Prof.
Clements Notes/Hints for Physics Courses
Chapter
8 Part 1 Open Stax College Physics or Most Introductory Physics
Courses
Momentum,
Impulse, Collisions
Terms
you should know at the end of this unit:
momentum, internal force, external force, impulse,
conservation of momentum, collision, KE, elastic collision, inelastic
collision
Linear
Momentum and Force
Your
studies in this chapter will inform you about the second great
conservation law of physics, conservation of momentum. Momentum is
a very useful quantity for the analysis of motion of objects. It is
calculated by multiplying the mass and velocity for an object. The
symbol for momentum is p.
p
= mv The units of momentum are kg m/sec. Momentum is a
vector that has the same direction as the velocity of the object.
The form of Newton’s Second Law that we have been using is actually
a special case of a more general law. Fnet
= Δ p/ Δ t
The
text shows how F = ma is derived from the general law.
F
= ma is not the form of the original “Newton’s Second Law.”
F
= change in momentum / time interval is the original form of
Newton’s second law.
When
a rocket is launched the equation F = ma is not directly useful for
finding the acceleration so we cannot just use the four kinematic
equations to determine the rocket’s motion. The mass will be
changing as fuel is consumed so the acceleration is not constant.
Also, the rate of use of fuel (and thus the force) is variable for
real rocket launches.
In
a system that consists of more than one object, the total momentum is
found by adding the separate momenta of each object.
Impulse
Impulse
is found by multiplying the average force by the length of time that
the force was acting. Impulse is equal to the change in momentum.
F Δt = Δp
The
Force will be the average, or effective, force during the time
interval.
e.g.
Two objects are thrown with a speed of 5 m/s toward a wall. Ball A
stops when it hits the wall and does not rebound. Ball B rebounds.
Does the wall exert the same force on the two balls?
Answer
The impulse on Ball A is greater than the impulse on Ball B. You may
assume the contact times are equal for the two collisions. The
momentum for Ball A went from some value to zero. The change in
momentum for ball B is twice as large (due to the rebound) as the
change in momentum for ball A so the force on B is larger by a factor
of 2. Also, consider the work done by the forces. The force on ball
A has just stopped the ball. The force on ball B has stopped the
ball and then accelerated it away from the wall.
The
use of “shock absorbers” in car bumpers is a safety and car care
device. In a low speed (e.g. 5 mile/hour or about 10 m/s) accident
the shock absorber creates a longer time of collision which leads to
a smaller contact force and less damage to the bumper.
Think
about the equation Fnet = Δ p/ Δ t
and try to write an explanation of the importance of bending your
knees when you land after jumping.
Describe
the hitting technique that causes the greatest change in momentum for
a tennis ball.
Answer:
Follow through the swing. This will lead to longer contact time and
more impulse.
Conservation
of Momentum
If
the external forces add to zero then the total momentum of the system
will be conserved (ptotal will be a constant).
Consider Fnet = Δ p/ Δ t . If
the net force is zero then Δ p is zero. The momentum has
not changed. We will use the expression pfinal -
pinitial = 0 or, pfinal = pinitial
The
total momentum of an isolated system of bodies remains constant.
Q6.
A halfback who weighs 180 pounds launches himself towards the goal
line with the football at 10 m/sec. A 220 pound linebacker launches
himself towards the halfback at 7 m/sec. They collide such that the
football is just a fraction of an inch away from the goal line.
Will
the ball cross the goal line?
Answer:
Each player has momentum. When they meet in the air, instead of
having their feet pushing on the ground, the external force is zero.
If one or both players push on the ground you cannot use conservation
of momentum to solve this problem. One of the players has positive
momentum and the other negative as one of the velocities will be in
the negative direction. You don’t need to convert units. Mass is
proportional to weight on the Earth. You must use the same units on
both sides of the conservation of momentum equation.
180
lb *10 m/s - 220 lb * 7 m/s = (180 lb + 220 lb) * Vfinal
Vfinal
= + 0.65 m/s . The velocity of the two person lump is positive so
the touchdown is scored.
Questions
for thought:
Suppose
that a car crash on a street occurs and both drivers apply their
brakes during the collision. Is momentum conserved for this
collision? Explain.
Answer:
No. The net external force is not zero due to friction of the road
on the cars.
Why
does a rifle recoil when it is shot?
Answer:
The momentum of the system before the rifle is shot is zero. The
explosion in the chamber that pushes the bullet out the rifle is an
internal force. Assume that the rifle is only lightly held so the
external force is about zero. The total momentum of the bullet and
the rifle must be zero after the shot is fired. The bullet goes in
one direction with positive momentum. The rifle moves in the other
direction with negative momentum.
Elastic
Collisions in One Dimension
There
are three types of collisions that occur between objects:
1)
elastic (also called perfectly elastic) KE of the system is
conserved eg. atomic collisions;
2)
inelastic some KE is lost e.g. kicking a football,
hitting a tennis ball;
3)
completely inelastic much KE is lost e.g. a lineman tackles a
halfback in football; two train cars link together; air track cars
with velcro strips stick together; a ball is caught by a pendulum.
If
the external forces can be ignored, momentum is conserved in all
types of collisions. This fact makes the momentum quantity very
useful in analyzing collisions.
In
a collision between a car and a truck, the force of the car on the
truck is the same value as the force of the truck on the car. The
forces have the same magnitude. This is a Newton’s Third Law
concept.
When
the equations of conservation of momentum and conservation of kinetic
energy are combined we find that
V1new
= V1 * (m1 – m2) / (m1
+ m2) + 2 * V2 * m2 / (m1
+ m2)
V2new
= 2*V1*m1 / (m1 + m2) -
V2 * (m1 – m2) / (m1 +
m2)
You
should deduce the answers to the following. Ask your instructor if
you have questions.
How
do these equations simplify for the case of V2 = 0 (target
at rest) and
a)
m1 = m2 ? b) m1 >>
m2 ? c) m1 << m2 ?
What
will happen if a pair of elastic balls are dropped in line onto a
hard surface? The balls are dropped at the same time and the top
object collides directly with the top of the bottom object as the
bottom object rebounds from the hard surface. Let the mass of the top ball be very small compared to the mass of the bottom ball. If the balls have a speed V when approaching the hard surface, what is the speed of the top ball as it leaves the collision?
Try
this demonstration on your own and share your observations with
someone in your physics class: Lay a marble in the groove of a
plastic ruler (or just on the table).
a)
Shoot a marble with similar mass at the stationary marble. Observe
the collision.
b)
Shoot a massive steel ball at a stationary marble. Observe the
collision.
c)
Shoot a marble at stationary, massive, steel ball. Observe the
collision.
Read
about the operation of “Newton’s Cradle” in your textbook or on
the Internet. This item is often seen in science gift shops at
museums.
Copyright©
2017 by Greg Clements Permission is granted to reproduce this
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