Transcript Linear Momentum - White Plains Public Schools

Linear Momentum
Unit 5
Lesson 1 : Linear Momentum and Its Conservation
F21 + F12 = 0
(Newton’s Third Law)
m1a1 + m2a2 = 0
(Newton’s Second Law)
m1
dv1
dv2
+ m2
=0
dt
dt
d(m1v1)
d(m2v2)
+
=0
dt
dt
d
dt
(m1v1 + m2v2) = 0
sum of linear momentum
is constant
p = mv
(linear momentum)
Linear momentum is a vector quantity
whose direction the same as the
direction of v. Its SI unit is kg . m/s.
dv
SF = ma = m
dt
d(mv)
dp
=
SF =
dt
dt
The time rate of change of the linear
momentum of a particle is equal to the
net force acting on the particle.
(This is the form in which Newton
presented his second law.)
Conservation of Linear Momentum
Whenever two or more particles in an isolated
system interact, the total momentum of the
system remains constant.
The total momentum of an isolated system at
all times equals its initial momentum.
Total p before = Total p after
Example 1
A 60. kg archer stands
at rest on frictionless
ice and fires a 0.50 kg
arrow horizontally at
50. m/s.
a) With what velocity does the archer move
across the ice after firing the arrow.
b) What if the arrow were shot in a direction
that makes an angle q with the
horizontal ? How will this change the
recoil velocity of the archer ?
Example 2
Two blocks of masses M
and 3M are placed on a
horizontal, frictionless
surface. A light spring is
attached to one of them,
and the blocks are
pushed together with the
spring between them. A
cord initially holding the
blocks together is
burned, and the 3M block
moves to the right with a
speed of 2.00 m/s.
a) What is the speed of the block of mass M ?
b) Find the original elastic potential energy in
the spring if M = 0.350 kg.
Example 3 : AP 1981 #2
A swing seat of mass M is connected to a fixed point P by a
massless cord of length L. A child also of mass M sits on the
seat and begins to swing with zero velocity at a position at
which the cord makes a 60o angle with the vertical as shown
in Figure I. The swing continues down until the cord is
exactly vertical at which time the child jumps off in a
horizontal direction.
The swing continues in the same direction until the cord
makes a 45o angle with the vertical as shown in Figure II; at
that point it begins to swing in the reverse direction. With
what velocity relative to the ground did the child leave the
swing ? (cos 45o = sin 45o = 2 , sin 30o = cos 60o = ½,
cos
30o
= sin
60o
=
3
2
)
2
Lesson 2 : Impulse and Momentum
d(mv)
dp
=
SF =
dt
dt
dp = F dt
Integrating F with respect to t,
tf
Dp = pf – pi =

F dt
ti
tf
I=
F dt
ti
Impulse (I)
Impulse is a vector quantity with the same direction
as the direction of the change in momentum.
Impulse has a magnitude equal to
the area under the force-time
graph.
Because the force imparting an impulse can
generally vary in time, we can express impulse as
I = FDt
Impulse – Momentum Theorem
tf
Dp = pf – pi =

F dt
ti
The impulse of the force F acting on a
particle equals the change in the
momentum of the particle.
I = Dp
Example 1
In a particular crash
test, a car of mass
1500 kg collides with
a wall, as shown. The
initial and final
velocities of the car
^
are vi = -15.0i m/s
^
and vf = 2.60i m/s,
respectively.
a) If the collision lasts for 0.150 s, find the impulse
caused by the collision and the average force
exerted on the car.
b) What if the car did not rebound from the wall ?
Suppose the final velocity of the car is zero
and the time interval of the collision remains
0.150 s. Would this represent a larger or a
smaller force by the wall on the car ?
Example 2
A 3.00 kg steel ball strikes a wall with a speed
of 10.0 m/s at an angle of 60.0o with the
surface. It bounces off with the same speed
and angle. If the ball is in contact with the wall
for 0.200 s, what is the average force exerted
by the wall on the ball ?
Example 3
An estimated force-time curve for a
baseball struck by a bat is shown above.
From this curve, determine
a) the impulse delivered to the ball
b) the average force exerted on the ball
c) the peak force exerted on the ball
Lesson 3 : Collisions in One-Dimension
Elastic Collisions
Inelastic Collisions
The total KE (as well as
total momentum) of the
system is the same before
and after the collision.
The total KE of the system
is not the same before and
after the collision (even
though the momentum of
the system is conserved).
KE is conserved
KE is not conserved
“perfectly inelastic”
“inelastic”
Perfectly Inelastic Collisions
when the colliding objects stick together
m1v1i + m2v2i = (m1 + m2)vf
Elastic Collisions
m1v1i + m2v2i = m1v1f + m2v2f
½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2
Example 1
An 1800 kg car stopped at a traffic light is struck
from the rear by a 900 kg car, and the two become
entangled, moving along the same path as that of
the originally moving car.
a) If the smaller car were moving at 20.0 m/s before
the collision, what is the velocity of the
entangled cars after the collision ?
b) Suppose we reverse the masses of the cars – a
stationary 900 kg car is struck by a moving
1800 kg car. Is the final speed the same as
before ?
Example 2
The ballistic pendulum is an apparatus used to measure
the speed of a fast-moving projectile, such as a bullet. A
bullet of mass m1 is fired into a large block of wood of
mass m2 suspended from some light wires. The bullet
embeds in the block, and the entire system swings
through a height h. How can we determine the speed of
the bullet from a measurement of h ?
Example 3
A block of mass m1 = 1.60 kg initially moving to the
right with a speed of 4.00 m/s on a frictionless
horizontal track collides with a spring attached to
a second block of mass m2 = 2.10 kg initially
moving to the left with a speed of 2.50 m/s. The
spring constant is 600 N/m.
a) Find the velocities of the two blocks after the
collision.
b) During the collision, at the instant block 1 is
moving to the right with a velocity of +3.00 m/s,
determine the velocity of block 2.
Example 4 : AP 1995 #1
A 5 kg ball initially at rest at the edge of a 2 m
long, 1.2 m high frictionless table, as shown
above. A hard plastic cube of mass 0.5 kg slides
across the table at a speed of 26 m/s and strikes
the ball, causing the ball to leave the table in the
direction in which the cube was moving.
The figure below shows a graph of the force exerted
on the ball by the cube as a function of time.
a) Determine the total impulse given to the ball.
b) Determine the horizontal velocity of the ball
immediately after the collision.
c) Determine the following for the cube
immediately after the collision.
i. Its speed
ii. Its direction of travel (right or left), if moving
d) Determine the kinetic energy dissipated in the
collision.
e) Determine the distance between the two
points of impact of the objects with the
floor.
Example 5 : AP 1994 #1
A 2 kg block and an 8 kg block are both attached to an
ideal spring (for which k = 200 N/m) and both are initially
at rest on a horizontal frictionless surface, as shown in
the diagram above.
In an initial experiment, a 100 g (0.1 kg) ball of clay is
thrown at the 2 kg block. The clay is moving horizontally
with speed v when it hits and sticks to the block. The 8 kg
block is held still by a removable stop. As a result, the
spring compresses a maximum distance of 0.4 m.
a) Calculate the energy stored in the spring at maximum
compression.
b) Calculate the speed of the clay ball and 2 kg block
immediately after the clay sticks to the block but
before the spring compresses significantly.
c) Calculate the initial speed v of the clay.
In a second experiment, an identical ball of clay is thrown
at another identical 2 kg block, but this time the stop is
removed so that the 8 kg block is free to move.
d) State whether the maximum compression of the spring
will be greater than, equal to, or less than 0.4 m.
Explain briefly.
e) State the principle or principles that can be used to
calculate the velocity of the 8 kg block at the instant
that the spring regains its original length. Write the
appropriate equation(s) and show the numerical
substitutions, but do not solve for the velocity.
Lesson 4 : Two-Dimensional Collisions
m1v1i = m1v1f cosq + m2v2f cosf
0 = m1v1f sinq - m2v2f sinf
Since this is an elastic collision,
KE is also conserved.*
½ m1v1i2 = ½ m1v1f2 + ½ m2v2f2
* If this were an inelastic collision,
KE would not be conserved, and
this equation does not apply.
Example 1
A 1500 kg car traveling east
with a speed of 25.0 m/s
collides at an intersection
with a 2500 kg van traveling
north at a speed of 20.0 m/s.
Find the direction and
magnitude of the velocity of
the wreckage after the
collision, assuming that the
vehicles undergo a perfectly
inelastic collision (that is,
they stick together).
Example 2
In a game of billiards, a player wishes to sink a target
ball in the corner pocket. If the angle to the corner
pocket is 35o, at what angle q is the cue ball deflected ?
Assume that friction and rotational motion are
unimportant and that the collision is elastic. Also
assume that all billiard balls have the same mass m.
Lesson 5 : The Center of Mass
Center of mass (CM) is the average
position of the system’s mass
Center of mass is located on the
line joining the two particles and
is closer to the particle having
the larger mass.
How to Locate the Center of Mass
System rotates clockwise
when F is applied between the
less massive particle and CM.
System rotates
counterclockwise when F is
applied between the more
massive particle and CM.
System moves in the direction
of F without rotating when F is
applied at CM.
xCM =
m1x1 + m2x2
m1 + m2
Center of mass for a system of many particles
xCM =
m1x1 + m2x2 + m3x3 + ……
m1 + m2 + m3 + ……
xCM =
Smixi
M
Y coordinates of CM
yCM =
Smiyi
M
Z coordinates of CM
zCM =
Smizi
M
Using a Position Vector (r) to locate CM of a
system of particles
^
^
^
Using a Position Vector (r) to locate CM of an
extended object
xCM = lim
Dmi  0
SxiDmi
M
xCM
1
=
M
x dm
(likewise for yCM and zCM)
rCM
1
=
M
r dm
Locating CM of an irregularly shaped object
Suspend the object first
from points A and C.
CM is where lines AB and
CD intersect
If wrench is hung freely from
any point, the vertical line
through this point must pass
through CM.
Example 1
A baseball bat is cut at the location of its center
of mass. The piece with smaller mass is
a) the piece on the right.
b) the piece on the left.
c) both have same mass.
d) impossible to determine.
Example 2
A system of three particles located as shown
above. Find the center of mass of the system.
Example 3
a) Show that the center of mass of a rod of
mass M and length L lies midway
between its ends, assuming the rod
has a uniform mass per unit length.
b) Suppose a rod is nonuniform such that
its mass per unit length varies
linearly with x according to the
expression l = ax, where a is a
constant. Find the x coordinate of
the center of mass as a fraction of L.
Lesson 6 : Motion of a System of Particles
Velocity of CM
vCM =
drCM
dt
Acceleration of CM
aCM =
dvCM
dt
Total momentum of a system of particles
MvCM = Smivi = Spi = ptot
(i is the ith particle)
total mass of
system
Newton’s Second Law for a system of particles
SFext = MaCM
CM of a system of particles of combined mass
M moves like an equivalent particle of mass M
would move under the influence of the net
external force on the system.
Example 1
Suppose you tranquilize a polar bear on a
smooth glacier as part of a research
effort. How might you estimate the bear’s
mass using a measuring tape, a rope, and
knowledge of your own mass ?
Example 2
Consider a system of two particles in the xy plane :
^
^
m1 = 2.00 kg is at r1 = (1.00 i + 2.00 j) m and
^
^
has a velocity of (3.00 i + 0.500 j) m/s
^
^
m2 = 3.00 kg is at r2 = (-4.00 i - 3.00 j) m and
^
^
has a velocity of (3.00 i – 2.00 j) m/s
a) Plot these particles on a grid or graph paper.
Draw their position vectors and show their
velocities.
b) Find the CM of the system and mark it on the
grid.
c) Determine the velocity of the CM and also
show it on the diagram.
d) What is the total linear momentum of the
system ?
Example 3
A rocket is fired vertically upward. At the
instant it reaches an altitude of 1000 m and
a speed of 300 m/s, it explodes into three
fragments having equal mass. One
fragment continues to move upward with a
speed of 450 m/s following the explosion.
The second fragment has a speed of 240
m/s and is moving east right after the
explosion. What is the velocity of the third
fragment right after the explosion ?