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) 
Conservation
of momentum gives:
Substituting
(1) into (2) gives:
b) Using equation (1)
c) After collision
At first glance the above proof is a little complicated. In line 3, the values of 
and 
have been substituted and
the m’s have been cancelled. In line 4, the u’s have been cancelled and the
fractions have been simplified. Lines 5 to the end involve multiplying out
two brackets and simplifying. Note that
e can’t be negative.
d) From part b we know:
Substituting
for e gives:
Hence
A is at rest after the collision.
Collisions
and Vector Notation
The only
thing to remember is to only work with like terms and that the magnitude of the
velocity vector is the speed.
Example
A
body A of mass 250g, is moving with velocity (-2i + 3j) 
when it collides with a
body B, of mass 750g, moving with velocity (5i + 8j) . Immediately after
the collision the velocity of A is (i +
9j). Find:
a) The velocity of B after the collision
b) The loss in kinetic energy of the system due
to the collision
c) The impulse of A on B due to the collision.
a) Conservation of momentum gives:
0.25 x (-2i
+ 3j) + 0.75 x (5i
+ 8j) = 0.25 x (i +
9j) + 0.75 x (xi + yj)
(3i
+ 4.5j) = 0.75(xi + yj)
Velocity of B = (4i + 6j)
b) Loss of kinetic energy
c) Impulse of A on B
Impulse =
change in momentum
= 0.75 x ((4i
+ 6j) - (5i + 8j))
= (-0.75i - 1.5j)Ns
Questions A
1 Two spheres A and B are of equal radii and
masses 2kg and 6kg
respectively. The spheres move towards each other
along the same straight line on a smooth horizontal surface with speeds 8and 4respectively. After the collision the particle A rebounds
with speed 7. Find the velocity of
B after the collision.
2 Two
spheres A and B of equal radii and masses 3kg and 1kg are travelling towards
each other along a straight line on a smooth horizontal surface. Initially, A has a speed of 6 and B has a speed of 2. After the collision
the direction of B is reversed and it is travelling at a speed of
4. Find the speed of A after the collision and the loss of
Kinetic energy due to the collision.
3 A
particle A, of mass 1.5kg, is moving with speed 2 on a smooth horizontal surface when it collides directly
with another particle B, of mass 1kg, which is moving with speed 1 in the same direction. After the collision A continues to
move in the same direction with speed 1.4ms-1. Find:
a) The speed of B after the impact,
b) The coefficient of restitution between
A and B.
4 Two
spheres A and B of equal radii and masses 175g and 100g are travelling towards
each other along a straight line on a smooth horizontal surface. Initially, A has a speed of 2 and B has a speed of 6. After the collision
both particles have reversed their original directions of motion and B now has
a speed of 3. Find the speed of A
after the collision and the loss of Kinetic energy due to the collision.
5 Two
spheres A and B of equal radii and masses m and 2m are travelling on a smooth
horizontal surface. Initially, A and B
are moving in the same direction with speeds 3u and 2u respectively. There is a direct collision between A and B,
the coefficient of restitution between them being e.
a) Find,
in terms of u and e, the speeds of A and B after the collision.
b) Show
that, whatever the value of e, the speed of B cannot exceed 
Given that e is 0.8,
c) Find
the magnitude of the impulse exerted by A on B in the collision.
6 A
body A of mass 5kg, is moving with velocity (-12i + 6j) when it collides with a body B, of mass 4kg, moving with
velocity (10i – 5j). Immediately after
the collision the velocity of A is (-8i +
4j). Find the velocity of
B immediately after the collision.
Impact of a particle with
a fixed surface.
It is
assumed that the collisions occur normally (i.e. along a perpendicular line)
Speed
of rebound = e
Speed
of approach
As well as
horizontal collisions with fixed surfaces one can also be asked to consider
particles falling from a height and rebounding to a new height.
Example
1 A small smooth ball falls from a height of
3m above a fixed smooth horizontal surface. It rebounds to a height 1.2m. Find
the coefficient of restitution between the ball and the plane.
The
first task is to calculate the time it takes to fall to the surface and the
speed at the time of contact.
Using 
with s = 3m, u
= 0 and a = 9.8
Then
using v = u +at
V = 9.8 x 
= 7.668
The particle
strikes the plane with a velocity of 7.668. After the rebound
the particle only manages to reach a height of 1.2m. The next task is to calculate the speed with
which the particle leaves the surface.
Using
with s =-1.2m, v = 0, a = -9.8
Therefore
using Speed
of rebound = e
Speed
of approach
Problems involving a range in the value of e
When attacking problems of this nature
students need to think about the fundamentals of the situation. Simply thinking of what is necessary for
further collisions to occur will lead to the right answer.
Example
1 A
small smooth sphere of mass 2.5kg moving on a smooth horizontal plane with
speed 7 collides directly with a sphere of mass 11kg which is at
rest. Given that the spheres move in
opposite directions after the collision, obtain the inequality satisfied by e.
Conservation
of momentum gives:
Newton’s
Law gives:
Substituting
equation (1) into (2) gives:
The
question states that the sphere A moves to the left, and the diagram has taken
this into consideration, therefore v1 must be positive.
Note also that e
< 1.
This question is
not mathematically difficult, there may be some tricky algebra and fraction
manipulation but the solution depends upon getting the expression for v1. All previous questions have involved finding
v1 and v2, so students shouldn’t worry too much when an
exam question asks for the range in values in e.
2 Two small smooth spheres, A and B, of equal radius, have masses m
and 5m respectively. The sphere A is moving with speed u on a smooth horizontal table when it
collides directly with B, which is at
rest on the table. As a result of the
collision the direction of motion of A is reversed The coefficient of
restitution between A and B is e.
a) Find the speed of A and B immediately after
the collision
b) Find the range of possible values of e.
a)
Conservation
of momentum gives:
Newton’s Law gives:
Substituting
equation (1) into (2) gives:
Substituting
back into equation (1) gives:
b) Once again v1 must be positive
therefore;
5e - 1
> 0
e > 0.2
Multiple
collisions and problems involving three particles.
Once
again these questions only require the same approach as before and students
should be able to set up the first set of simultaneous equations and find the
values for and .
Example
1 A uniform sphere A of mass m is moving
with speed u on a smooth horizontal
table when it collides directly with another uniform sphere B of mass 2m which is at rest on the table. The spheres are of equal radius
and the coefficient of restitution between them is e. The direction of motion of A
is unchanged by the collision.
(a) Find
the speeds of A and B immediately after the collision.
(b) Find
the range of possible values of e.
After being struck by A, the sphere B collides
directly with another sphere C of mass 4m and of the same size as B. The
sphere C is at rest on the table
immediately before being struck by B. The
coefficient of restitution between B and
C is also e.
(c) Show that, after B has struck C, there
will be a further collision between A and
B.
a) Remember to always draw a diagram
Conservation
of momentum gives:
Newton’s Law gives:
Adding
the two equations gives:
And
by substituting back into equation (1) gives:
b) Range of values of e
The
question states that is positive,
therefore:
1 - 2e
> 0
e < 0.5
c) Considering B going on to strike C
Before
attempting to find a value for 
one needs to consider
where the question is going. For A to
collide with B, must be greater than w2.
Conservation
of momentum gives:
Newton’s Law gives:
Equation
(1) - 2 x Equation (1) gives:
Substituting
into the above
equation gives:
Remembering
to note that the aim is to show that for A to collide with B, must be greater than and that
Equation
(3) becomes:
e
< 0.5 
Hence
and therefore there will be a further collision between A and
B.
2 Three identical spheres A, B and C lie
on a smooth horizontal surface with their centres in a straight line and with B
between A and C. Given that A is
projected towards B with speed u, show that, after impact, B moves with speed
0.5u(1+e), where e is the coefficient of restitution between each of the
spheres. When C first moves, it is found
to have speed 
.
a) Find e.
b) Show that when C begins to move, 
of the initial kinetic
energy has been lost.
a) A
strikes B
Conservation
of momentum gives:
Newton’s
Law gives:
Adding
the two equations gives:
b) Considering B colliding with C
Conservation
of momentum gives:
(1)
Newton’s Law gives:
Adding
together equations (1) and (2) gives:
We
know v2 from part a)
Therefore
equation (3) becomes:
The only
valid solution is e = 0.5
Recoil
of a Gun
A
bullet is fired from a gun with a horizontal velocity of 400
. The mass of the gun
is 3Kg and the mass of the bullet is 60g. Find the initial speed of recoil of
the gun and the gain in kinetic energy of the system.
The
rifle is brought to rest by a horizontal force exerted by the soldier’s shoulder,
against which the rifle is pressed. This force is assumed to be constant. Given
that the rifle recoils a distance of 3cm before coming to rest, find the magnitude of the horizontal force
exerted on the rifle by the soldier in bringing it to rest.
Conservation
of momentum gives:
Initial
Ke = 0
The initial
speed of recoil of the gun is 8ms-1 and the gain in kinetic energy
due to the explosion is 4896J.
The
work done by the horizontal force must equal the kinetic energy of the rifle
(96J).
Work done = force x distance moved.
96J = F x 0.03
F = 3200N
What does this force mean in real terms?
Questions B
1 A pile
driver of mass 1000kg falls from a height of 1.8m on to a pile of mass
300kg. After the impact the pile and the
driver move on together. Given that the
pile is driven a distance of 0.35m into the ground, find:
a) The speed at which the pile starts to move
into the ground,
b) The magnitude of the resistance of the
ground, in Kn (assumed constant).
2 Three identical spheres A, B and C each
of mass 2.5Kg lie at rest on a smooth horizontal table. Sphere A is projected with speed 20 to strike sphere B directly. Sphere B then goes on to strike
sphere C directly. Given that the
coefficient of restitution between any two spheres is 
a) Find the speeds of the spheres after these
two collisions.
b) Find the total loss of kinetic energy due to
these two collisions
3 A particle P of mass 0.6kg is moving
with a speed 2.5 on a smooth horizontal table of height 1m. Another particle Q of mass Mkg is at rest on
the edge of the top of the table. Particle P strikes particle Q and they
coalesce into a single particle R. The
particle R then falls to the floor. The
horizontal displacement of R is 0.4m.
a) Find the time it takes for R to reach the
floor.
b) Find the value of M
4 A
uniform sphere A of mass m is moving with speed v on a smooth horizontal table when it
collides directly with another uniform sphere B of mass 2m which is at
rest on the table. The spheres are of equal radius and the coefficient of
restitution between them is e.
a) Find the speeds of the two spheres after the
impact.
Given that one
half of the kinetic energy is lost in the impact,
b) Find the value of e.
5 A
particle A, of mass m, is moving with speed u on a smooth horizontal surface
when it collides directly with a stationary particle B, of mass 3m. The coefficient of restitution between A
and B is e. The direction of motion of is reversed by
the collision.
a) Show that the speed of B after the collision
is 
b) Find the speed of A after the collision.
Subsequently
B hits a wall fixed at right angles to the direction of motion of A and B. The
coefficient of restitution between B and the wall is 0.5. After B rebounds from the wall there is
another collision between A and B.
c) Show that 
d) In the case where e = 0.5, find the magnitude
of the impulse exerted on B by the wall.
6 Two particles P and Q each of mass 3.5kg
are at rest 1.5m apart on a rough horizontal table. The coefficient of friction between the
particles and the surface is 0.45. Particle P is projected towards Q with a
speed of 8.5.
a) Calculate the speed of P just before it
strikes Q.
Particle
P collides with Q and the two particles coalesce into a single particle S.
b) Calculate the distance that S travels before
coming to rest.