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The linear momentum of an object is defined
as the product of its mass and its velocity.
p = mv
The direction of the momentum is the
direction of the velocity. Because velocity
depends on the reference frame, so does
momentum; thus, the reference frame must be
specified.

According to the equation, a fast-
moving car has more momentum
than a slow-moving car of the
same mass; a heavy truck has
more momentum than a small car
moving with the same speed.
Thus, the more momentum an
object has, the harder it is to stop
it, and the greater effect it will
have if it is brought to rest by
striking another object.

A force is required to change the momentum
of an object, whether it is to increase the
momentum, to decrease it, or to change its
direction. Newton originally stated his second law
of motion in terms of momentum:
Σ𝑭 =
Δ𝒑
Δ𝑡
“The rate of change of momentum of an object is
equal to the net force applied to it.”

The preceding equation is more
general than the more familiar version
(F = ma) because it includes the
situation in which the mass may change.
A change in mass occurs in certain
circumstances, such as for rockets which
lose mass as they burn fuel.

Example #1:
For a top player, a tennis ball may
leave the racket on the serve with a
speed of 55 m/s. If the ball has a mass
of 0.06 kg and is in contact with the
racket for about 4 ms, estimate the
average force on the ball. Would this
force be large enough to lift a 60-kg
person? The tennis ball is hit when its
initial velocity is very nearly zero at
the top of the throw.

Example #2:
Water leaves a hose at a
rate of 1.5 kg/s with a speed
of 20 m/s and is aimed at
the side of a car, which
stops it. Ignoring any
splashing back, what is the
force exerted by the water
on the car?

Assuming that the net
external force on the
system in the figure is
zero, the only significant
forces during collision are
the forces that each ball
exerts on the other.
Conservation of Momentum

Although the momentum of
each of the two balls changes
as a result of the collision, the
sum of their momenta is found
to be the same before as after
the collision.
Total momentum before = Total momentum after
mA1vA1 + mB1vB1 = mA2vA2 + mB2vB2

The general statement of the law of conservation
of momentum is
“The total momentum of an isolated system
of objects remains constant.”
By a system, we simply mean a set of objects
that we choose, and which may interact with
each other. An isolated system is one in which
the only significant forces are those between the
objects in the system.

Example #3:
A 10,000-kg railroad car, A, traveling at a speed of
24.0 m/s strikes an identical car, B, at rest. If the cars
lock together as a result of the collision, what is their
common speed just afterward?

Example #4:
Calculate the recoil velocity of a 5.0-
kg rifle that shoots a 0.020-kg bullet at a
speed of 620 m/s.

During a collision
of two ordinary
objects, both objects
are deformed, often
considerably, because
of the large forces
involved.
Collisions and Impulse

The force usually
jumps from zero at the
moment of contact to a
very large force within a
very short time, and then
rapidly returns to zero
again. The time interval is
usually very distinct and
very small.

The product of the force times the time over which the force
acts is called the impulse.
Σ𝑭Δ𝑡 = Δ𝒑
Σ𝑭Δ𝑡 = 𝑱 = 𝑖𝑚𝑝𝑢𝑙𝑠𝑒
The concept of impulse is useful mainly when dealing with
forces that act during a short time interval, as when a ball hits
a baseball. The force is generally not constant, thus we use
the average force acting during the time interval.

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Conservation of Energy and Momentum in Collisions
A collision, in which the total kinetic energy is
conserved, is called an elastic collision.
1
2
𝑚𝐴1𝑣𝐴1
2 +
1
2
𝑚𝐵1𝑣𝐵1
2 =
1
2
𝑚𝐴2𝑣𝐴2
2 +
1
2
𝑚𝐵2𝑣𝐵2
2
At the atomic level, the collisions of atoms and
molecules are often elastic. But in the “macroscopic”
world of ordinary objects, an elastic collision is an
ideal that is never quite reached.

We do need to
remember that even
when the kinetic
energy is not
conserved, the total
energy is always
conserved.

Then, deriving the equation for elastic collisions in one
dimension, we use the conservation of momentum and
energy:
mAvA1 + mBvB1 = mAvA2 + mBvB2
1
2
𝑚𝐴𝑣𝐴1
2
+
1
2
𝑚𝐵𝑣𝐵1
2
=
1
2
𝑚𝐴𝑣𝐴2
2
+
1
2
𝑚𝐵𝑣𝐵2
2
Then, if we know the masses and velocities before collision,
we can solve for the their respective velocities after collision:
𝑣𝐴1 − 𝑣𝐵1 = − 𝑣𝐴2 − 𝑣𝐵2
This is called a head-on collision. This tells us that the
difference of speed of the two objects after collision has the
same magnitude as before, no matter what the masses are.

We can also derive each final velocity due to
their masses:
𝑣𝐴2 =
𝑚𝐴 − 𝑚𝐵
𝑚𝐴 + 𝑚𝐵
𝑣𝐴1 +
2𝑚𝐵
𝑚𝐴 + 𝑚𝐵
𝑣𝐵1
𝑣𝐵2 =
2𝑚𝐴
𝑚𝐴 + 𝑚𝐵
𝑣𝐴1 +
𝑚𝐵 − 𝑚𝐴
𝑚𝐴 + 𝑚𝐵
𝑣𝐵1

Example #5:
Billiard ball A of mass m moving with
speed v collides head-on with ball B of
equal mass at rest. What are the speeds of
the two balls after collision, assuming it is
elastic?

Example #6:
Block 1 approaches a line of two stationary blocks with a
velocity of v1i = 10 m/s. It collides with block 2, which then
collides with block 3, which has mass m3 = 6.0 kg. After the
second collision, block 2 is again stationary and block 3 has
velocity v3f = 5.0 m/s. Assume that the collisions are elastic
and momentum is conserved. What are the masses of blocks
1 and 2? What is the final velocity v1f of block 1?

Collisions in which kinetic energy is not conserved
are said to be inelastic collisions. The kinetic energy
that is lost is changed into other forms of energy, often
thermal energy, so that the total energy (as always) is
conserved. Then,
KEA1 + KEB1 = KEA2 + KEB2 + thermal and other forms of
energy
If two objects stick together as a result of a
collision, the collision is said to be completely inelastic.

Example #7:
For the completely inelastic
collision of two railroad cars that we
considered in example #3, calculate
how much of the initial kinetic energy
is transformed to thermal or other
forms of energy.

Example #8:
Figure shows a ballistic
pendulum, a system for measuring
the speed of a bullet. The bullet
with mass mB=5.00 g, is fired into a
block of wood with mass mW=2.00
kg, suspended like a pendulum, and
makes a completely inelastic
collision with it. After the impact of
the bullet, the block swings up to a
max height of 3.00 cm. What is the
initial speed of the bullet?

Collisions in Two Dimensions
Common type of non-head-on collision is that
a moving object (called the “projectile”) strikes a
second object initially at rest (called the “target”).

Example #9:
Billiard ball A moving with
speed va = 3.0 m/s in the positive
x direction strikes an equal-mass
ball B initially at rest. The two
balls are observed to move off at
45o to the x axis, where ball A
above the x axis and ball B below
(as shown in the figure). What
are the speeds of the two balls
after collision?

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