Download Lesson 05

Conservation of Linear
Momentum
Book Sections 9.1 – 9.3
Momentum and
Force
Momentum and Its Relation to Force
Momentum is a vector symbolized by the symbol p ,
and is defined as
The rate of change of momentum is equal to the net
force:
This can be shown using Newton’s second law.
ConcepTest 9.3a Momentum and Force
A net force of 200 N acts on a 100-kg
boulder, and a force of the same
magnitude acts on a 130-g pebble.
How does the rate of change of the
boulder’s momentum compare to
the rate of change of the pebble’s
momentum?
1) greater than
2) less than
3) equal to
ConcepTest 9.3a Momentum and Force
A net force of 200 N acts on a 100-kg
boulder, and a force of the same
magnitude acts on a 130-g pebble.
How does the rate of change of the
boulder’s momentum compare to
the rate of change of the pebble’s
momentum?
1) greater than
2) less than
3) equal to
The rate of change of momentum is, in fact, the force.
Remember that F = Dp/Dt. Because the force exerted
on the boulder and the pebble is the same, then the
rate of change of momentum is the same.
ConcepTest 9.3b Velocity and Force
A net force of 200 N acts on a 100-kg
boulder, and a force of the same
magnitude acts on a 130-g pebble.
How does the rate of change of the
boulder’s velocity compare to the
rate of change of the pebble’s
velocity?
1) greater than
2) less than
3) equal to
ConcepTest 9.3b Velocity and Force
A net force of 200 N acts on a 100-kg
boulder, and a force of the same
magnitude acts on a 130-g pebble.
How does the rate of change of the
boulder’s velocity compare to the
rate of change of the pebble’s
velocity?
1) greater than
2) less than
3) equal to
The rate of change of velocity is the acceleration.
Remember that a = Dv/Dt. The acceleration is related
to the force by Newton’s 2 Second Law (F = ma), so
the acceleration of the boulder is less than that of
the pebble (for the same applied force) because the
boulder is much more massive.
Conservation of
Momentum
Momentum in Collisions
During a collision, measurements show that the
total momentum does not change:
Momentum and Newton’s Laws
Conservation of
momentum can also be
derived from Newton’s
laws. A collision takes a
short enough time that
we can ignore external
forces. Since the internal
forces are equal and
opposite, the total
momentum is constant.
Momentum and Newton’s Laws
For more than two objects,
Or, since the internal forces cancel,
Conservation of Momentum
This is the law of conservation of linear
momentum:
when the net external force on a system of
objects is zero, the total momentum of the
system remains constant.
Equivalently,
the total momentum of an isolated system
remains constant.
Conservation of Momentum
Momentum conservation works for a rocket as long as
we consider the rocket and its fuel to be one system,
and account for the mass loss of the rocket.
ConcepTest 9.1 Rolling in the Rain
An open cart rolls along a
frictionless track while it is
raining. As it rolls, what
happens to the speed of the
cart as the rain collects in it?
(Assume that the rain falls
vertically into the box.)
1) speeds up
2) maintains constant speed
3) slows down
4) stops immediately
ConcepTest 9.1 Rolling in the Rain
An open cart rolls along a
frictionless track while it is
raining. As it rolls, what
happens to the speed of the
cart as the rain collects in it?
(Assume that the rain falls
vertically into the box.)
1) speeds up
2) maintains constant speed
3) slows down
4) stops immediately
Because the rain falls in vertically, it
adds no momentum to the box, thus
the box’s momentum is conserved.
However, because the mass of the
box slowly increases with the added
rain, its velocity has to decrease.
Follow-up: What happens to the cart when it stops raining?
ConcepTest 9.4 Collision Course
A small car and a large truck
collide head-on and stick
together. Which one has the
larger momentum change?
1) the car
2) the truck
3) they both have the same
momentum change
4) can’t tell without knowing the
final velocities
ConcepTest 9.4 Collision Course
A small car and a large truck
collide head-on and stick
together. Which one has the
larger momentum change?
1) the car
2) the truck
3) they both have the same
momentum change
4) can’t tell without knowing the
final velocities
Because the total momentum of the is
conserved, that means that Dp = 0 for
the car and truck combined.
Therefore, Dpcar must be equal and
opposite to that of the truck (–Dptruck) in
order for the total momentum change
to be zero. Note that this conclusion
also follows from Newton’s Third Law.
Follow-up: Which one feels
the larger acceleration?
Impulse
Collisions and Impulse
During a collision, objects are
deformed due to the large
forces involved.
Since
write
Integrating,
, we can
Collisions and Impulse
This quantity is defined as the impulse, J:
The impulse is equal to the change in
momentum:
Collisions and Impulse
Since the time of the collision is often very short, we may
be able to use the average force, which would produce
the same impulse over the same time interval.
ConcepTest 9.5a Two Boxes I
Two boxes, one heavier than the
other, are initially at rest on a
horizontal frictionless surface. The
same constant force F acts on each
one for exactly 1 second. Which
box has more momentum after the
force acts ?
F
1) the heavier one
2) the lighter one
3) both the same
light
F
heavy
ConcepTest 9.5a Two Boxes I
Two boxes, one heavier than the
other, are initially at rest on a
horizontal frictionless surface. The
same constant force F acts on each
one for exactly 1 second. Which
box has more momentum after the
force acts ?
We know:
Dp ,
Fav =
Dt
so impulse Dp = Fav Dt.
In this case F and Dt are the
same for both boxes!
Both boxes will have the
same final momentum.
F
1) the heavier one
2) the lighter one
3) both the same
light
F
heavy
ConcepTest 9.5b Two Boxes II
In the previous question,
1) the heavier one
which box has the larger
2) the lighter one
velocity after the force acts?
3) both the same
ConcepTest 9.5b Two Boxes II
In the previous question,
1) the heavier one
which box has the larger
2) the lighter one
velocity after the force acts?
3) both the same
The force is related to the acceleration by Newton’s
Second Law (F = ma). The lighter box therefore has
the greater acceleration and will reach a higher
speed after the 1-second time interval.
Follow-up: Which box has gone a larger distance after the force acts?
ConcepTest 9.6 Watch Out!
You drive around a curve in a narrow
one-way street at 30 mph when you see
an identical car heading straight toward
you at 30 mph. You have two options: hit
the car head-on or swerve into a massive
concrete wall (also head-on). What
should you do?
1) hit the other car
2) hit the wall
3) makes no difference
4) call your physics prof!!
5) get insurance!
ConcepTest 9.6 Watch Out!
You drive around a curve in a narrow
one-way street at 30 mph when you see
an identical car heading straight toward
you at 30 mph. You have two options: hit
the car head-on or swerve into a massive
concrete wall (also head-on). What
should you do?
1) hit the other car
2) hit the wall
3) makes no difference
4) call your physics prof!!
5) get insurance!
In both cases your momentum will decrease to zero in the collision.
Given that the time Dt of the collision is the same, then the force
exerted on YOU will be the same!!
If a truck is approaching at 30 mph, then you’d be better off hitting
the wall in that case. On the other hand, if it’s only a mosquito, well,
you’d be better off running him down...
ConcepTest 9.7 Impulse
A small beanbag and a bouncy
rubber ball are dropped from the
same height above the floor.
They both have the same mass.
Which one will impart the greater
impulse to the floor when it hits?
1) the beanbag
2) the rubber ball
3) both the same
ConcepTest 9.7 Impulse
A small beanbag and a bouncy
rubber ball are dropped from the
same height above the floor.
They both have the same mass.
Which one will impart the greater
1) the beanbag
2) the rubber ball
3) both the same
impulse to the floor when it hits?
Both objects reach the same speed at the floor. However, while
the beanbag comes to rest on the floor, the ball bounces back
up with nearly the same speed as it hit. Thus, the change in
momentum for the ball is greater, because of the rebound.
The impulse delivered by the ball is twice that of the beanbag.
For the beanbag:
For the rubber ball:
Dp = pf – pi = 0 – (–mv ) = mv
Dp = pf – pi = mv – (–mv ) = 2mv
ConcepTest 9.9a Going Bowling I
A bowling ball and a Ping-Pong ball
are rolling toward you with the
same momentum. If you exert the
same force to stop each one, which
takes a longer time to bring to rest?
1) the bowling ball
2) same time for both
3) the Ping-Pong ball
4) impossible to say
p
p
ConcepTest 9.9a Going Bowling I
A bowling ball and a Ping-Pong ball
are rolling toward you with the
same momentum. If you exert the
same force to stop each one, which
takes a longer time to bring to rest?
We know:
Dp
Fav =
Dt
1) the bowling ball
2) same time for both
3) the Ping-Pong ball
4) impossible to say
so Dp = Fav Dt
p
Here, F and Dp are the same for both balls!
It will take the same amount of time
to stop them.
p
ConcepTest 9.11 Golf Anyone?
You tee up a golf ball and drive it
down the fairway. Assume that the
collision of the golf club and ball is
elastic. When the ball leaves the
tee, how does its speed compare to
the speed of the golf club?
1) greater than
2) less than
3) equal to
ConcepTest 9.11 Golf Anyone?
You tee up a golf ball and drive it
down the fairway. Assume that the
collision of the golf club and ball is
elastic. When the ball leaves the
tee, how does its speed compare to
the speed of the golf club?
1) greater than
2) less than
3) equal to
If the speed of approach (for the golf club and ball) is v,
then the speed of recession must also be v. Because the
golf club is hardly affected by the collision and it continues
with speed v, then the ball must fly off with a speed of 2v.
ConcepTest 9.10b Elastic Collisions II
Carefully place a small rubber ball (mass
m) on top of a much bigger basketball
(mass M) and drop these from some
height h. What is the velocity of the
smaller ball after the basketball hits the
ground, reverses direction, and then
collides with the small rubber ball?
1) zero
2) v
3) 2v
4) 3v
5) 4v
ConcepTest 9.10b Elastic Collisions II
Carefully place a small rubber ball (mass
m) on top of a much bigger basketball
(mass M) and drop these from some
height h. What is the velocity of the
smaller ball after the basketball hits the
ground, reverses direction, and then
collides with the small rubber ball?
• Remember that relative
velocity has to be equal
before and after collision!
Before the collision, the
basketball bounces up
with v and the rubber ball
is coming down with v,
so their relative velocity is
–2v. After the collision, it
therefore has to be +2v!!
m
v
1) zero
2) v
3) 2v
4) 3v
5) 4v
3v
v
v
v
M
(a)
v
(b)
(c)
ConcepTest 9.12a Inelastic Collisions I
A box slides with initial velocity 10 m/s on
a frictionless surface and collides
inelastically with an identical box. The
boxes stick together after the collision.
What is the final velocity?
1) 10 m/s
2) 20 m/s
3) 0 m/s
4) 15 m/s
5) 5 m/s
vi
M
M
vf
M
M
ConcepTest 9.12a Inelastic Collisions I
A box slides with initial velocity 10 m/s on
a frictionless surface and collides
inelastically with an identical box. The
boxes stick together after the collision.
What is the final velocity?
1) 10 m/s
2) 20 m/s
3) 0 m/s
4) 15 m/s
5) 5 m/s
The initial momentum is:
M vi = (10) M
vi
M
M
The final momentum must be the same!!
The final momentum is:
Mtot vf = (2M) vf = (2M) (5)
vf
M
M
ConcepTest 9.14a Recoil Speed I
Amy (150 lbs) and Gwen (50 lbs) are
standing on slippery ice and push
off each other. If Amy slides at 6
m/s, what speed does Gwen have?
1) 2 m/s
2) 6 m/s
3) 9 m/s
4) 12 m/s
5) 18 m/s
150 lbs
50 lbs
ConcepTest 9.14a Recoil Speed I
Amy (150 lbs) and Gwen (50 lbs) are
standing on slippery ice and push
off each other. If Amy slides at 6
m/s, what speed does Gwen have?
1) 2 m/s
2) 6 m/s
3) 9 m/s
4) 12 m/s
5) 18 m/s
The initial momentum is zero,
so the momenta of Amy and
Gwen must be equal and
opposite. Because p = mv,
then if Amy has three times
more mass, we see that
Gwen must have three times
more speed.
150 lbs
50 lbs
Problems and
Case Studies
Question 4.24 – Bear Wants my Lunch
Case Studies 01
Lessons 01 - 05
Example A – Particle in a Plane
A particle of mass m moves along the xy-plane
with a velocity function given by
tˆ
t ˆ
v (t ) = ae i  be j
for 0  t  1
where a and b are constants. Set any constants of
integration to zero, and use (x0, y0) as the
particle’s initial position (at t = 0 s).
Example A – Particle in a Plane
t ˆ
ˆ
v (t ) = ae i  be j
for 0  t  1
t
(a) Find the position function r(t)
We can apply the definition of velocity:
dr
v(t ) =
dt
r (t ) =  v(t )dt
= [aet ˆi  bet ˆj]dt

=  ae dtˆi   be
t
t ˆ
ˆ
r (t ) = ae i  be j
t
t
dt ˆj
Example A – Particle in a Plane
t ˆ
ˆ
v (t ) = ae i  be j
t
for 0  t  1
(b) Find the particle’s initial position in terms of
the other constants.
The particle was at its initial position at t = 0, so we
need the value of the position function at that time;
that will give us the initial position.
r (t ) = ae ˆi  be t ˆj
t
0ˆ
ˆ
r (0) = ae i  be j
= aˆi  bˆj
So the particle’s initial position is:
( x0 , y0 ) = (a, b)
0
Example A – Particle in a Plane
t ˆ
ˆ
v (t ) = ae i  be j
t
for 0  t  1
(c) Find the acceleration function a(t)
We can apply the definition of acceleration:
dv d
a(t ) =
= (aet ˆi  bet ˆj)
dt dt
a(t ) = aet ˆi  be t ˆj
Example A – Particle in a Plane
t ˆ
ˆ
r (t ) = ae i  be j
t
(d) Use r(t) to find y(x).
From r(t):
x(t ) = ae
t
t
y(t ) = be
We can solve for t in terms of x, and then substitute
this into y(t).
x
e =
a
x
ln e = ln
a
y = b exp(  ln x )
a
= b exp(ln a )
x
t
t
x
t = ln
a
ab
y ( x) =
x
Example A – Particle in a Plane
t ˆ
ˆ
r (t ) = ae i  be j
t
(e) Plot the trajectory of the particle in the xyplane from t = 0 to t = 1. Assume a = 2, b = 1.
From part (b) we know the particle’s initial position,
now we just need its final position, at t = 1, to plot its
y(x) trajectory. Using r(t), we get:
 1ˆ
ˆ
r (1) = 2e i  e j
1
( x f , y f )  (5.4,0.37 )
Example A – Particle in a Plane
2
y ( x) =
x
( x0 , y0 ) = (2,1)
( x f , y f )  (5.4,0.37 )
Example B – Hanging Weight
A block of mass m is hanging from two walls from
two taut strings. The strings make angles q, and
2q with respect the horizontal (as shown on the
figure below). What is the tension on each of the
strings?
Example B – Hanging Weight
We can start solving this problem by drawing a nice
free-body diagram depicting all the forces involved.
y
T1
T2
2q
q
w = mg ( ˆj) =  mgˆj
Example C – Clay vs Asteroid
A small asteroid of mass 5m is floating motionlessly.
Suddenly, a ball of clay of mass m hits the asteroid with a
speed of v0. The two objects collide head-on, and remain
stuck together after the collision. Assume no external
forces act on this system.
vf
v0
m
5m
6m
x
Example C – Clay vs Asteroid
a) What is the final velocity of the clay-asteroid ball in terms of v0?
b) If the actual collision between the two objects lasted T, what
was the average force on the clay? What about on the asteroid?
c) During the collision, what was the average acceleration of the
clay? How about the asteroid? (Ignore the deformation effects
of the clay).


 2 
t ,
d) Given that the impulsive force is given by | F |= A1  cos
 T 

find the coefficient A for the asteroid.
e) Use the form of the impulse force given in part d) to find the
magnitude of the maximum force experienced by objects
during the collision.
f) Using T = 0.13 s, m = 1.0 kg, v0 = 15 m/s, what is the numerical
value of the force on part e)?
Example C – Clay vs Asteroid
Example C – Clay vs Asteroid
Example C – Clay vs Asteroid
Example C – Clay vs Asteroid