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AP Physics – Mechanics – Chapter 6 – Momentum and Collisions
Text chapter 6 - Reading pp. 141-161 - textbook HW -#1,3,4,6,9,15,16,20,21,23,26,27,25,34,63,70,71
6.1 Momentum and Impulse
A. What is momentum? Newton defined momentum as
“the quantity of motion”
Momentum is a VECTOR quantity.
The instantaneous momentum of an object is defined as
p = mv
Since momentum is a VECTOR quantity, we can break it into
its components:
px = mvx
and
py = mvy
#1) A 5-kg flowerpot falls out of a window 50 m above the ground. Find
the momentum of the flowerpot at the instant it hits the ground. (158 kgm/s)
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B. What is Impulse? An impulse applied to an object will
change the momentum of the object
Before deriving the equation for impulse, think about this one:
What happens when a ball (or a child cruising in his car) hits a wall? Is
the force on the ball constant during impact?(Note: a ball, like a kid’shead, is
compressible and, similar to a spring, the force will change from a value of zero to a maximum, then back to
zero as the head bounces back.
_
_
Thus, F = m a
(in this course, since we only deal with uniform acceleration, the average acceleration will be
equivalent to the uniform, or constant acceleration)
_
_
from F = m a and a = Δv/Δt, the impulse momentumtheorem can be derived.
Note: on the graph above: impulse is the area under the
curve of a force-time graph.
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#2) A 50-g golf ball sits on a tee and is struck by Tiger Wood’s golf club.
The balls leaves the club at 44 m/s.
a) What is the ball’s impulse due to the collision? (2.2 kgm/s)
b) What is the average force on the ball if the club is in contact with the
ball for 1 x 10-3 s? (2200 N)
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#3) A ball of 0.1 kg is dropped from a height 2 m above the ground. It
rebounds to a height of 1.5 m after colliding with the floor.
a) Find the momentum of the ball at the instant before it hits the ground.
b) Find momentum of the ball at the instant after it leaves the ground.
c) Find impulse of the ball during the collision.
d) If average force on the ball during the collision is 200 N, find the time
ball is in contact with ground.
e) Calculate the mechanical energy lost during the collision.
2m
1.5m
[(a) 0.632 kgm/s down; (b) 5.48 kgm/s up; (c) 1.18 kgm/s up; (d) .0059 sec; (e) 0.5 J ]
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#4) An 822-kg Smart Car collides and rebounds off of a wall as shown.
The car is in contact with the wall for .15s.
a) Find the impulse during the collision. (14500 kgm/s)
b) Find average force the wall exerts on the car. (96500 N)
Before Collision
V= 15 m/s
After Collision
V= 2.6 m/s
Question: Two identical twins are dropped from the same height at the
Anne Hutch Playground. One is dropped over and faceplants on the
concrete sidewalk, the other faceplants on the green foam pad. Which
child is more likely to break his face? (Explain in terms of impulse.)
Now go home and try HW -- #1,3,4,6,9
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6.2 Conservation of Momentum
When two objects collide, the force that object 1 exerts
on object 2 is equal in magnitude and opposite in
direction to the force that object 2 exerts on object 1.
F12 = -F21 (Newton’s 3rd law)
Also, the time that object 1 is in contact with object 2 is
is equal to the time object 2 is in contact with object 1.
Δt1 =Δt2
Therefore, the impulse (or change in momentum) given
to each object by the other is THE SAME.
Derivation of the Conservation of Momentum formula
for a two-object collision
F12Δt1= -F21Δt2
(m1v1f – m1v1i) = -(m2v2f – m2v2i) which becomes:
m1v1f – m1v1i = -m2v2f + m2v2i rearranging yields:
m1v1i + m2v2i = m1v1f + m2v2f
This equation works for an isolated system of two objects
where no external forces are applied to the system.
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#5) A 822-kg car traveling 10 m/s collides with a 2740-kg truck at rest.
a) If the car bounces back from the truck at 2 m/s, what is the velocity of
the truck after the collision? (3.6 m/s fwd)
b) How much momentum did the truck gain in collision? (9864 kgm/s)
c) How much momentum did the car lose? (9864 kgm/s)
Before Collision
V= 10 m/s
V= 0 m/s
After Collision
V= ?
V= 2 m/s
Smart Car Crash pics and video of mini car crash tests.
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#6) What is the recoil velocity of a 4-kg m16 rifle that shoots a 3.2-g
bullet at 850 m/s? (0.68 m/s to the right)
v= 850 m/s
v= ?
Since: m1v1i + m2v2i =
0, we use the Recoil shortcut:
m1v1f = - m2v2f
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#7) A 65-kg boy and a 40-kg girl, both wearing ice skates face each other
in the middle of the skating rink. The boy pushes the girl sending her
away from him at 4 m/s. What happens to the boy? (Neglect friction)
(moves at 2.46 m/s in opposite direction)
#8) A boy is standing at rest on a frozen pond. He fires a 50-g bullet at
400 m/s from a gun that has a mass of 5 kg. If the boy has a mass of 60
kg and the coefficient of kinetic friction is 0.01, does the boy reach the
shore? If not, where does he stop? (0.54 to right of x0)
v= 400 m/s
m bullet = 50 g
m boy = 60 kg
m gun = 5 kg
µ k = .01
50m
Now go home and try HW -- #15,16,20,21
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6.3 Types of Collisions (or explosions)
In some collisions, although momentum IS conserved,
kinetic energy MAY NOT BE conserved.
Before Collision
After Collision
In this example use conservation of momentum to find V2f
V1i = 3 m/s
V2i = -3 m/s
V1f = -3 m/s
V2f =
Now find:
KE1i =
KE2i =
Total KEi=
KE1f =
KE2f =
Total KEf=
Is the total energy before the collision equal to the total
energy after the collision? If the answer is yes, then this
collision was perfectly elastic
A.PERFECT ELASTIC COLLISION: both p and KE are conserved
∑Pi = ∑Pf and ∑KEi = ∑KEf
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#8) In a game of pool, if the cue ball hits the 8 ball ELASTICALLY, find
the speed of each ball after the collision. (ball1=0 m/s, ball2= 2m/s right)
(see pg 152-153 text for convenient derivation)
Before Collision
Both of the following equations have to be satisfied:
m1v1i + m2v2i = m1v1f + m2v2f
½m1v1i2 + ½m2v2i2 = ½m1v1f2 + ½m2v2f2
This is always the same in a head-on elastic collision between two objects of the
same mass: they EXCHANGE VELOCITIES. DEMO: Newton’s balls/ pool balls on rod
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Try the same problem, using the shortcut equation from the text. Remember that the shortcut equation is
only for elastic head on collisions only. Do not assume a collision is elastic and use it anyway.
#8) In a game of pool, if the cue ball hits the 8 ball ELASTICALLY, find the
speed of each ball after the collision.
Before Collision
(note: standard cue balls are .17 kg (6 oz) and numbered balls are .16 kg (5.5 oz)
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#10) (now m1 ≠ m2) If the cue ball hits the 8 ball ELASTICALLY, find the
speed of each ball after the collision. (ball1= .667 m/s left, ball2= 1.33m/s right)
Before Collision
Note: Back in the 70’s there was a 1000-fold increase in 8 ball thefts,
thanks to the Magic 8 Ball craze. As a result many pool halls replaced
their 8 balls with steel 8 balls with the same volume and double the
mass.
DEMO: BIG BALL LITTLE BALL – Then do next problem
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#12) Two balls are held touching each other, one of mass m
directly above the one of mass M = 5m. The lower ball is
dropped from a height h. They are dropped together.
Assume all collisions are elastic.
a) How fast are the balls moving when the
lower strikes the ground?
b) At what speed does the lower ball rebound off the ground? (ignore
the upper ball)
c) What speed does the upper ball now rebound off the lower ball?
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#11)
V1i = Vo
V2i = 0 m/s
(V1f = ½ Vo )
2m
3m
A car of mass 2m has a speed of Vo as it strikes another car of mass 3m
initially at rest. If after the collision, the first car slows to half of its speed,
a) Find V2f in terms of V0. (V2f =V0/3)
b) Is this collision elastic? Show why/why not. (No.)
c) If the collision is not elastic, how much energy was lost in the collision?
(7/12 of the KE was lost)
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B. INELASTIC COLLISION ∑Pi = ∑Pf and ∑KEi ≠ ∑KEf
Special Case: Perfect Inelastic Collision
A collision in which the objects couple during the collision
and go off together with a common speed.
m1v1i + m2v2i = (m1+ m2)vf
∑KEi = ∑KEf + Q or
∑KEi - ∑KEf = Q
#13) A Big Wheel® of mass 2m has a speed of Vo as it strikes a lemonade
V1i = Vo
V2i = 0 m/s
stand on wheels of mass 3m initially
at rest. If after the collision, the two
vehicles couple (stick together) :
a) Find the common speed Vcom that
the two vehicles go off with, after the
2m
3m
collision in terms of V0. (Vcom =2V0/5)
b) Is this collision elastic/inelastic?
Show why/why not. (Inelastic)
c) If the collision is not elastic, how much
energy was lost in the collision? (6/10 KE is lost)
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#14) The Ballistic Pendulum
m bullet = .05 kg
Vi = 850 m/s
5 kg
A 0.50 Caliber bullet from an M2 machine gun has mass of approximately
50 g. It is fired into a 5-kg ballistic block at 850 m/s and gets stuck in the
5-kg block that is initially at rest and suspended by a 10.5 m rope.
a) Find height that system rises to. (3.54 m)
b) Find Energy lost in collision. (17883 J)
In a ballistic Pendulum, KE is not conserved, but the
mechanical energy of the bullet-pendulum system after the
collision is conserved.
m1v1i + m2v2i = (m1+ m2)vf then, using vf and
KElost = PEgain
½(m1+ m2)vf2 = (m1+ m2)gh
Demo: Happy Sad Ball Pendulum. Gotta find it.
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#15) A steel ball of mass m is fastened to a light cord of length L and
released when the cord is horizontal. At the bottom of the path, the ball
ELASTICALLY strikes a hard plastic block of mass 4m, initially at rest on a
frictionless horizontal surface.
L
m
4m
M
a) Find the speed of block immediately after the collision.
b) To what height h will the ball rebound after the collision?
Now go home and try HW -- #23, 26, 27
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C. Glancing Collisions
Since conservation of momentum says that ∑Pi = ∑Pf and
since momentum is a vector, we can break the momentum
into its components.
Hence ∑Pix = ∑Pfx and ∑Piy = ∑Pfy
#16) A 1976 Ford Pinto equipped with optional exploding gas tank
collides with a 1970 Dodge Dart in a perfect inelastic collision.
a) Find the magnitude and direction (from east) of the velocity after
collision. (15.5 m/s, 49.3° NE)
b) Find energy lost in collision. (339000 J)
m1 = 1100 kg
V1 = 25 m/s
m2 = 1600 kg
V2 = 20 m/s
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#17) A cue ball hits the 8 ball in billiards as shown. The mass of each
ball is 160 g.
Before Collision
After Collision
a) Find angle and speed of the 8 ball after
the collision. (13.26 m/s, 35.4° NE)
b) Determine the type of collision this is. (Q=6.416 J - Inelastic)
Now go home and try HW -- #25, 34, 63, 70, 71
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#18) Find velocity of the 2000-kg car after the collision.
Before Collision
After Collision
V1f = 15 m/s
V1i = 20 m/s
θ = 30°
V2i = 30 m/s
(32.78 m/s, 15.3° NE)
Chapter 6: Done.
θ = 45°
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Calculus definitions:
1. The time rate of change of linear momentum of a particle is equal to
the net force acting on the particle.




dv dp
F  ma  m

dt dt
(Net force is the slope of the line tangent to a momentum-time graph)
2. The integral of force with respect to time is a change in momentum

 dp

becomes:
Fdt  dp
F
dt
Integration yields:  Fdt   dp
or:
 tf 

I   Fdt  p
ti
(impulse is the area under the curve of a force-time graph)