Points to remember
from earlier work on collisions
- In the
absence of any other forces acting on the two bodies, the changes in
momentum of A and B will be equal in magnitude, but opposite in direction.
- Momentum
is a vector.
- The gain
in momentum of a body will equal the loss in momentum of the other
body. Hence the sum of momentum
before the impact will equal the sum of momentum after the impact (Conservation of Linear Momentum)
- The impulse of a force = change in
momentum produced
I = mv
- mu
The
two bodies shown collide on a horizontal surface. Find the speed v1 of the lighter
body after the impact.
Always draw two diagrams, one
highlighting the situation before the collision and the other showing the
situation after the collision.
Assuming that velocities
to the right are positive, then;
Momentum before
collision = (2 x 6) + (5 x (-4))
Momentum
after collision = 
Therefore
by the Conservation of Momentum 
The speed of the lighter body after the impact is 
, and
its direction is reversed.
More realistic collisions.
In
M2 the time of collision is split into two parts
- The
period of compression
- The
period of restitution (shape recovery)
The
property that allows for compression and restitution is called the elasticity. For perfectly elastic collisions there is
no loss of Kinetic Energy and for inelastic collisions the particles
coalesce. Obviously in the real world
there is some loss in Kinetic Energy by way of noise or heat. Collisions between two particles have a
coefficient of restitution e and this determines the speed of
separation
of the particles. This leads to a
further
Speed
of separation of particles = e
Speed
of approach of particles
Obviously;
, for
perfectly elastic collisions e = 1 and for inelastic collisions e = 0
Another assumption
is that the surface is smooth. The presence
of friction would be accompanied by spinning of the particles.
Examples
1 Two spheres A and B, are of equal radii
and masses 2kg and 1.5kg
respectively.
The spheres A and B move towards each other along the same straight line on a
smooth horizontal surface with speeds 
and 
respectively.
If the coefficient of restitution between
the spheres is 0.4, find their speeds and directions after the impact.
Once again remember to draw two diagrams.
It is
worth remembering that at this point one could not be certain of the direction
of v1.
By
Conservation of Momentum (assuming that speeds to the right are positive)
By
adding equation (2) to twice equation (1) this gives:
By
substituting v2 into equation (1) gives:
(note
that particle A has reversed its direction).
Questions involving the collision of two
particles should always be approached in the same manner. Students appear to struggle with collisions questions
on exam papers but there should be no excuse for them not starting the
question. Setting up the equations for
Conservation of Momentum and
2 Two
spheres A and B of equal radii and masses 200g and 150g are travelling towards
each other along a straight line on a smooth horizontal surface. Initially, A has a speed of 4 ms-1
and B has a speed of 
. The spheres collide
and the collision reduces particle A to rest. Find the coefficient of
restitution between the spheres.
Once
again the standard approach is to set up the two equations.
Conservation
of momentum gives:
Loss
of Mechanical Energy
When particles collide there is no loss of
momentum but there is a loss of Kinetic Energy. Some of the KE is transformed
into other forms of energy at the impact, e.g. heat and sound energy.
Examples
1 Two
spheres A and B of equal radii and masses 125g and 300g are travelling towards
each other along a straight line on a smooth horizontal surface. Initially, A has speed of 
and B has speed of . After the collision
the direction of B is reversed and it is travelling at a speed of 
. Find the speed of A
after the collision and the loss of Kinetic energy due to the collision.
Conservation
of momentum gives:
Note
again that the direction for v1 in the second diagram is incorrect as v1 is negative.
Loss
of Kinetic Energy
2 A smooth sphere A of mass m is moving with
speed u on a smooth horizontal table
when it collides directly with another smooth sphere B of mass 3m, which is at
rest on the table. The coefficient of restitution between A and B is e. The spheres have the same radius and
are modelled as particles.
(a) Show
that the speed of B immediately after
the collision is 
(1 + e)u.
(b) Find
the speed of A immediately after the
collision.
Immediately after the collision the total kinetic energy of the spheres is 
mu2.
(c) Find
the value of e.
(d) Hence
show that A is at rest after the
collision.
a)