Define an oblique collision with a surface.

An oblique collision is one in which the object meets the surface at an angle other than a right angle. The motion then has two dimensions to consider, because the velocity has a component along the surface as well as one into it.

Which part of a particle's velocity is changed by an oblique collision with a surface?

Only the component perpendicular to the surface , because the impulse from a smooth surface acts perpendicular to it. The component parallel to the surface passes through the collision completely unchanged.

A sphere bounces off one wall and then off a second. What connects the two collisions?

The velocity of rebound from the first wall is the velocity of approach to the second, since everything is smooth and nothing acts in between. Each impact is then treated as a separate oblique collision, with its own coefficient of restitution.

Define the line of centres of two colliding spheres.

The line of centres is the line through the centres of both spheres at the moment they touch. It is perpendicular to their common tangent , which is the line in which their surfaces meet.

In which direction does the impulse act when two smooth spheres collide?

Along the line of centres . Smooth spheres touch at a single point and so cannot push each other along their common tangent, which means the tangential component of each velocity passes through the collision unchanged.

Why do you need conservation of momentum for two colliding spheres but not for a sphere hitting a fixed wall?

Because two spheres form a closed system , so their total momentum is unchanged by the impact. A fixed wall is external and supplies an impulse from outside, so the sphere's momentum is not conserved and restitution alone has to carry the perpendicular direction.

You have found both components of a sphere's velocity after an oblique collision. How do you get its speed and direction?

Combine them with Pythagoras for the speed, and with right-angled trigonometry for the angle to the line of centres. The unchanged tangential component and the new line-of-centres component are perpendicular, so together they form a right-angled triangle.

True or False? The equations for an oblique collision only work if the line of centres is horizontal.

False. The relationships hold whatever direction the line of centres points, because they are statements about components along it and perpendicular to it. It is usually drawn horizontally simply because that makes the components easy to read off.

Define the angle of deflection of a sphere in a collision.

The angle of deflection is the angle between the path the sphere was on before the collision and the path it takes afterwards. It measures how far the direction of travel has been turned, which is not the same as the angle the sphere makes with the surface.

True or False? One of two colliding spheres can gain kinetic energy even though the pair together lose it.

True. The impulse transfers kinetic energy from one sphere to the other as well as removing some from the system. A sphere that was at rest therefore ends up with kinetic energy, which is why a question has to say whether it wants the loss for one sphere or for both.

Why does the expression for the speed of separation depend on how you drew the arrows?

Because a speed of separation is a difference of velocities, and the sign of each velocity follows the direction its arrow points. Arrows drawn away from each other give a sum, while arrows drawn both the same way give a difference.

A LevelFurther MathsEdexcelFurther Mechanics 1FlashcardsElastic CollisionsElastic Collisions in 2D

Elastic Collisions in 2D (Edexcel A Level Further Maths: Further Mechanics 1): Flashcards

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FrontOblique Collisions with a Surface
Define an oblique collision with a surface.

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Cards in this collection (22)

Define an oblique collision with a surface.
An oblique collision is one in which the object meets the surface at an angle other than a right angle.
The motion then has two dimensions to consider, because the velocity has a component along the surface as well as one into it.
Which part of a particle's velocity is changed by an oblique collision with a surface?
Only the component perpendicular to the surface, because the impulse from a smooth surface acts perpendicular to it.
The component parallel to the surface passes through the collision completely unchanged.
A particle approaches a surface at angle $α$ with speed $u$ and leaves at angle $β$ with speed $v$ . Fill in the right-hand side of each component equation:
$v \cos β =______$
$v \sin β =______$
The completed equations are:
$v \cos β = u \cos α$
$v \sin β = e u \sin α$
The components along the surface match exactly, while those into the surface are related by the coefficient of restitution.
What single equation links the angles of approach and rebound?
Dividing the perpendicular equation by the parallel one eliminates both speeds and leaves $\tan β = e \tan α$ .
Since $e \leq 1$ this forces $β \leq α$ , so a particle always rebounds at an angle no larger than the one it arrived at.
True or False?
A smooth surface changes the direction of a particle but not its speed.
False.
The perpendicular component is multiplied by $e$ , so unless $e = 1$ the speed after the collision is smaller than the speed before it.
Only the parallel component survives untouched, and that is not the whole of the velocity.
A collision is given in vector form with the surface not parallel to an axis. Which two scalar product equations do you use?
Perpendicular to the surface, $v \cdot P = - e u \cdot P$ , where $P$ points perpendicular to it.
Parallel to the surface, $v \cdot W = u \cdot W$ , where $W$ points along it.
A surface lies in the direction $x i + y j$ . What is a vector perpendicular to it?
Swap the two components and change one sign, giving $y i - x j$ .
So is $- y i + x j$ , which points the opposite way along the same line, and a diagram tells you which of the two you want.
A question gives the velocities before and after a collision but not the direction of the wall. How do you find it?
Use $I = m (v - u)$ to get the impulse, which always acts perpendicular to the wall.
The wall then lies in the direction perpendicular to that impulse.
A sphere bounces off one wall and then off a second. What connects the two collisions?
The velocity of rebound from the first wall is the velocity of approach to the second, since everything is smooth and nothing acts in between.
Each impact is then treated as a separate oblique collision, with its own coefficient of restitution.
Define the line of centres of two colliding spheres.
The line of centres is the line through the centres of both spheres at the moment they touch.
It is perpendicular to their common tangent, which is the line in which their surfaces meet.
In which direction does the impulse act when two smooth spheres collide?
Along the line of centres.
Smooth spheres touch at a single point and so cannot push each other along their common tangent, which means the tangential component of each velocity passes through the collision unchanged.
Why do you need conservation of momentum for two colliding spheres but not for a sphere hitting a fixed wall?
Because two spheres form a closed system, so their total momentum is unchanged by the impact.
A fixed wall is external and supplies an impulse from outside, so the sphere's momentum is not conserved and restitution alone has to carry the perpendicular direction.
Two spheres approach along the line of centres with components $u_{A} \cos α$ and $u_{B} \cos β$ , and separate with components $x_{A}$ and $x_{B}$ . Fill in the numerator:
$e = \frac{______}{u_{A} \cos α + u_{B} \cos β}$
The completed equation is:
$e = \frac{x_{A} + x_{B}}{u_{A} \cos α + u_{B} \cos β}$
Only the line-of-centres components appear, and both pairs are added here because the spheres approach from opposite directions and separate in opposite directions.
You have found both components of a sphere's velocity after an oblique collision. How do you get its speed and direction?
Combine them with Pythagoras for the speed, and with right-angled trigonometry for the angle to the line of centres.
The unchanged tangential component and the new line-of-centres component are perpendicular, so together they form a right-angled triangle.
True or False?
The equations for an oblique collision only work if the line of centres is horizontal.
False.
The relationships hold whatever direction the line of centres points, because they are statements about components along it and perpendicular to it.
It is usually drawn horizontally simply because that makes the components easy to read off.
Define the angle of deflection of a sphere in a collision.
The angle of deflection is the angle between the path the sphere was on before the collision and the path it takes afterwards.
It measures how far the direction of travel has been turned, which is not the same as the angle the sphere makes with the surface.
A sphere hits a wall at angle $α$ to it and leaves at angle $β$ . What is the angle of deflection?
It is $α + β$ .
Continuing the original path as a dashed line and using vertically opposite angles shows that the turn is made up of those two angles together.
How can you find an angle of deflection directly from the two velocity vectors?
Take the scalar product of the velocity before and the velocity after, then use $θ = \arccos (\frac{a \cdot b}{| a | | b |})$ .
The angle between those two vectors is precisely the angle through which the path has turned.
True or False?
The angle of deflection between two spheres is always $180 ° - α - β$ .
False.
That expression comes from one particular arrangement of the marked angles, in which each is measured from the common tangent.
Mark the angles differently and the deflection is a different combination of them, so the geometry has to be worked out afresh for each diagram.
A sphere has velocity $(3 i - 4 j) {m s}^{- 1}$ . How do you find its kinetic energy?
Square the two components and add them to get the speed squared, which is $25$ here, then put that straight into the kinetic energy formula.
There is no need to take the square root first, since the formula would only square the speed again.
True or False?
One of two colliding spheres can gain kinetic energy even though the pair together lose it.
True.
The impulse transfers kinetic energy from one sphere to the other as well as removing some from the system.
A sphere that was at rest therefore ends up with kinetic energy, which is why a question has to say whether it wants the loss for one sphere or for both.
Why does the expression for the speed of separation depend on how you drew the arrows?
Because a speed of separation is a difference of velocities, and the sign of each velocity follows the direction its arrow points.
Arrows drawn away from each other give a sum, while arrows drawn both the same way give a difference.

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