AP Physics 1 – Unit 5
IMPULSE, MOMENTUM, AND COLLISIONS
Learning Objectives:
BIG IDEA 3: The interactions of an object with other objects can be described by forces.
3.D.1.1: I can justify the selection of data needed to determine the relationship between the direction of
the force acting on an object and the change in momentum caused by that force. [SP 4.1]
3.D.2.1: I can justify the selection of routines for the calculation of the relationships between changes in
momentum of an object, average force, impulse, and time of interaction. [SP 2.1]
3.D.2.2: I can predict the change in momentum of an object from the average force exerted on the object
and the interval of time during which the force is exerted. [SP 6.4]
3.D.2.3: I can analyze data to characterize the change in momentum of an object from the average force
exerted on the object and the interval of time during which the force is exerted. [SP 5.1]
3.D.2.4: I can design a plan for collecting data to investigate the relationship between changes in
momentum and the average force exerted on an object over time. [SP 4.2]
Learning Objectives
BIG IDEA 4: Interactions between systems can result in changes in those systems.
4.B.1.1: I can calculate the change in linear momentum of a two-object system with
constant mass in linear motion from a representation of the system (data, graphs,
etc.). [SP 1.4, 2.2]
4.B.1.2: I can analyze data to find the change in linear momentum for a constantmass system using the product of the mass and the change in velocity of the center
of mass. [SP 5.1]
4.B.2.1: I can apply mathematical routines to calculate the change in momentum of
a system by analyzing the average force exerted over a certain time on the
system. [SP 2.2]
4.B.2.2: I can perform analysis on data presented as a force-time graph and predict
the change in momentum of a system. [SP 5.1]
Learning Objectives
BIG IDEA 5: Changes that occur as a result of interactions are constrained by conservation laws.
5.A.2.1: I can define open and closed systems for everyday situations and apply conservation concepts for energy,
charge, and linear momentum to those situations. [SP 6.4, 7.2]
5.D.1.1: I can make qualitative predictions about natural phenomena based on conservation of linear momentum and
restoration of kinetic energy in elastic collisions. [SP 6.4, 7.2]
5.D.1.2: I can apply the principles of conservation of momentum and restoration of kinetic energy to reconcile a
situation that appears to be isolated and elastic, but in which data indicate that linear momentum and kinetic energy
are not the same after the interaction, by refining a scientific question to identify interactions that have not been
considered. Students will be expected to solve qualitatively and/or quantitatively for one-dimensional situations and
only qualitatively in two-dimensional situations. [SP 2.2, 3.2, 5.1, 5.3]
5.D.1.3: I can apply mathematical routines appropriately to problems involving elastic collisions in one dimension and
justify the selection of those mathematical routines based on conservation of momentum and restoration of kinetic
energy. [SP 2.1, 2.2]
5.D.1.4: I can design an experimental test of an application of the principle of the conservation of linear momentum,
predict an outcome of the experiment using the principle, analyze data generated by that experiment whose
uncertainties are expressed numerically, and evaluate the match between the prediction and the outcome. [SP 4.2, 5.1,
5.3, 6.4]
Learning Objectives
5.D.1.5: I can classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear
momentum and restoration of kinetic energy as the appropriate principles for analyzing an elastic collision, solve for
missing variables, and calculate their values. [SP 2.1, 2.2]
5.D.2.1: I can qualitatively predict, in terms of linear momentum and kinetic energy, how the outcome of a collision
between two objects changes depending on whether the collision is elastic or inelastic. [SP 6.4, 7.2]
5.D.2.2: I can plan data collection strategies to test the law of conservation of momentum in a two-object collision that is
elastic or inelastic and analyze the resulting data graphically. [SP 4.1, 4.2, 5.1]
5.D.2.3: I can apply the conservation of linear momentum to a closed system of objects involved in an inelastic collision to
predict the change in kinetic energy. [SP 6.4, 7.2]
5.D.2.4: I can analyze data that verify conservation of momentum in collisions with and without an external friction
force. [SP 4.1, 4.2, 4.4, 5.1, 5.3]
5.D.2.5: I can classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear
momentum as the appropriate solution method for an inelastic collision, recognize that there is a common final velocity for
the colliding objects in the totally inelastic case, solve for missing variables, and calculate their values. [SP 2.1, 2.2]
5.D.3.1: I can predict the velocity of the center of mass of a system when there is no interaction outside of the system but
there is an interaction within the system (i.e., the student simply recognizes that interactions within a system do not affect
the center of mass motion of the system and is able to determine that there is no external force). [SP 6.4]
Impulse = Change in Momentum
Consider Newton’s 2nd Law and the
definition of acceleration
Units of Impulse:
Ns
Units of Momentum: Kg x m/s
Momentum is defined as “Inertia in Motion”
Example:
A 100 g ball is dropped from a height of h = 2.00 m above the floor. It rebounds vertically to a height of
h'= 1.50 m after colliding with the floor. (a) Find the momentum of the ball immediately before it collides
with the floor and immediately after it rebounds, (b) Determine the average force exerted by the floor on
the ball. Assume that the time interval of the collision is 0.01 seconds.
EB  E A
Uo  K
mgho  1 mv 2
2
v  2 gho  2 * 9.8 * 2  6.26 m / s


p  mv
pbefore  0.100(6.26)  0.626 kg * m / s
pafter  0.100(5.4)  0.54 kg * m / s
EB  E A
Ko  U
v  2 gh  2 * 9.8 *1.5  5.4 m / s
Ft  mv  m(v  vo )
F (0.01)  0.100(5.4  (6.26))
F  116.6 N
Impulse is the Area
Since J=Ft, Impulse is the AREA of a Force vs. Time graph.
How about a collision?
Consider 2 objects speeding toward
each other. When they collide......
Due to Newton’s 3rd Law the FORCE
they exert on each other are EQUAL
and OPPOSITE.
The TIMES of impact are also equal.
Therefore, the IMPULSES of the 2
objects colliding are also EQUAL
F1   F2
t1  t 2
( Ft )1  ( Ft ) 2
J1   J 2
How about a collision?
If the Impulses are equal then the
MOMENTUMS are also equal!
J1   J 2
p
before
  p after
m1vo1  m2 vo 2  m1v1  m2 v2
p1   p2
m1v1  m2 v2
m1 (v1  vo1 )  m2 (v2  vo 2 )
m1v1  m1vo1  m2 v2  m2 vo 2
Momentum is Conserved!
The Law of Conservation of Momentum: “In the absence of an external
force (gravity, friction), the total momentum before the collision is equal
to the total momentum after the collision.”
po ( truck)  mvo  (500)(5)  2500kg * m / s
po ( car )  (400)( 2)  800kg * m / s
po ( total)  3300kg * m / s
ptruck  500 * 3  1500kg * m / s
pcar  400 * 4.5  1800kg * m / s
ptotal  3300kg * m / s
Several Types of Collisions
Sometimes objects stick together or blow apart. In this case, momentum is
ALWAYS conserved.
p
before
  p after
m1v01  m2 v02  m1v1  m2 v2
When 2 objects collide and DON’T stick
m1v01  m2 v02  mtotalvtotal
When 2 objects collide and stick together
mtotalvo (total)  m1v1  m2 v2
When 1 object breaks into 2 objects
Elastic Collision = Kinetic Energy is Conserved
Inelastic Collision = Kinetic Energy is NOT Conserved
Example:
A bird perched on an 8.00 cm tall swing has a mass of 52.0 g, and the base of the swing has a
mass of 153 g. Assume that the swing and bird are originally at rest and that the bird takes off
horizontally at 2.00 m/s. If the base can swing freely (without friction) around the pivot, how
high will the base of the swing rise above its original level?
How many objects due to have BEFORE the action?
How many objects do you have AFTER the action?
1
2
EB  E A
K o ( swing )  U swing
pB  p A
mT vo T  m1v1  m2 v2
(0.205)(0)  (0.153)v1( swing )  (0.052)( 2)
vswing 
-0.680 m/s
1 mvo2  mgh
2
vo2
(0.68) 2 0.024 m
h

2g
19.6
Example:
Granny (m=80 kg) whizzes around the rink with a velocity of
6 m/s. She suddenly collides with Ambrose (m=40 kg)
who is at rest directly in her path. Rather than knock him
over, she picks him up and continues in motion without
"braking." Determine the velocity of Granny and
Ambrose.
pb  pa
How many objects do I have before the collision?
2
How many objects do I have after the collision?
1
m1vo1  m2 vo 2  mT vT
(80)(6)  (40)(0)  120vT
vT 
4 m/s
Collisions in 2 Dimensions
The figure to the left shows a
collision between two pucks on
an air hockey table. Puck A has
a mass of 0.025-kg and is
vA
vAsinq
moving along the x-axis with a
velocity of +5.5 m/s. It makes a
collision with puck B, which has
vAcosq
a mass of 0.050-kg and is
initially at rest. The collision is
vBcosq
NOT head on. After the
vBsinq
collision, the two pucks fly
apart with angles shown in the
drawing. Calculate the speeds
of the pucks after the collision.
Collisions in 2 Dimensions
p
ox
  px
m AvoxA  mB voxB  m AvxA  mB vxB
(0.025)(5.5)  0  (.025)(v A cos 65)  (.050)(vB cos 37)
vA
0.1375  0.0106vA  0.040vB
vAsinq
vAcosq
vBcosq
p
oy
  py
0  m Av yA  mB v yB
vBsinq
0  (0.025)(v A sin 65)  (0.050)( vB sin 37)
0.0300vB  0.0227v A
vB  0.757v A
Collisions in 2 Dimensions
0.1375  0.0106vA  0.040vB
vB  0.757vA
0.1375  0.0106v A  (0.050)(0.757v A )
0.1375  0.0106v A  0.03785v A
0.1375  0.04845v A
v A  2.84m / s
vB  0.757(2.84)  2.15m / s