Unit #4 Momentum – Impulse &
Momentum
http://www.aplusphysics.com/courses/honors/momentum/impulse.html
Unit #4 Momentum

Objectives and Learning Targets
 Define and calculate the momentum of an object.
 Determine the impulse given to an object.
 Use impulse to solve a variety of problems.
 Interpret and use force vs. time graphs.
 Apply conservation of momentum to solve a variety of
problems.
 Distinguish between elastic and inelastic collisions.
 Calculate the center of mass for a system of point particles.
Unit #4 Momentum
Defining Momentum

Let’s assume there’s a car speeding toward you, out of control without its brakes, at a
speed of 27 m/s (60 mph). Can you stop it by standing in front of it and holding out
your hand? Why not?

Unless you're Superman, you probably don't want to try stopping a moving car by
holding out your hand. It's too big, and it's moving way too fast. Attempting such a
feat would result in a number of physics demonstrations upon your body, all of which
would hurt.

We can't stop the car because it has too much momentum. Momentum is a vector
quantity, given the symbol p, which measures how hard it is to stop a moving object.
Of course, larger objects have more momentum than smaller objects, and faster
objects have more momentum than slower objects. We can therefore calculate
momentum using the equation:
Unit #4 Momentum
Defining Momentum
 Momentum is the product of an object's mass times its
velocity, and its units must be the same as the units of mass
[kg] times velocity [m/s], therefore the units of momentum
must be [kg*m/s], which can also be written as a Newtonsecond [N*s].
Unit #4 Momentum
Sample Problem #1
Question: Two trains, Big Red and
Little Blue, have the same velocity.
Big Red, however, has twice the mass
of Little Blue. Compare their
momenta.
Answer: Because Big Red has twice
the mass of Little Blue, and Big Red
must have twice the momentum of
Little Blue.
Unit #4 Momentum
Sample Problem #2
***Because momentum is a vector, the direction of
the momentum vector is the same as the direction
of the object's velocity.
Question: An A-5 Vigilante supersonic bomber, with a
mass of 21000 kg, departs from its home airbase with
a velocity of 400 m/s due east. What is the jet's
momentum?
Answer:
Unit #4 Momentum
Sample Problem #2
Now, let's assume the jet drops its payload and has
burned up most of its fuel as it continues its journey
to its destination air field.
Question: If the jet's new mass is 16,000 kg, and due
to its reduced weight the pilot increases the cruising
speed to 550 m/s, what is the jet's new momentum?
Answer:
Unit #4 Momentum
Impulse
As you can see, momentum can change, and a change in
momentum is known as an impulse. In Regents Physics, the vector
quantity impulse is represented by a capital J, and since it's a
change in momentum, its units are the same as those for
momentum, [kg*m/s], and can also be written as a Newtonsecond [N*s].
Unit #4 Momentum
Sample Problem #3
Question: Assume the A-5 bomber, which had a
momentum of 8.8*106 kg*m/s, comes to a halt on
the ground. What impulse is applied?
Answer: Let's assume that east is the positive
direction:
Unit #4 Momentum
Impulse-Momentum Theorem
Since momentum is equal to mass times velocity, we can write
that
. We also know that impulse is a change in momentum,
so impulse can be written as
. If we combine these
equations, we find:
Since the mass of a single object is constant, a change in the
product of mass and velocity is equivalent to the product of mass
and change in velocity. Specifically:
Unit #4 Momentum
Impulse-Momentum Theorem
So we're talking about changes in velocity... but what do we call changes in
velocity? Of course, acceleration! And what causes an acceleration? A force!
And does it matter if the force is applied for a very short time or a very long
time? Absolutely it does -- common sense tells us the longer the force is
applied, the longer the object will accelerate, the greater the object's change
in momentum!
We can prove this by using an old mathematician's trick -- if we multiply the
right side of our equation by 1, we of course get the same thing. And if we
multiply the right side of our equation by Δt/Δt, which is 1, we still get the same
thing. Take a look:
Unit #4 Momentum
Impulse-Momentum Theorem
If you look carefully at this equation, you can find a Δv/Δt, which is, by definition
acceleration. Let's replace Δv/Δt with acceleration a in the equation below.
One last step... perhaps you can see it already. On the right-hand side of this
equation, we have maΔt. Utilizing Newton's 2nd Law, I can replace the product
of mass and acceleration with force F, giving us the final form of our equation,
often-times referred to as the Impulse-Momentum Theorem:
This equation relates impulse to change in momentum to force applied over a
time interval.
To summarize: When an unbalanced force acts on an object for a period of
time, a change in momentum is produced, known as an impulse.
Unit #4 Momentum
Sample Problem #4
Question: A tow-truck applies a force of 2000N
on a 2000-kg car for a period of 3 seconds. What
is the magnitude of the change in the car's
momentum?
Answer:
Unit #4 Momentum
Sample Problem #5
Question: If the car starts at rest,
what will be its speed after 3s?
Answer:
Unit #4 Momentum
Sample Problem #6
Unit #4 Momentum
Sample Problem #6
Unit #4 Momentum
Non-Constant Forces
But not all forces are constant... what do you do if a changing
force is applied for a period of time? In that case, we can make a
graph of the force applied on the y-axis vs. time on the x-axis.
The area under the Force-Time curve is the impulse, or change in
momentum.
Unit #4 Momentum
Non-Constant Forces
For the case of the sample graph at the right, we could determine
the impulse applied by calculating the area of the triangle under
the curve. In this case:
Unit #4 Momentum