Dependent on the two previous M marks

M4 June 2013 Q3

EdexcelOld spec10 marksCollisions: 2 Spheres Oblique Impulse & Momentum

Figure 2: sphere A (3m) moving right with speed u, sphere B (2m) moving left with speed 2u, paths of centres 1.6r apart — Figure 2

Two smooth uniform spheres \(A\) and \(B\), of equal radius \(r\), have masses \(3m\) and \(2m\) respectively. The spheres are moving on a smooth horizontal plane when they collide. Immediately before the collision they are moving with speeds \(u\) and \(2u\) respectively. The centres of the spheres are moving towards each other along parallel paths at a distance \(1.6r\) apart, as shown in Figure 2.

The coefficient of restitution between the two spheres is \(\dfrac{1}{6}\).

Find, in terms of \(m\) and \(u\), the magnitude of the impulse received by \(B\) in the collision. (10)

Spheres at impact: centres 2r apart, 1.6r perpendicular to paths, 1.2r along paths, angle alpha — Mark scheme

Velocity components: A before 0.6u and 0.8u, A after a and 0.8u; B before 1.2u and 1.6u, B after b and 1.6u — Mark scheme

Scheme	Marks
\(0.6u\) or \(u\cos\alpha\)	B1
\(1.2u\) or \(2u\cos\alpha\)	B1
\(2m \times 1.2u - 3m \times 0.6u = 3ma + 2mb\)	M1
\((3a + 2b = 0.6u)\)	A1ft
\(e(1.2u + 0.6u) = a - b\)	M1
\((a - b = 0.3u)\)	A1ft
	DM1
\(a = 0.24u\) or \(b = -0.06u\)	A1
\((1.2u - (-0.06u)) \times 2m = 2.52mu\)	M1
or \((0.24u - (-0.6u)) \times 3m = 2.52mu\)	A1
	(10 marks)

M4 June 2012 Q1

EdexcelOld spec13 marksCollisions: 2 Spheres Oblique

1. A smooth uniform sphere \(S\), of mass \(m\), is moving on a smooth horizontal plane when it collides obliquely with another smooth uniform sphere \(T\), of the same radius as \(S\) but of mass \(2m\), which is at rest on the plane. Immediately before the collision the velocity of \(S\) makes an angle \(\alpha\), where \(\tan\alpha = \dfrac{3}{4}\), with the line joining the centres of the spheres. Immediately after the collision the speed of \(T\) is \(V\). The coefficient of restitution between the spheres is \(\dfrac{3}{4}\).

(a) Find, in terms of \(V\), the speed of \(S\)

(i) immediately before the collision,

(ii) immediately after the collision.

(9)

(b) Find the angle through which the direction of motion of \(S\) is deflected as a result of the collision. (4)

Diagram: S moving with speed u at angle alpha to the line of centres; after the collision S has components w along and u sin alpha perpendicular to the line of centres, T has V along and 0 perpendicular — Mark scheme (a)

Scheme	Marks

\(mu\cos\alpha = mw + 2mV\)	M1 A1
\(eu\cos\alpha = -w + V\)	M1 A1
\(u\cos\alpha(e + 1) = 3V \Rightarrow\) (i) \(u = \dfrac{15V}{7}\)	M1 A1
\(\Rightarrow w = -\dfrac{2V}{7}\)	A1
(ii) speed of \(S = \sqrt{\left(\frac{-2V}{7}\right)^2 + (u\sin\alpha)^2} = \dfrac{V\sqrt{85}}{7}\)	M1 A1
	(9)

Mark scheme (b)

Scheme	Marks
\(\tan\theta = \dfrac{\frac{9V}{7}}{\frac{2V}{7}} = \dfrac{9}{2}\)	M1 A1
defln angle \(= 180^\circ - (\theta + \alpha)\)	DM1
\(= 65.7^\circ\) (3 sf)	A1
	(4)
	(13 marks)

Notes

M1 Direction of \(S\) after the collision. Condone \(\dfrac{2}{9}\)

A1 \(77.5^\circ\) or \(12.5^\circ\) seen or implied

DM1 Combine their \(\theta\) and \(\alpha\) to find the required angle. e.g. \(12.5^\circ + \tan^{-1}\left(\dfrac{4}{3}\right)\)

A1 Accept \(66^\circ\)

M4 June 2011 Q1

EdexcelOld spec10 marksCollisions: 2 Spheres Oblique

Figure 1: spheres A (2m kg) and B (3m kg) in contact with their line of centres parallel to j; A moving with velocity (3i - 4j) m/s and B with velocity (2i + 3j) m/s — Figure 1

Two smooth uniform spheres \(A\) and \(B\) have masses \(2m\) kg and \(3m\) kg respectively and equal radii. The spheres are moving on a smooth horizontal surface. Initially, sphere \(A\) has velocity \((3\mathbf{i} - 4\mathbf{j})\) m s\(^{-1}\) and sphere \(B\) has velocity \((2\mathbf{i} - 3\mathbf{j})\) m s\(^{-1}\). When the spheres collide, the line joining their centres is parallel to \(\mathbf{j}\), as shown in Figure 1. The coefficient of restitution between the spheres is \(\dfrac{3}{7}\). Find, in terms of \(m\), the total kinetic energy lost in the collision. (10)

Diagram: A (2m) with velocity (3i - 4j) and B (3m) with velocity (2i + 3j) before; after, A has components a along i and v along j, B has b along i and w in the minus j direction — Mark scheme

Scheme	Marks

\(\leftrightarrow a = 3\ \&\ b = 2\)	B1
b Conservation of linear momentum: \(\ -4 \times 2 + 3 \times 3 = 2v - 3w\ (= 1)\)	M1 A1
Restitution: \(\ v + w = e \times 7\ \ (= 3)\)	M1 A1
Solve the simultaneous equations	DM1
giving \(v = 2\) and \(w = 1\)	A1
KE lost \(= \dfrac{1}{2} \times 2m \times ((16 + 9) - (4 + 9)) + \dfrac{1}{2} \times 3m \times ((9 + 4) - (1 + 4))\)	M1 A1
\(= 24m\) (J)	A1
	(10 marks)

M4 June 2010 Q2

EdexcelOld spec14 marksCollisions: 1 Sphere Oblique Collisions: 2 Spheres Oblique

2. Two smooth uniform spheres \(S\) and \(T\) have equal radii. The mass of \(S\) is 0.3 kg and the mass of \(T\) is 0.6 kg. The spheres are moving on a smooth horizontal plane and collide obliquely. Immediately before the collision the velocity of \(S\) is \(\mathbf{u}_1\) m s\(^{-1}\) and the velocity of \(T\) is \(\mathbf{u}_2\) m s\(^{-1}\). The coefficient of restitution between the spheres is 0.5. Immediately after the collision the velocity of \(S\) is \((-\mathbf{i} + 2\mathbf{j})\) m s\(^{-1}\) and the velocity of \(T\) is \((\mathbf{i} + \mathbf{j})\) m s\(^{-1}\). Given that when the spheres collide the line joining their centres is parallel to \(\mathbf{i}\),

(a) find

(i) \(\mathbf{u}_1\),

(ii) \(\mathbf{u}_2\).

(6)

After the collision, \(T\) goes on to collide with a smooth vertical wall which is parallel to \(\mathbf{j}\). Given that the coefficient of restitution between \(T\) and the wall is also 0.5, find

(b) the angle through which the direction of motion of \(T\) is deflected as a result of the collision with the wall, (5)

Diagram: after the collision S has components 1 to the left and 2 up, T has 1 to the right and 1 up; before, S has v to the right and 2 up, T has w to the left and 1 up — Mark scheme (a)

Scheme	Marks

\(0.3v - 0.6w = 0.3\)	M1 A1
\(v - 2w = 1\)
\(\tfrac{1}{2}(v + w) = 2\)	M1 A1
\(v + w = 4\)
\(w = 1,\ v = 3\)
(i) \(\mathbf{u}_1 = 3\mathbf{i} + 2\mathbf{j}\) (ii) \(\mathbf{u}_2 = -\mathbf{i} + \mathbf{j}\)	A1 A1
	(6)

Scheme	Marks
\(v = 0.5\)	B1

\(\tan\theta = 0.5\) \(\tan\theta =\) their \(v\)	M1
\(\theta = 26.6\)	A1
their \(\theta + 45^\circ\)	M1
Defln angle \(= 45 + 26.6 = 71.6^\circ\)	A1
	(5)

Scheme	Marks
KE Loss \(= \tfrac{1}{2} \times 0.6 \times \left\{(1^2 + 1^2) - (1^2 + v^2)\right\}\)	M1 A1
\(= 0.225\) J	A1
	(3)
	(14 marks)

M4 June 2009 Q5

EdexcelOld spec13 marksCollisions: 2 Spheres Oblique

5. Two small smooth spheres \(A\) and \(B\), of mass 2 kg and 1 kg respectively, are moving on a smooth horizontal plane when they collide. Immediately before the collision the velocity of \(A\) is \((\mathbf{i} + 2\mathbf{j})\) m s\(^{-1}\) and the velocity of \(B\) is \(-2\mathbf{i}\) m s\(^{-1}\). Immediately after the collision the velocity of \(A\) is \(\mathbf{j}\) m s\(^{-1}\).

(a) Show that the velocity of \(B\) immediately after the collision is \(2\mathbf{j}\) m s\(^{-1}\). (3)

(b) Find the impulse of \(B\) on \(A\) in the collision, giving your answer as a vector, and hence show that the line of centres is parallel to \(\mathbf{i} + \mathbf{j}\). (4)

Scheme	Marks
CLM: \(\ 2(\mathbf{i} + 2\mathbf{j}) + -2\mathbf{i} = 2\mathbf{j} + \mathbf{v}\)	M1 A1
\(\mathbf{v} = 2\mathbf{j}\) m s\(^{-1}\)	A1
	(3)

Mark scheme (b)

Scheme	Marks
\(\mathbf{I} = 2(\mathbf{j} - (\mathbf{i} + 2\mathbf{j}))\)	M1 A1
\(= (-2\mathbf{i} - 2\mathbf{j})\) Ns	A1
Since \(\mathbf{I}\) acts along l.o.c.c., l.o.c.c is parallel to \(\mathbf{i} + \mathbf{j}\)	B1
	(4)

Mark scheme (c)

Scheme	Marks
Before \(A\): \(\ (\mathbf{i} + 2\mathbf{j})\cdot\dfrac{1}{\sqrt{2}}(\mathbf{i} + \mathbf{j}) = \dfrac{3}{\sqrt{2}}\) \(B\): \(\ -2\mathbf{i}\cdot\dfrac{1}{\sqrt{2}}(\mathbf{i} + \mathbf{j}) = \dfrac{-2}{\sqrt{2}}\) After \(A\): \(\ \mathbf{j}\cdot\dfrac{1}{\sqrt{2}}(\mathbf{i} + \mathbf{j}) = \dfrac{1}{\sqrt{2}}\) \(B\): \(\ 2\mathbf{j}\cdot\dfrac{1}{\sqrt{2}}(\mathbf{i} + \mathbf{j}) = \dfrac{2}{\sqrt{2}}\)	M1 A3
NIL: \(\ e = \dfrac{\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{\frac{3}{\sqrt{2}} - \frac{-2}{\sqrt{2}}} = \dfrac{1}{5}\)	DM1 A1
	(6)
	(13 marks)

Notes

(Corrected from the printed mark scheme: the component of \(B\)’s velocity before the collision is printed as \(-2\mathbf{j}\cdot\dfrac{1}{\sqrt{2}}(\mathbf{i} + \mathbf{j})\); the velocity of \(B\) is \(-2\mathbf{i}\).)

M4 June 2008 Q2

EdexcelOld spec5 marksCollisions: 2 Spheres Oblique

2. Two small smooth spheres \(A\) and \(B\) have equal radii. The mass of \(A\) is \(2m\) kg and the mass of \(B\) is \(m\) kg. The spheres are moving on a smooth horizontal plane and they collide. Immediately before the collision the velocity of \(A\) is \((2\mathbf{i} - 2\mathbf{j})\) m s\(^{-1}\) and the velocity of \(B\) is \((-3\mathbf{i} - \mathbf{j})\) m s\(^{-1}\). Immediately after the collision the velocity of \(A\) is \((\mathbf{i} - 3\mathbf{j})\) m s\(^{-1}\).

Find the speed of \(B\) immediately after the collision. (5)

Mark scheme

Scheme	Marks
\(2m(2\mathbf{i} - 2\mathbf{j}) + m(-3\mathbf{i} - \mathbf{j}) = 2m(\mathbf{i} - 3\mathbf{j}) + m\mathbf{v}\)	M1 A1
\((\mathbf{i} - 5\mathbf{j}) = (2\mathbf{i} - 6\mathbf{j}) + \mathbf{v}\)
\((-\mathbf{i} + \mathbf{j}) = \mathbf{v}\)	A1
\(\|\mathbf{v}\| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}\) m s\(^{-1}\) cwo	DM1 A1
	(5 marks)

M4 June 2007 Q5

EdexcelOld spec13 marksCollisions: 2 Spheres Oblique Impulse & Momentum

5. A smooth uniform sphere \(A\) has mass \(2m\) kg and another smooth uniform sphere \(B\), with the same radius as \(A\), has mass \(m\) kg. The spheres are moving on a smooth horizontal plane when they collide. At the instant of collision the line joining the centres of the spheres is parallel to \(\mathbf{j}\). Immediately after the collision, the velocity of \(A\) is \((3\mathbf{i} - \mathbf{j})\) m s\(^{-1}\) and the velocity of \(B\) is \((2\mathbf{i} + \mathbf{j})\) m s\(^{-1}\). The coefficient of restitution between the spheres is \(\tfrac{1}{2}\).

(a) Find the velocities of the two spheres immediately before the collision. (7)

(b) Find the magnitude of the impulse in the collision. (2)

(c) Find, to the nearest degree, the angle through which the direction of motion of \(A\) is deflected by the collision. (4)

Before: B (mass m) has components 2 across and v1 down the line of centres, A (mass 2m) has 3 across and v2 up; after: B has 2 across and 1 up, A has 3 across and 1 down; impulses I on each — Mark scheme (a)

Scheme	Marks
CLM: \(\quad 2v_2 - v_1 = 1 - 2 = -1\)	M1A1
NIL: \(\quad 1 + 1 = \dfrac{1}{2}(v_1 + v_2)\)	M1A1
\(\therefore v_2 = 1,\ v_1 = 3\qquad\) Dependent on both M’s above	DM1
Horizontal components unchanged (i.e. 2 & 3)\(\qquad\) Independent of all other marks	A1
\(\mathbf{v}_A = 3\mathbf{i} + \mathbf{j};\ \mathbf{v}_B = 2\mathbf{i} - 3\mathbf{j}\)	A1
	(7)

Mark scheme (b)

Scheme	Marks
For B: I = m(1-(-3)) = 4m	M1A1
(Or For A: -I = 2m(-1 – 1) \(\ \therefore\) I = 4m)
	(2)

Notes

M1 Impulse = change in momentum for one sphere. Condone order of subtraction.

A1 Magnitude correct.

Triangle with sides 3 and 1 and angle theta; the required angle is 2 theta — Mark scheme (c)

Scheme	Marks
\(\begin{pmatrix}3\\1\end{pmatrix}.\begin{pmatrix}3\\-1\end{pmatrix} = \sqrt{3^2 + 1^2}.\sqrt{3^2 + (-1)^2}\cos\theta\)	M1A1
\(\Rightarrow 8 = 10\cos\theta\)	M1
\(\theta = 37^o\)	A1
	(4)
	(13 marks)

M4 June 2006 Q6

EdexcelOld spec14 marksCollisions: 2 Spheres Oblique

Figure 2: spheres A (mass m) and B (mass 2m) in contact with centres on line L, each moving with speed U at 45 degrees to L towards the other — Figure 2

Two small smooth spheres \(A\) and \(B\), of equal size and of mass \(m\) and \(2m\) respectively, are moving initially with the same speed \(U\) on a smooth horizontal floor. The spheres collide when their centres are on a line \(L\). Before the collision the spheres are moving towards each other, with their directions of motion perpendicular to each other and each inclined at an angle of 45\(^\circ\) to the line \(L\), as shown in Figure 2. The coefficient of restitution between the spheres is \(\tfrac{1}{2}\).

(a) Find the magnitude of the impulse which acts on \(A\) in the collision. (9)

Figure 3: line L parallel to a vertical wall at distance d — Figure 3

The line \(L\) is parallel to and a distance \(d\) from a smooth vertical wall, as shown in Figure 3.

(b) Find, in terms of \(d\), the distance between the points at which the spheres first strike the wall. (5)

Before and after components: A (m) has U/root 2 along and across the line of centres before, v2 along and U/root 2 across after, impulse I; B (2m) has U/root 2 along and across before, v1 along and U/root 2 across after — Mark scheme (a)

Mark scheme (b)

Scheme	Marks
\(v_2 - v_1 = \dfrac{U}{\sqrt{2}}\qquad\) (Separation speed)	M1
time to wall \(= \dfrac{d}{U/\sqrt{2}} = \dfrac{d\sqrt{2}}{U}\)	M1 A1
\(\therefore\) Separation \(= \dfrac{d\sqrt{2}}{U} \times \dfrac{U}{\sqrt{2}} = d\)	M1 A1
	(5)
	(14 marks)

M4 January 2006 Q5

EdexcelOld spec16 marksCollisions: 2 Spheres Oblique

5. Two smooth uniform spheres \(A\) and \(B\) have equal radii. Sphere \(A\) has mass \(m\) and sphere \(B\) has mass \(km\). The spheres are at rest on a smooth horizontal table. Sphere \(A\) is then projected along the table with speed \(u\) and collides with \(B\). Immediately before the collision, the direction of motion of \(A\) makes an angle of 60\(^\circ\) with the line joining the centres of the two spheres. The coefficient of restitution between the spheres is \(\tfrac{1}{2}\).

(a) Show that the speed of \(B\) immediately after the collision is \(\dfrac{3u}{4(k + 1)}\). (6)

Immediately after the collision the direction of motion of \(A\) makes an angle \(\arctan\left(2\sqrt{3}\right)\) with the direction of motion of \(B\).

(b) Show that \(k = \tfrac{1}{2}\). (6)

Before: A (mass m) has components u cos 60 along and u sin 60 perpendicular to the line of centres, B (mass km) at rest; after: A has v along and u sin 60 perpendicular, B has w along the line of centres — Mark scheme (a)

Velocity of A after impact: v along the line of centres and u sin 60 perpendicular, at angle theta to the line of centres — Mark scheme (b)

Mark scheme (c)

Scheme	Marks
\(k = \dfrac{1}{2} \Rightarrow v = \dfrac{u}{4},\ w = \dfrac{u}{2}\)	B1
KE loss \(= \dfrac{1}{2}mu^2 - \left(\dfrac{1}{2}m.\dfrac{u^2}{16} + \dfrac{1}{2}m.\dfrac{3u^2}{4} + \dfrac{1}{2}.\dfrac{1}{2}m.\dfrac{u^2}{4}\right)\)	M1 A1
\(= \dfrac{1}{2}mu^2\left(1 - \dfrac{1}{16} - \dfrac{3}{4} - \dfrac{1}{8}\right)\)
\(= \dfrac{1}{32}mu^2\)	A1
	(4)
	(16 marks)

M4 June 2005 Q3

EdexcelOld spec11 marksCollisions: 2 Spheres Oblique

Figure 1: sphere Q touching sphere P along the line of centres; Q approaches at angle alpha to the line of centres and leaves at angle beta to it — Figure 1

A smooth sphere \(P\) lies at rest on a smooth horizontal plane. A second identical sphere \(Q\), moving on the plane, collides with the sphere \(P\). Immediately before the collision the direction of motion of \(Q\) makes an angle \(\alpha\) with the line joining the centres of the spheres. Immediately after the collision the direction of motion of \(Q\) makes an angle \(\beta\) with the line joining the centres of spheres, as shown in Figure 1. The coefficient of restitution between the spheres is \(e\).

Show that \((1 - e)\tan\beta = 2\tan\alpha\). (11)

Spheres Q and P, each of mass m: before, Q has components u cos alpha along and u sin alpha perpendicular to the line of centres, P at rest; after, Q has v1 along and v3 perpendicular, at angle beta, P has v2 along and 0 perpendicular — Mark scheme

M4 January 2005 Q1

EdexcelOld spec7 marksCollisions: 2 Spheres Oblique

1. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal perpendicular unit vectors.]

Two smooth uniform spheres \(A\) and \(B\) have equal radius but masses \(m\) and \(5m\) respectively. The spheres are moving on a smooth horizontal plane when they collide. Immediately before the collision, the velocities of \(A\) and \(B\) are \((\mathbf{i} + 2\mathbf{j})\) m s\(^{-1}\) and \((-\mathbf{i} + 3\mathbf{j})\) m s\(^{-1}\) respectively. Immediately after the collision, the velocity of \(A\) is \((-2\mathbf{i} + 5\mathbf{j})\) m s\(^{-1}\).

(a) By considering the impulse on \(A\), find a unit vector parallel to the line joining the centres of the spheres when they collide. (4)

(b) Find the velocity of \(B\) immediately after the collision. (3)

Mark scheme (a)

Scheme
Impulse on \(A\) is in the direction of the line of centres.
Impulse on \(A = \Delta(mv) = m(-2\mathbf{i} + 5\mathbf{j}) - m(\mathbf{i} + 2\mathbf{j}) = m(-3\mathbf{i} + 3\mathbf{j})\).
Therefore direction of line of centres is \((-\mathbf{i} + \mathbf{j})\). A unit vector in this direction is \(\dfrac{(-\mathbf{i} + \mathbf{j})}{\sqrt{2}}\).

Notes

The published mark scheme for this paper is a set of worked answers: no mark allocation is printed.

Mark scheme (b)

Scheme
Let velocity of \(b\) after collision be \(v_1\mathbf{i} + v_2\mathbf{j}\)
Momentum conserved: \(m(\mathbf{i} + 2\mathbf{j}) + 5m(-\mathbf{i} + 3\mathbf{j}) = m(-2\mathbf{i} + 5\mathbf{j}) + 5m(v_1\mathbf{i} + v_2\mathbf{j})\)
Cancel \(m\) and equate coefficients: \(\quad\mathbf{i}\colon\ -4 = -2 + 5v_1 \qquad v_1 = -\tfrac{2}{5}\)
\(\mathbf{j}\colon\ 17 = 5 + 5v_2 \qquad v_2 = \tfrac{12}{5}\)
Velocity of \(B\) after collision \(= -\tfrac{2}{5}\mathbf{i} + \tfrac{12}{5}\mathbf{j}\).

For those who prefer everything with trig:
\(v_A\sin\alpha = 3u\sin\theta,\quad v_B\sin\beta = u\sin\theta\)	B1
\(m.3u\cos\theta - 2m.u\cos\theta = mv_A\cos\alpha + 2mv_B\cos\beta\)	M1
\(\left(u\cos\theta = v_A\cos\alpha + 2v_B\cos\beta\right)\)	A1
\(\dfrac{1}{8} \times (3u\cos\theta + u\cos\theta) = v_B\cos\beta - v_A\cos\alpha\)	M1
\(\left(\dfrac{u}{2}\cos\theta = v_B\cos\beta - v_A\cos\alpha\right)\)	A 1
\(\dfrac{u}{2}\cos\theta = v_B\cos\beta,\quad 0 = v_A\cos\alpha\ (\Rightarrow \sin\alpha = 1)\)	DM1 A1
\(v_A\sin\alpha = v_A = 2v_B = 3u\sin\theta\)	M1
\(v_B\sin\beta = u\sin\theta \Rightarrow \dfrac{3u\sin\theta}{2}\sin\beta = u\sin\theta\)	A1
\(\sin\beta = \dfrac{2}{3}\)	A1
\(2v_B = 3u\sin\theta\ \ \&\ \ \dfrac{u}{2}\cos\theta = v_B\cos\beta\) \(\Rightarrow 6\tan\theta = \dfrac{2}{\cos\beta}\left(= 2 \times \dfrac{3}{\sqrt{5}}\right)\)	M1
\(\tan\theta = \dfrac{1}{\sqrt{5}},\quad \theta = 24.1(^\circ)\ \ (0.421\text{ radians})\)	A1

Collisions: 2 Spheres Oblique

Edexcel

Edexcel · Old spec

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5(a) alt1

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Alt using CLM

Alternative (non-vector form)

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3 alt

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Special case

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