Problem Solving with Oblique Collisions (Edexcel A Level Further Maths: Further Mechanics 1)

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Jamie W

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Jamie W

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Energy in 2D Collisions

How do I find the kinetic energy loss from a collision?

  • A common question, usually as a follow up to a collisions question, is to find the kinetic energy lost as a result of the impact
  • If e equals 1 kinetic energy is conserved in the collision
    • If e less than 1 there will be a decrease in the total kinetic energy of the two particles
  • Recall that kinetic energy can be calculated using K. E. space equals 1 half m v squared
  • As the velocity is squared, its sign does not affect the kinetic energy
    • Hence, v squared here is effectively the speed squared
  • When dealing with motion in two dimensions, the velocity may be described in two components, e.g. bold italic v equals stretchy left parenthesis 3 bold i space minus space 4 bold j stretchy right parenthesis space ms to the power of negative 1 end exponent
    • To use this with  K. E. space equals 1 half m v squared, the magnitude of the vector must be found, using Pythagoras
      • In this case open vertical bar v close vertical bar equals square root of 3 squared plus 4 squared end root equals 5 space ms to the power of negative 1 end exponent
      • If m equals 7 space kg then K. E. space equals 1 half cross times 7 cross times 5 squared space equals space 87.5 space straight J 
  • To find the loss in kinetic energy due to a collision, find the difference between the kinetic energy before the collision, and the kinetic energy after the collision
    • The question may ask to find the loss in kinetic energy for one particular particle,
      • or it could ask to find the total loss in kinetic energy for both particles

Exam Tip

  • When finding the speed from two components, your working will look like this: v equals square root of x squared plus y squared end root
    • When finding kinetic energy, the speed is squared, so your working may look like this: 1 half m open parentheses square root of x squared plus y squared end root close parentheses squared
    • It can be quicker to not do the square rooting part if you know you are only using it to find v squared
    • e.g. 1 half m open parentheses x squared plus y squared close parentheses
  • You can also use the scalar product to find the kinetic energy:
    • 1 half m open parentheses bold italic v bold times bold italic v close parentheses where bold italic v is the vector form of the velocity

Worked example

A smooth sphere A of mass 4 kg is moving on a smooth horizontal surface with velocity open parentheses 3 bold i plus 2 bold j close parentheses space ms to the power of negative 1 end exponent. Another smooth sphere B of mass 3 kg and the same radius as A is moving on the same surface with velocity open parentheses negative 5 bold i plus 4 bold j close parentheses space ms to the power of negative 1 end exponent. The spheres collide when their line of centres is parallel to bold i. The coefficient of restitution between the spheres is 3 over 4.
Find the kinetic energy lost in the impact in total.

worked example showing how to find the loss in kinetic energy after 2 spheres collide, with velocities in 2 dimensions

Angles of Deflection

How do I find the angle of deflection after collision with a surface?

  • Once the speed and direction of a sphere after a collision have been calculated, a common follow up question is to find the angle of deflection
  • The angle of deflection is the angle by which the path of the object has changed from its original trajectory
  • To find the angle of deflection it is helpful to sketch a new diagram, with only the angles marked on it
    • The diagram below shows a particle which collides with the surface at angle alpha degree and leaves the surface at angle beta degree
  • Draw a dashed line showing the continuing path of the object, if it had not collided with the surface
  • Use vertically opposite angles to mark the other angle which is equal to alpha degree
  • It can now be seen that the angle of deflection, the angle by which the path of the object has turned,
    • is equal to open parentheses alpha plus beta close parentheses degree

diagram of a rebound with a wall, showing the angle of deflection

How do I find the angle of deflection after a collision between two spheres?

  • Exactly the same concept described for finding the angle of deflection after colliding with a surface, applies when two spheres collide
    • The "surface" when two spheres collide, is the common tangent line between the two spheres
  • The angle of deflection is still the angle by which the path of the object has changed from its original trajectory
  • The main difference with spheres, is that the angles may be marked in varying ways on the diagram depending on the problem
  • To deal with this draw a diagram clearly showing:
    • the velocity of the sphere before and after the collision,
    • a dashed line showing the continuing path of the object, if it had not collided with the other sphere,
    • any angles that were given or have been calculated,
    • and the common tangent line of the spheres
  • Remembering that the angle of deflection is the angle between the continued original path of the sphere, and its new path, find any missing angles on the diagram using:
    • vertically opposite angles are equal,
    • angles on a straight line sum to 180°,
    • and angles in a right angle sum to 90°

diagram of an oblique collision of two spheres, showing the angle of deflection

  • In the above diagram, a sphere collides at an angle alpha degree above the horizontal, and after the collision has a direction which forms angle beta degree below the horizontal
    • By drawing the common tangent line, the angle 90 minus alpha can be filled in, adjacent to alpha
    • Similarly, 90 minus beta can be filled in, adjacent to beta
    • By using vertically opposite angles, another 90 minus alpha angle can be marked on the diagram
    • This then shows (in green) that the angle of deflection, the angle by which the path of the object has turned, is equal to open parentheses 90 minus alpha close parentheses plus open parentheses 90 minus beta close parentheses
    • This simplifies to 180 minus alpha minus beta
  • Note that this answer does not always apply, as it depends which angles are given and marked on the diagram

How do I use the scalar (dot) product to find the angle of deflection?

  • The scalar product (or dot product) can also be used to find the angle of deflection
  • We can use the property: bold italic a bold times bold italic b space equals space open vertical bar bold italic a close vertical bar space open vertical bar bold italic b close vertical bar space cos theta
  • If bold italic a is the vector describing the velocity of the sphere before the collision,
    • and bold italic b is the vector describing the velocity of the sphere after the collision,
    • then theta will be the angle of deflection
  • This can then be rearranged to theta equals cos to the power of negative 1 end exponent open parentheses fraction numerator bold italic a bold times bold italic b over denominator open vertical bar bold italic a close vertical bar space open vertical bar bold italic b close vertical bar end fraction close parentheses

Exam Tip

  • Drawing a separate diagram focusing on only the angles can be helpful, without the algebra and masses etc that you may have used earlier in the question
    • You should include the velocities both before and after on the same diagram to do this

Worked example

Two small smooth spheres A and B collide when the line joining their centres is parallel to bold i. Before the collision the velocity of A is open parentheses 3 bold i plus 2 bold j close parentheses space ms to the power of negative 1 end exponent and the velocity of B is open parentheses negative 4 bold i plus 5 bold j close parentheses space ms to the power of negative 1 end exponent. After the collision the velocity of A is open parentheses negative 5.4 bold i plus 2 bold j close parentheses space ms to the power of negative 1 end exponent and the velocity of B is open parentheses 1.6 bold i plus 5 bold j close parentheses space ms to the power of negative 1 end exponent.

Find the angle of deflection of sphere A.

worked example showing how to find the angle of deflection using geometry or the scalar product

Problem Solving with Oblique Collisions

Tips for problem solving with oblique collisions

  • Drawing a large, clear diagram will always help and prevent working from becoming squashed together
    • Establish a routine of how you lay out your diagrams and working, so you can do it routinely and quickly in an exam
  • Adding an arrow, or pair of arrows, to show the direction of the impulse will help remind you which components will be affected
  • Fill in the things you can work out straight away
    • For example the components of the velocity perpendicular to the impulse are unaffected, so you can usually fill this in immediately
  • Use sensible labels for velocities and components of velocities
    • e.g. x subscript A for the unknown horizontal component of the velocity of sphere A, y subscript A for an unknown vertical component, and v subscript A for the unknown final speed open parentheses square root of x subscript A squared plus y subscript A squared end root close parentheses of sphere A.
  • If you are not sure how to find a piece of missing information, and have filled in everything you can work out by inspection, follow the usual processes to form equations which may help you
    • Apply conservation of linear momentum in the direction parallel to the impulse
      • m subscript 1 u subscript 1 space end subscript plus space m subscript 2 u subscript 2 space equals space m subscript 1 v subscript 1 space plus space m subscript 2 v subscript 2
    • Apply Newton's law of restitution in the direction of the impulse
      • e space equals space fraction numerator Speed space of space separation space of space the space objects over denominator Speed space of space approach space of space the space objects end fraction
    • This will often lead to a pair of simultaneous equations which can be solved
    • Be careful with positive and negative signs when forming these equations, the signs of the velocities will depend on how they are modelled in the diagram
      • e.g. If the arrows are drawn pointing away from each other in the "after" diagram, the speed of separation will be x subscript A plus x subscript B
      • If the arrows were drawn both pointing to the right, the speed of separation would be x subscript B minus x subscript A
  • The equation I italic equals m stretchy left parenthesis v minus u stretchy right parenthesis, where I is the impulse, can also sometimes be useful with 2D collisions
    • It can be used with vectors for bold Ibold v, and bold u:
      • bold I equals m open parentheses bold v minus bold u close parentheses
    • This is particularly helpful when a question does not describe the direction of a wall, as it can be used to find the direction of the impulse, which is always perpendicular to the wall
  • Make use of the scalar (dot) product formulae for collisions with a surface
    • bold v bold times bold P equals negative e space bold u bold times bold P where bold P is the direction perpendicular to the surface
    • bold v bold times bold W equals bold u bold times bold W where bold W is the direction of the surface (or wall)
      • Remember that the direction of the wall and the direction of the impulse will be perpendicular, e.g. open parentheses 3 bold i plus 1 bold j close parentheses and open parentheses negative 1 bold i plus 3 bold j close parentheses  
    • These formulae are most useful when a surface is not in a "nice" direction
      • e.g. when it is not parallel to the x or y axis when the sphere is moving in the xy-plane
  • You may need to consider kinetic energy before and after the collision for one or both of the objects using,
    • K. E. space equals 1 half m v squared

Exam Tip

  • Sometimes questions will be entirely algebraic,
    • in this case apply the same procedures and methods as above,
    • and form equations using conservation of linear momentum and Newton's law of restitution
    • Label any angles on the diagram clearly and use simple angle geometry to help you
      • e.g. Vertically opposite angles are equal, angles in a right angle sum to 90°, and angles on a straight line sum to 180°
  • Sometimes you may need to use trigonometric identities from your pure maths knowledge to help simplify or rearrange to reach a required expression

Worked example

A smooth uniform sphere A collides with an identical sphere B which is at rest. When the spheres collide A is moving such that it forms an angle of 30° to the line joining the centres of the spheres, as shown in the diagram. The coefficient of restitution between the two spheres is e. Sphere A is deflected by angle alpha as a result of the collision, where 0 degree less than alpha space less than 60 degree.

Show that tan space alpha equals fraction numerator square root of 3 open parentheses 1 plus e close parentheses over denominator 5 minus 3 e end fraction.

worked example question image showing incident angle of a sphere in an oblique collision of two spheres

worked example for an algebraic problem solving question for oblique collisions

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Jamie W

Author: Jamie W

Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.