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Impulse and Momentum
So far we have talked about how things move when we apply a force. While we are
going to continue to do that, we are going to add an additional feature. Now we want
to investigate what happens when two objects collide with each other.
During a collision, the two objects are often in contact with each other for a short but
measurable time (i.e. the collision takes time to occur and is not instantaneous).
During the time the two objects are in contact, they are usually both changing their
velocity (one speeds up while the other slows down).
Impulse (FDt) is defined as being the product of the average force (F) on an object
times the time interval (Dt) over which the force acts (the time that the two objects
are in contact with each other) and has units of N∙s.
From Newton’s laws: FDt = mD
Momentum (ρ) is defined as the mass of an object times its velocity: ρ = m
and has units of kg∙m/s. Remember that D is f - i so that mD will become
Dρ or ρf - ρi so that the Impulse on an object is equal to the change in an object’s
momentum.
FDt = ρf - ρi which is known as the Impulse-Momentum theorem
Imagine that a pitcher throws a baseball and the ball is hit by the bat. How can we use
the Impulse-Momentum theorem to evaluate the collision?
For now we are assuming that the ball is traveling only in the –x or +x direction (is
remains level throughout the throw and hit).
Let’s assume that we carefully measure the impact (see pages 229-231 for this
discussion) and find that the impulse is 13.1 N∙s in the +x direction. Let’s also assume
that the pitcher threw the 0.145 kg ball with a velocity of -38 m/s (- means in the
negative x direction).
Therefore the initial momentum is ρi = mi = (0.145 kg)(- 38 m/s) = - 5.5 kg∙m/s
From the Impulse-Momentum Theorem:
FDt = ρf - ρi
so
ρf = ρi + FDt
ρf = ρi + FDt = - 5.5 kg∙m/s + 13.1 N∙s = 7.6 kg∙m/s or 7.6 N∙s
since N∙s and kg∙m/s are the same units.
We can extend the problem by asking what the velocity of the baseball is after impact.
ρf = mf
so
f = (7.6 N∙s)/(0.145 kg)= 52 m/s
Impulse-Momentum theorem can be used to understand how collisions work. In
addition, we can use it to understand how important the time interval can be in
determining what happens in a collision.
Let’s look at a car coming to a stop in three different ways (note that the change in
momentum will be the same in all three examples because the mass and change in
velocity is the same).
A 1550 kg car is traveling at 105 km/h (ρi = 4.52X104 N∙s) and will come to a stop
(ρf = 0 N∙s). In case a, the cars comes to a stop over 18.5 s (normal breaking)
In case b, the car comes to a stop in 2.95 s (emergency breaking). In case c, the car
comes to a stop in 0.194 s (collision with an immovable object). What force was
exerted on the car in each of these cases?
ρf - ρi = - 4.52X104 N∙s
so
FDt = - 4.52X104 N∙s
Therefore F = (- 4.52X104 N∙s)/Dt for each case.
In case a: F = - 2.44X103 N
In case b: F = - 1.53X104 N
In case c: F = - 2.33X105 N
See Example problem on page 232 and practice problems on page 233.
In case a: F = - 2.44X103 N
In case b: F = - 1.53X104 N
In case c: F = - 2.33X105 N
If we look at these three forces of “deceleration” we can see that a collision may cause
a force that is about 100 times greater than normal breaking.
Imagine being in the car when that kind of deceleration occurs.
What could happen to you?
This type of understanding of physics is why airbags have been invented.
In a crash where airbags deploy, your body will have a little more time to decelerate
because of the airbags. As a result, the force exerted on your body will be less than it
otherwise would have been. This can make many collisions survivable that would not
have been before the invention of airbags.
Angular Impulse and Angular Momentum
The arguments we have used for impulse and momentum for linear motion can also be
applied to rotating systems to create the Angular Impulse and Angular Momentum
theorem.
Angular Impulse (tDt) is defined as being the product of the average torque (t) on
a rotating object times the time interval (Dt) over which the torque acts.
From Newton’s laws:
tDt = ID
but note that I is defined as angular momentum
Angular Momentum (L) is defined as the moment of inertia of an object times its
angular velocity: L = I 
Remember that D  is  f -  i so that ID  will become
DL or Lf - Li so that the Angular Impulse on an object is equal to the change in an
object’s angular momentum.
tDt = Lf - Li which is known as the Angular Impulse-Angular Momentum theorem
If no force is acting on an object, then the its linear momentum (r) is constant. Since
the mass of an object can not change if the linear momentum is constant, then the
linear velocity must also be constant.
If no torque is acting on an object, then the its angular momentum (L) is constant.
While the mass of an object can not change, the angular momentum depends not just
on mass, but also on the way the mass is distributed about the axis of rotation (i.e. on
the moment of inertia). Therefore, even though no torque is acting on an object, an
object’s angular velocity can change which is not true for linear movement.
If you have ever watched ice skaters, then you have seen this idea in action. As an ice
skater spins on their axis with their arms a spread out wide, they rotate more slowly.
When they bring their arms in tight, they spin much faster. They did not apply any
additional torque to make themselves spin faster, they just used this idea of angular
momentum. They decreased their moment of inertia when they moved their arms in
close to their body.
DL = ID
so
D = DL/I
Bringing their arms in tight lowers a skater’s moment of inertia which will result in a
greater angular velocity (remember that the change in angular momentum is constant
here because no torque is being applied).
The Law of Conservation of Momentum
The Law of Conservation of Momentum states that for any closed, isolated system the
momentum remains constant.
Consider two round objects colliding as illustrated below (one Red and one Black).
rRi
rBi
FB on R
FR on B
rRf
rBf
Prior to the collision, both objects have their own independent momenta.
In the collision, they exert equal but opposite forces on each other (Newton’s 3rd Law)
FB on R = - FR on B
FB on RDt = - FR on BDt
The time interval (Dt) for the collision is the same for both objects. Therefore, the
impulse for both objects are also equal but opposite which means that the change in
momentum for each object is also equal but opposite.
rRf - rRi = - (rBf - rBi)
rRf + rBf
= rRi + rBi
Rearranging this equation proves that the total momentum before and after the
collision will be the same.
Imagine that we throw one Velcro covered ball (mB = 0.25 kg, Bi = 15 m/s) at another
stationary one (mR = 0.18 kg, Ri = 0 m/s) so that they stick together after the collision.
What will be the final velocity of the two balls moving together?
Bi
BRf = ?
Ri = 0
rRf + rBf
= rRi + rBi
mRBRf + mBBRf
= mRRi + mBBi Click to see solution
BRf (mR + mB) = mRRi + mBBi
BRf (mR + mB) =
BRf =
BRf =
0 + mBBi
mBBi
(mR + mB)
(0.25 kg)(15 m/s)
= 8.7 m/s
(0.18 kg + 0.25 kg)
See example problem on page 237 and practice problems on page 238
Anyone that has fired a gun knows something about recoil.
Recoil is motion created when an internal force causes an object of lower mass to be
projected in one direction while the force simultaneously causes an object of greater
mass to recoil in the opposite direction.
When a gun is fired, an explosion inside the casing of the bullet causes the bullet
(smaller mass object, mB) to be accelerated down the barrel of the gun. At the same
time, the force of the explosion causes the mass of the gun (mG) to “kick” back in the
opposite direction to the bullet. Since nothing was moving prior to firing the gun, the
total momentum of the system was zero initially. Due to the Law of Conservation of
Momentum, the momentum after firing is also zero.
After Firing
Before Firing
rGi + rBi
=
rGf + rBf
0
=
rGf + rBf
- rGf
=
rBf
- mGGf
=
mBBf
Gf =
− mBBf
mG
or
Bf =
− mGGf
mB
An astronaut experiencing weightlessness on the international space station needs to
put out a fire. She uses a pressurized CO2 extinguisher. She releases 54 g of CO2
which will be traveling at 750 m/s. If the astronaut has a total mass of 73.5 kg (herself
and gear). How fast will the astronaut be moving after using the extinguisher?
After Using
Before Using
=
rAf + rCO2f
0
=
rAf + rCO2f
- rAf
=
rCO2f
- mAAf
=
mCO2CO2f
rAi + rCO2i
Click to see solution
− mCO2CO2f
mA
− (0.054 g)(750 m/s)
=
(73.5 kg)
Af =
What is the meaning
of the negative sign
in the answer?
Af
Af = - 0.55 m/s
See example and practice problems on page 240
What would happen to the astronaut if she did not hold the extinguisher directly in
front of her center of mass?
Two Dimensional Collisions: when objects collide but do not end up traveling along
only one axis, we may need to use the idea of component vectors to solve the problem.
Imagine a red pool ball colliding with a stationary green pool ball (a dotted arrow
indicates direction and speed of a ball).
Before Collision
After Collision
rRi + rGi
rGi = 0
so
rRix = rRf x + rGfx
rRiy = 0
rRf + rGf
=
so
rRi = rRf + rGf
and
rRiy = rRf y + rGfy
rRf y + rGfy = 0
Depending upon the situation, we will need to remember our geometry:
For a right triangle: h = (x2 + y2)1/2
and tan() = y/x
x = h(cos)
and y = h(sin)
A red Velcro covered sphere (mR = 0.420 kg) is moving in the +x direction at 3.87 m/s.
The red sphere collides with a green Velcro covered sphere (mG = 0.190 kg) that was
moving in the +y direction at 1.68 m/s. What will the velocity of the spheres be after
the collision (assume they stick together).
Before Collision
rf
After Collision
rfy
rfx
Clickrto see
solution
Ri + rGi = rf
rf = rfx + rfy
mR = 0.420 kg
mRG = 0.610 kg
Rix =3.87 m/s
f = ? m/s
mG = 0.190 kg
=?
Giy =1.68 m/s
Riy = 0 m/s
Gix = 0 m/s
rRi = rRix = rfx = mRRix = 1.6254 kgm/s
rGi = rGiy = rfy = mGGiy = 0.3192 kgm/s
rf = mRGf
f =
1.66 kgm/s
rf
=
mRG
(0.610 kg)
f = 2.72 m/s
rf = ((rfx)2 + (rfy)2)1/2 = ((1.6254)2 + (0.3192)2)1/2 = 1.66 kgm/s
See example and practice problems
-1
-1
o
 = tan (rfy/rfx) = tan (0.3192/1.6254) = 11.1 on pages 242, 243
Law of Conservation of Angular Momentum
If no outside torque acts on an object then its angular momentum does not change.
This is known as the Law of Conservation of Angular Momentum
In equation form:
L i = Lf
an objects initial angular momentum is equal to its final angular momentum
Using the definition of angular momentum (L = I) and plugging it into the law of
conservation of angular momentum gives:
Iii = If f
Rearranging gives:
f
I
= i note that this makes it clear that I and  are
i
If
inversely proportional (i.e. the faster an object’s angular velocity, the lower its
moment of inertia).
Using the definition of frequency (f = /2p) and substituting for  gives:
ff
I
= i
If
fi
Based on the Law of Conservation of Angular Momentum the direction of rotation of a
spinning object can only be changed if torque is applied.
Applying a torque to a spinning object will cause the object to precess. Precession is
where the axis of rotation slowly revolves around a pivot point.
If you have ever watched a spinning top, then you have seen precession.
Initially, the top is spinning fast enough to remain upright (axis of rotation is vertical).
However, as the top slows down (due to friction effects), at some point, the effect of
gravity on the top will start to tip the top over (this is an applied torque). At this point,
you will see the top start to wobble around in a circle (the axis of rotation is no longer
vertical and it will trace out an imaginary circle).
As the red top in the diagram precesses, the axis of
rotation of the top (the dotted lines) will trace out a
circle as illustrated. The direction of precession will be
in the same direction as the rotation of the top.
A gyroscope is based on this same idea.
A gyroscope is created when we have a disk spinning rapidly around its axis and we
mount the spinning disk in a case that can freely move in any direction.
Due to the spinning disk, the axis of rotation of the disk does not want to change its
orientation. If we pick up the gyroscope and turn it sideways, the axis of rotation will
remain pointing in its original direction (since we built the case so that the disk and its
axis can move in any direction). As the case moves, the spinning disk does not.
Gyroscopes are used to always know the relative orientation of a vehicle. For
example, a submarine that has a gyroscope can determine its orientation in the
water by seeing how the ship is oriented relative to the gyroscope.