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

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 $e=1$ kinetic energy is conserved in the collision

  • If $e<1$ there will be a decrease in the total kinetic energy of the two particles

• Recall that kinetic energy can be calculated using $K.E.=\frac{1}{2}m{v}^2$

• As the velocity is squared, its sign does not affect the kinetic energy

  • Hence, $v^2$ here is effectively the speed squared

• When dealing with motion in two dimensions, the velocity may be described in two components, e.g. $\mathit{v}=(3\mathbf{i}-4\mathbf{j}){\mathrm{ms}}^{-1}$

  • To use this with  $K.E.=\frac{1}{2}m{v}^2$, the magnitude of the vector must be found, using Pythagoras

    • In this case $\left|v\right|=\sqrt{3^2+4^2}=5{\mathrm{ms}}^{-1}$

    • If $m=7\mathrm{kg}$ then $K.E.=\frac{1}{2}\times 7\times 5^2=87.5\mathrm{J}$ 

• 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

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 $\alpha °$ and leaves the surface at angle $\beta °$

• 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 $\alpha °$

• It can now be seen that the angle of deflection, the angle by which the path of the object has turned,

  • is equal to $\left(\alpha +\beta \right)°$

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°

• In the above diagram, a sphere collides at an angle $\alpha °$ above the horizontal, and after the collision has a direction which forms angle $\beta °$ below the horizontal

  • By drawing the common tangent line, the angle $90-\alpha$ can be filled in, adjacent to $\alpha$

  • Similarly, $90-\beta$ can be filled in, adjacent to $\beta$

  • By using vertically opposite angles, another $90-\alpha$ 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 $\left(90-\alpha \right)+\left(90-\beta \right)$

  • This simplifies to $180-\alpha -\beta$

• 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: $\mathit{a}\mathbf{·}\mathit{b}=\left|\mathit{a}\right|\left|\mathit{b}\right|\cos \theta$

• If $\mathit{a}$ is the vector describing the velocity of the sphere before the collision,

  • and $\mathit{b}$ 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 $\theta ={\cos }^{-1}\left(\frac{\mathit{a}\mathbf{·}\mathit{b}}{\left|\mathit{a}\right|\left|\mathit{b}\right|}\right)$

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. $x_A$ for the unknown horizontal component of the velocity of sphere A, $y_A$ for an unknown vertical component, and $v_A$ for the unknown final speed $\left(\sqrt{{x_A}^2+{y_A}^2}\right)$ 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

    • $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

  • Apply Newton's law of restitution in the direction of the impulse

    • $e=\frac{\mathrm{Speed}\mathrm{of}\mathrm{separation}\mathrm{of}\mathrm{the}\mathrm{objects}}{\mathrm{Speed}\mathrm{of}\mathrm{approach}\mathrm{of}\mathrm{the}\mathrm{objects}}$

  • 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 $x_A+x_B$

    • If the arrows were drawn both pointing to the right, the speed of separation would be $x_B-x_A$

• The equation $I\mathit{=}m\left(v-u\right)$, where $I$ is the impulse, can also sometimes be useful with 2D collisions

  • It can be used with vectors for $\mathbf{I}$, $\mathbf{v}$, and $\mathbf{u}$:

    • $\mathbf{I}=m\left(\mathbf{v}-\mathbf{u}\right)$

  • 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

  • $\mathbf{v}\mathbf{·}\mathbf{P}=-e\mathbf{u}\mathbf{·}\mathbf{P}$ where $\mathbf{P}$ is the direction perpendicular to the surface

  • $\mathbf{v}\mathbf{·}\mathbf{W}=\mathbf{u}\mathbf{·}\mathbf{W}$ where $\mathbf{W}$ 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. $\left(3\mathbf{i}+1\mathbf{j}\right)$ and $\left(-1\mathbf{i}+3\mathbf{j}\right)$  

  • 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,

  • $K.E.=\frac{1}{2}m{v}^2$