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Oblique impactEdexcel A-Level Further Maths: Revision notes

Section 1

What makes an impact oblique

In an oblique impact the velocity of a sphere is not along the line of centres (sphere-sphere) or not perpendicular to the surface (sphere-wall). The spheres have equal radii, are smooth, and are modelled as particles. Because the surfaces are smooth, the impulse acts only along the line of centres (or along the normal to the wall). So resolve every velocity into a component along this line and a component perpendicular to it.

Key termsoblique impactline of centres
Exam tip

Draw the line of centres or the normal first, then resolve each velocity along and perpendicular to it.

Section 2

Perpendicular components are unchanged

There is no impulse perpendicular to the line of centres (or parallel to a smooth wall), so each sphere's component in that direction is the same before and after the impact. For a particle with speed 1010 m s−1^{-1} making 60∘60^\circ with a wall the parallel component is 10cos⁡60∘=510\cos60^\circ=5 m s−1^{-1} both before and after the impact.

Key termssmooth
Common mistake

Applying the coefficient of restitution to the whole speed or to the parallel component.

Section 3

Components along the line of centres

Along the line of centres apply the same two laws as in a direct impact: m1u1+m2u2=m1v1+m2v2,v2−v1=e(u1−u2).m_1u_1+m_2u_2=m_1v_1+m_2v_2,\qquad v_2-v_1=e(u_1-u_2). For a fixed smooth surface the normal component reverses and is multiplied by ee, and the component along the surface does not change. Example: a ball hits a wall at 1313 m s−1^{-1} with components 55 (parallel) and 1212 (perpendicular) and e=0.5e=0.5. After impact the components are 55 and 66, so the speed is 61=7.81\sqrt{61}=7.81 m s−1^{-1} at tan⁡−165=50.2∘\tan^{-1}\frac65=50.2^\circ to the wall.

Key termscoefficient of restitution

Section 4

Speed and direction after impact

Recombine the two components: speed =(parallel)2+(normal)2=\sqrt{(\text{parallel})^2+(\text{normal})^2} and the angle θ\theta with the wall or line of centres from tan⁡θ=normalparallel\tan\theta=\frac{\text{normal}}{\text{parallel}}. Because e<1e<1 reduces the normal component, a particle leaves the wall at a smaller angle to the wall than the angle at which it arrived: tan⁡θafter=etan⁡θbefore\tan\theta_{\text{after}}=e\tan\theta_{\text{before}}. If e=1e=1 the angles are equal.

Key termsangle of rebound

Section 5

Vector form and energy loss

In vector problems choose the line of centres to be along i\mathbf{i} (as stated in the question). Apply momentum and restitution to the i\mathbf{i}-components only; the j\mathbf{j}-components stay the same. Impulse is change of momentum: the impulse on a sphere is along the line of centres, so its j\mathbf{j}-component is zero. Kinetic energy lost == KE before −- KE after, using the full speeds. Only the components along the line of centres contribute to the loss, so it is zero when e=1e=1.

Key termsimpulse
Common mistake

Forgetting that the impulse has no component perpendicular to the line of centres.

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Exam questions on Oblique impact

  1. A smooth particle of mass 0.40.4 kg moves on a smooth horizontal surface and strikes a fixed smooth vertical wall with speed 1010 m s−1^{-1}. Its direction of motion makes an angle of 60∘60^\circ with the wall. The coefficient of restitution between the particle and the wall is 0.50.5.
    Find the kinetic energy lost by the particle in the impact.2 marks
  2. A smooth ball of mass 0.20.2 kg hits a smooth horizontal floor. The unit vectors i\mathbf{i} and j\mathbf{j} are horizontal and vertically upwards respectively. Immediately before the impact the velocity of the ball is (6i−8j)(6\mathbf{i}-8\mathbf{j}) m s−1^{-1}. The coefficient of restitution between the ball and the floor is 0.750.75.
    Find the speed of the ball immediately after impact, and the angle between its direction of motion and the floor.2 marks
  3. Two smooth spheres AA and BB of equal radii and equal mass mm lie on a smooth horizontal surface, with BB at rest. AA moves with speed 1010 m s−1^{-1} and strikes BB. At the instant of impact the direction of motion of AA makes an angle α\alpha with the line of centres, where cos⁡α=0.8\cos\alpha=0.8. The coefficient of restitution between the spheres is 0.50.5.
    Find the speed of BB immediately after the impact.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).