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M2 Revision

This document discusses the coefficient of restitution and how to calculate the speed of objects after collisions with walls and other objects. It contains the following key points: 1) The coefficient of restitution e measures how bouncy objects are during collisions, ranging from 0 (no bounce) to 1 (perfectly elastic). 2) For a ball hitting a wall at 90 degrees, its speed after the collision eu away from the wall. 3) For a ball hitting a wall at an angle, its velocity must be split into components parallel and perpendicular to the wall, with only the perpendicular component's speed changing during the collision. 4) To calculate speeds after collisions between two objects, set up two

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Jack Fielding
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70 views3 pages

M2 Revision

This document discusses the coefficient of restitution and how to calculate the speed of objects after collisions with walls and other objects. It contains the following key points: 1) The coefficient of restitution e measures how bouncy objects are during collisions, ranging from 0 (no bounce) to 1 (perfectly elastic). 2) For a ball hitting a wall at 90 degrees, its speed after the collision eu away from the wall. 3) For a ball hitting a wall at an angle, its velocity must be split into components parallel and perpendicular to the wall, with only the perpendicular component's speed changing during the collision. 4) To calculate speeds after collisions between two objects, set up two

Uploaded by

Jack Fielding
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Impulse for M2

The coefficient of restitution e is a measure of how bouncy an object is


when it collides with some other specific object. (So a ball say doesnt have
a coefficent of restitution, but the system of a ball colliding with a wall does
have a coefficient of restitution.) It ranges from 0 e 1. If e = 1 collision
is perfectly elastic, ie no loss of kinetic energy. If e = 0 the object does not
bounce at all, if it hits a wall all kinetic energy is lost in the collision.
Hitting a wall at 90 If a ball hits a wall with initial speed u and the
coefficient of restitution between the wall and ball is e the speed of the ball
after the collision is eu away from the wall. So the chage in momentum (if
the ball has mass m) is mu (meu) = mu(1 + e) normal to the wall.

Hitting a wall not at 90 If a ball is travelling towards a wall with speed


u and at an angle to the plane of the wall, to work out the speed after the
collision the initial velocity MUST be split into its components parallel and
perpendicular to the wall.
speed bef ore = u sin
speed bef ore = u cos
The wall only produces a normal force on the ball, hence only its component
perpendicular to the wall is changed.
speed af ter = eu sin
speed af ter = u cos

The overall speed of the particle after the impact v (by pythagorus) is
q

u2 cos2 + e2 u2 sin2

This looks pretty rank but in the exam you would have numbers to substitute
in, so it would be much nicer. The angle v makes with the plane is , so
tan =

eu sin
=
= e tan

u cos

Collision of Two Non-fixed Particles at 90 This is the hardest bit


of collisions, but really it is just writing two simultaneous equations then
solving them for 6+ marks. Suppose there are two particles of mass m1 and
m2 travelling at speeds of v1 and v2 respectively as in the diagram below.

Take right to be positive, it doesnt matter which way you take to be


positive, but it is easier to choose the way that most of the balls are moving.
(If some balls are moving against the direction you have selected as positive
then they have negative values for their velocity.) First draw a diagram
showing the speeds and masses before and after. To find the velocities of
the balls after they collide (V1 and V2 in the above diagram) we need two
eqations, one for the conservation of linear momentum and one for Newtons
experimental law. We assume they both travel in our positive direction, after
(the sign of our answer will indicate which direction they acually end up
travelling)
(1) by COLM m1 v1 + m2 v2 = m1 V1 + m2 V2
(2) by N EL e(speed of approach) = V2 V1
By speed of approach I mean the relative speed of the two particles before
the collison, so eg if both are travelling in the same direction at 8ms1 and
6ms1 the relative speed is 2ms1 . If these two particles were travelling
towards each other the relative speed would be 14ms1 . V2 V1 in the
equation above represents the speed of seperation (since we assume they are
both travelling to the right) In a question most of the data would be given
to you, it would look much more friendly than the general equations above.
So for example if m1 = 2 m2 = 6 v1 = 7 v2 = 3 and e = 14 . We would have,
2

(1) 2 7 + 6 3 = 2V1 + 6V2


(2)

1
(7 3) = V2 V1
4

so
(1) 32 = 2V1 + 6V2
(2) 1 = V2 V1
17
Solving simultaneously we get V1 = 13
4 and V2 = 4 If we were to get a
negative answer this would just indicate that a ball was travelling to the left
instead.

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