AP PHYSICS C
Collisions
Collisions in One Dimension
Consider an object of mass m1 with initial velocity v1i approaching a second object of mass m2
that is moving in the same direction with initial velocity v2i.
If v2i < v1i, the objects collide. Let v1f and v2f are their final velocities after the collision.
Conservation of momentum gives us the following relationship:
m1v1i + m2 v2i = m1v1 f + m 2 v2 f
Perfectly Inelastic Collisions
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In a perfectly inelastic collision, the particles stick together after the collision. The final
velocities are equal to each other and to the velocity of the center of mass:
v1 f = v2 f = vcm
This result combined with conservation of momentum gives:
†
m1v1i + m2 v2i = (m1 + m2 )vcm
Example 1:
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An astronaut of mass 60 kg is on a space walk to repair a communications satellite. Suddenly she
needs to consult her physics book. You happen to have it with you, so you throw it to her with
speed 4 m/sec relative to your spacecraft. She is at rest relative to the spacecraft just before
catching the 3.0 kg book.
Find
(a) her velocity just after she catches the book
(b) the initial and final mechanical energy of the book-astronaut system
(c) the impulse exerted by the book on the astronaut
It is useful to express the kinetic energy K of a particle in terms of its momentum p. For a mass
moving with speed v, we have
1 2 (mv) 2
K = mv =
2
2m
Since p = mv,
p2
K=
2m
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We can apply this to a perfectly inelastic collision where one object is initially at rest. The
momentum of the system is that of the incoming object:
†
P = m1v1i
The initial kinetic energy is:
†
†
P2
Ki =
2m1
After colliding, the objects move together as a single mass m1 + m2 with vcm. Momentum is
conserved so the final momentum equals P. The final kinetic energy is then:
P2
Kf =
2(m1 + m 2 )
Example 2:
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In a feat of marksmanship, you fire a bullet into a hanging target. The target, with bullet
embedded, swings upward. Noting the height reached at the top of the swing, you immediately
inform the crowd of the bullet’s speed. For arbitrary masses m1 and m2 and height h, how
would you calculate the speed?
Elastic Collisions in One Dimension
For elastic collisions, the initial and final kinetic energies are equal:
1
1
1
1
m1v1i2 + m2 v2i2 = m1v12f + m 2 v22 f
2
2
2
2
This, together with conservation of momentum, is sufficient to determine the final velocities of
the two objects. However, because of the quadratic nature of the equations, the solution of an
elastic collision problem is more complicated. They can be treated more easily if the relative
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velocities of the two particles after the collision is expressed in terms of the relative velocity
before collision. We can then write:
m1 (v1i2 - v12f ) = m 2 (v22 f - v22i )
or
†m 2 (v2 f - v2i )(v2 f + v2 i ) = m1 (v1i - v1 f )(v1i + v1 f )
From conservation of momentum, we know:
m1v1i + m2 v2i = m1v1 f + m 2 v2 f
†
so that
m1 (v1i - v1 f ) = m 2 (v2 f - v2 i )
†
Dividing these equations, we get
v1i + v1 f = v2 i + v2 f
†
or:
†
†
-(v2 i - v1i ) = v2 f - v1 f
If the two objects are to collide, v2i - v1i must be negative, making their speed of approach -(v2i v1i). After colliding, the objects’ speed of recession is v2f - v1f which is positive.
In elastic collisions, the speed of recession equals the speed of approach.
Example 3:
A 4kg block moving right at 6 m/sec collides elastically with a 2 kg block moving right at 3 m/sec.
Find their final velocities.
Example 4:
A neutron of mass m1 and speed v1i collides elastically with a carbon nucleus of mass m2 at rest.
(a) What are the final velocities of both particles?
(b) What fraction of its initial energy does the neutron lose?
Collisions in Three Dimensions
For collisions in three dimensions, the total initial momentum is the sum of the initial momentum
vectors of each object involved in the collision. For perfectly inelastic collisions, the objects stick
together, and since their final momentum equals the initial momentum, they move off in the
direction of the resultant total momentum with velocity vcm given by:
vcm =
P
m1 + m2
where P = p1 + p2 is the total momentum of the system. Since P is in the plane formed by p1
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and p2, the collision takes place in this plane.
Example 5:
A small car of mass 1.2 x 10 3 kg traveling east at 60 km/h collides at an intersection with a truck
of mass 3 x 10 3 kg traveling north at 40 km/h. The car and the truck stick together. Find the
velocity of the wreckage just after the collision.
Elastic Collisions in Three Dimensions
These collisions are highly complicated. The figure shows an off-center collision between an
object of mass m1 moving with velocity v1i parallel to the x axis toward an object of mass m2
that is initially at rest at the origin.
The distance b between the centers measured perpendicular to the direction of v1i is called the
impact parameter. After the collision, object 1 moves off with velocity v1f, making an angle Q1
with its initial velocity and object 2 moves with velocity v2f, making an angle Q2 with v1i.
Conservation of momentum gives:
P = m1v1i = m1v1 f + m 2 v2 f
†
From this equation, the vector v2f must lie in the plane formed by v1i and v1f, which we will
take to be the xy plane. Assuming that we know the initial velocity v1i, we have four unknowns:
the x and y components of both final velocities; or the final speedos and the two angles of
deflection.
The x and y components of the conservation-of-momentum equation gives us two of the needed
relations among these quantities. Conservation of energy gives a third relation.
The fourth relation needed depends on the impact parameter b and on the type of interacting
force exerted by the bodies on each other. This is found experimentally by measuring the angle of
deflection or the angle of recoil.
Let us focus on the special case of the off-center elastic collision of two objects of equal mass
when one is initially at rest. If v1i and v1f are the initial and final velocities of object 1 and v2f is
the final velocity of object 2, conservation of momentum gives:
mv1i = mv1 f + mv2 f
or
†
v1i = v1 f + v2 f
These vectors form a triangle shown below. Since energy is conserved in the collision,
†
1 2 1 2 1 2
mv1i = mv1 f + mv2 f
2
2
2
or:
†
v1i2 = v12f + v22 f
This equation is the Pythagoren theorem for a right triangle formed by the vectors v1f, v2f and
v1i with the hypoteneuse of the triangle being v1i.
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