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Physics
Chapter 6:
Momentum and Collisions
Momentum and Collisions
Force is Not Always Constant
Application of Force May Vary with Time
Momentum and Collisions
Impulse
The Product of the Average Force and the
Time Interval During Which the Force Acts
A Change in Momentum
Impulse  Ft
Momentum and Collisions
Impulse
Units
Force = Newton
Time = Second
Impulse = Newton * Second (N*s)
Impulse  Ft
Momentum and Collisions
Linear Momentum (p)
The Reaction of an Object to a Force Exerted
Upon It
The Product of an Object’s Mass and Velocity
p  mv
Momentum and Collisions
Linear Momentum (p)
Vector Quantity
Direction is Always the Same as the Direction
of the Velocity Upon the Object
p  mv
Momentum and Collisions
Linear Momentum (p)
Units
Mass = kilogram
Velocity = meters / second
Linear Momentum = kilogram*meter/second
(kg*m/s)
p  mv
Momentum and Collisions
Impulse – Momentum Theorem
a
v f  vi
t
 v f  vi  mv f  mvi
 
F  ma  m
t
 t 
Momentum and Collisions
Impulse – Momentum Theorem
 v f  vi  mv f  mvi
 
F  ma  m
t
 t 
Ft  mv f  mvi
Momentum and Collisions
Impulse – Momentum Theorem
Impulse = Final Momentum – Initial
Momentum
Impulse = Change in Momentum
Ft  mv f  mvi
Momentum and Collisions
Impulse – Momentum Theorem
Impulse Time is Usually Very Short
Experimental Measurement of Impulse Force is
Usually Difficult
I – M Theorem Allows Calculation of Impulse
and Subsequently the Force Applied
Ft  mv f  mvi
Momentum and Collisions
Impulse – Momentum Theorem
The Impulse Produced by a Net Force is Equal
to the Change in the Object’s Momentum
Ft  mv f  mvi
Momentum and Collisions
Problem
– Hilary strikes a 0.058kg golf ball with a force of
272N and gives it a velocity of 62.0m/s. How long
was Hilary’s club in contact with the ball?
Momentum and Collisions
Solution
m = 0.058kg
F = 272N
vi = 0m/s
vf = 62m/s
Ft  mv
Ft  mv f  mvi  m(v f  vi )
t 
m(v f  vi )
F
(0.058kg)(62m / s  0m / s )
t 
 0.013s
272 N
Momentum and Collisions
Problem
– A 6.0g bullet is fired at a velocity of 350m/s into
a container of ballistic gelatin that stops the
bullet in 1.8ms. What is the average force that
stops the bullet?
Momentum and Collisions
Solution
m = 6g = 0.006kg
 t = 1.8ms
vi = 350m/s
vf = 0m/s
Ft  mv
m v
F
t
(0.006kg)( 350m / s)
3
F
 1.2 x10 N
3
1.8 x10 s
Momentum and Collisions
Problem
– A hockey puck has a mass of 0.115kg and is at
rest. A hockey player makes a shot, exerting a
constant force of 30.0N on the puck for 0.16s.
With what speed does it head toward the goal?
Momentum and Collisions
Solution
m = 0.115kg
F = 30N
 t = 0.16s
Ft  mv
Ft
v 
m
(30 N )(0.16s)
v 
 42m / s
0.115kg
Momentum and Collisions
Problem
– A 0.150kg ball, moving in the positive direction
at 12m/s, is acted on by the impulse shown in the
graph. What is the ball’s speed at 4.0 s?
Momentum and Collisions
Solution
m = 0.150kg
v = 12m/s
 t = 4.0s
Ft  mv
1
Ft  (y )( x)  mv
2
1
Ft  (2.0 N )( 2.0s)  2.0 Ns  mv
2
2.0 Ns  m(v f  vi )
2.0kg * m / s
vf 
 12m / s  25m / s
0.150kg
Momentum and Collisions
Problem
– Small rockets are used to make tiny adjustments
in the speeds of satellites. One such rocket has a
thrust of 35N. If it is fired to change the velocity
of a 72,000kg spacecraft by 63cm/s, how long
should it be fired?
Momentum and Collisions
Solution
F = 35N
m = 7.2x104kg
 v = 63cm/s
4
Ft  mv
m v
t 
F
(7.2 x10 kg)(0.63m)
t 
 1296s(21.6 min)
35 N
Momentum and Collisions
Homework
Page 232 - 233
Problems
14 (160N Right)
16 (0.010s, 0.13m)
24 (a, 2.43m/s right b, 7.97x10-2m/s right)
26 (a, 2.5m/s right b, 3.0m/s right)
Momentum and Collisions
Conservation of Momentum
Work – Energy Theorem
The Work Done by a Net Force is Equal to the
Change in the Object’s Kinetic Energy
W  KE f  KEi
 Impulse – Momentum Theorem
The Impulse Produced by a Net Force is Equal to the
Change in the Object’s Momentum
Ft  mv
Momentum and Collisions
Conservation of Momentum
Work – Energy Theorem
Conservation of Mechanical Energy
 Impulse – Momentum Theorem
Conservation of Linear Momentum
Momentum and Collisions
Conservation of Linear Momentum
“System”
Internal Forces
Forces that Objects Within the System Exert on
Each Other
External Forces
Forces Exerted on the Objects in the System by
Agents Outside the System
Momentum and Collisions
Conservation of Linear Momentum
Collisions
Object A
Mass (mA)
Velocity (vA)
Weight (WA)
Object B
Mass (mB)
Velocity (vB)
Weight (WB)
Momentum and Collisions
Conservation of Linear Momentum
Collisions
Object A
Exerts Force on Object B (FA-B) (Internal)
Weight
Fg=mg (External)
Object B
Exerts Force on Object A (FB-A) (Internal)
Weight
Fg=mg (External)
Momentum and Collisions
Conservation of Linear Momentum
Collisions
Newton’s 3rd Law Tells Us that These
Forces are Equal and Opposite
FAB  (FB A )
Momentum and Collisions
Conservation of Linear Momentum
Collisions
In an Isolated System
No External Forces
p f  pi
Momentum and Collisions
Conservation of Linear Momentum
Collisions
The Total Linear Momentum of an Isolated
System Remains Constant. An Isolated
System is One for Which the Vector Sum of
the External Forces Acting on the System is
Zero
p f  pi
Momentum and Collisions
Collisions
Recoil
The momentum of a baseball changes when the
external force of a bat is exerted on it. The baseball,
therefore, is not an isolated system.
On the other hand, the total momentum of two
colliding balls within an isolated system does not
change because all forces are between the objects
within the system.
p f  pi
Momentum and Collisions
Collisions
Propulsion in Space
Recoil Principle
p f  pi
Momentum and Collisions
Conservation of Linear Momentum
Two-Dimensional Collisions (Non-Linear)
p f  pi
Momentum and Collisions
Conservation of Linear Momentum
Two-Dimensional Collisions (Non-Linear)
Initial momentum of the moving ball is pCi and the
momentum of the stationary ball is zero.
The momentum of the system before the collision is
equal to pCi
p f  pi
Momentum and Collisions
Conservation of Linear Momentum
Two-Dimensional Collisions (Non-Linear)
After the collision, both billiard balls are moving
and have momenta.
Ignoring friction with the tabletop, the system is
closed.
Conservation of momentum can be used. The initial
momentum equals the vector sum of the final
momenta.
pCi  pCf  pDf
Momentum and Collisions
Conservation of Linear Momentum
Two-Dimensional Collisions (Non-Linear)
The sum of the components of the vectors before
and after the collision must be equal.
Suppose the x-axis is defined to be in the direction of
the initial momentum, then the y-component of the
initial momentum is equal to zero.
Therefore, the sum of the final y-components also
must be zero.
pCf  y  pDf  y  0
Momentum and Collisions
Conservation of Linear Momentum
Two-Dimensional Collisions (Non-Linear)
The y-components are equal in magnitude
but are in the opposite direction and, thus,
have opposite signs. The sum of the
horizontal components also is equal.
pCi  pCf  x  pDf  x
Momentum and Collisions
Collisions
The Total Linear Momentum is
Conserved When Two Objects Collide,
Provided They Constitute an Isolated
System
Momentum and Collisions
Collisions
Sometimes, Total KE of the Objects is
Maintained
ex. Electrons
Sometimes, Total KE of the Objects is
Transferred into Other Types of Energy
ex. Trains
Momentum and Collisions
Collisions
Elastic Collision
Total KE and Momentum of the System After the
Collision is Equal to the Total KE and Momentum
of the System Before the Collision
Inelastic Collision
Total KE and Momentum of the System After the
Collision is NOT Equal to the Total KE and
Momentum of the System Before the Collision
Momentum and Collisions
Collisions
Elastic Collision
m1v1i  m2 v2i  m1v1 f  m2v2 f
1
1
1
1
2
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v2 f
2
2
2
2
Momentum and Collisions
Collisions
Examples (Elastic or Inelastic)
Pool Balls
Car Crash
Newtonian Swing
Photon
Punch
Basketball
Superball
Snowball
Momentum and Collisions
Collisions
Most Collisions are Neither
Completely Elastic nor Completely
Inelastic
Momentum and Collisions
Problem
Ben and Erika have a combined mass of 168kg.
At an ice skating rink they stand at rest and Ben
pushes Erika. Ben is sent in one direction at a
velocity of 0.90m/s while Erika is sent the other
direction at 1.2m/s. What is Ben’s mass?
Momentum and Collisions
mB viB  mE viE  mB v fB  mE v fE
Solution
0  mB v fB  mE v fE
mB = ?
mB v fB  (168kg  mB )v fE
mE = 168kg - mB
vE = -1.2m/s
mB v fB  (168kg)(v fE )  (mB v fE )
vB = 0.90m/s
(mB v fB )  (v fE mB )  (168kg)(v fE )
mB (v fB  v fE )  (168kg)(v fE )
mB 
(168kg)(v fE )
v fB  v fE

(168kg)(1.2m / s)
 96kg
0.90m / s  1.2m / s
Momentum and Collisions
Problem
A car moving at 10.0m/s crashes into a barrier
and stops in 0.050s. There is a 20.0kg child in
the car. Assume that the child’s velocity is
changed by the same amount as that of the car,
and in the same time period. What is the
impulse needed to stop the child?
Momentum and Collisions
Solution
v = 10.0m/s
t = 0.050s
m = 20.0kg
Ft  mv
Ft  (20.0kg)( 10.0m / s)  200.0 Ns
Momentum and Collisions
Problem
A car moving at 10.0m/s crashes into a barrier
and stops in 0.050s. There is a 20.0kg child in
the car. Assume that the child’s velocity is
changed by the same amount as that of the car,
and in the same time period. What is the
average force on the child?
Momentum and Collisions
Solution
v = 10.0m/s
t = 0.050s
m = 20.0kg
Ft  mv
F
m(v f  vi )
t
(20.0kg)(0m / s  10.0m / s)
3
F
 4.0 x10 N
0.050s
Momentum and Collisions
Problem
Marble C, with mass 5.0g, moves at speed of
20.0cm/s. It collides with second marble, D,
with mass 10.0g, moving 10.0cm/s in the same
direction. After the collision, marble C
continues with a speed of 8.0cm/s in the same
direction. What are the marbles’ momenta
before the collision?
Momentum and Collisions
Solution
mC = 5.0x10-3kg
vCi = 0.20m/s
mD = 1.0x10-2kg
vDi = 0.10m/s
vCf = 0.08m/s
p  mv
3
3
2
3
mC vCi  (5.0 x10 kg)(0.20m / s)  1x10 kg * m / s
mD vDi  (1.0 x10 kg)(0.10m / s)  1x10 kg * m / s
Momentum and Collisions
Problem
Marble C, with mass 5.0g, moves at speed of
20.0cm/s. It collides with second marble, D,
with mass 10.0g, moving 10.0cm/s in the same
direction. After the collision, marble C
continues with a speed of 8.0cm/s in the same
direction. What is the momentum of marble C
after the collision?
Momentum and Collisions
Solution
mC = 5.0x10-3kg
vCi = 0.20m/s
mD = 1.0x10-2kg
vDi = 0.10m/s
vCf = 0.08m/s
3
p  mv
4
mC vCf  (5.0 x10 kg)(0.08m / s)  4 x10 kg * m / s
Momentum and Collisions
Problem
Marble C, with mass 5.0g, moves at speed of
20.0cm/s. It collides with second marble, D,
with mass 10.0g, moving 10.0cm/s in the same
direction. After the collision, marble C
continues with a speed of 8.0cm/s in the same
direction. What is the momentum of marble D
after the collision?
Momentum and Collisions
p  mv
Solution
mC = 5.0x10-3kg
vCi = 0.20m/s
mD = 1.0x10-2kg
vDi = 0.10m/s
vCf = 0.08m/s
pCi  pDi  pCf  pDf
pDf  pCi  pDi  pCf
pDf  (1.0 x10 3 kg * m / s )  (1.0 x10 3 kg * m / s ) 
4
3
(4.0 x10 kg * m / s )  1.6 x10 kg * m / s
Momentum and Collisions
Problem
Marble C, with mass 5.0g, moves at speed of
20.0cm/s. It collides with second marble, D,
with mass 10.0g, moving 10.0cm/s in the same
direction. After the collision, marble C
continues with a speed of 8.0cm/s in the same
direction. What is the speed of marble D after
the collision?
Momentum and Collisions
p  mv
Solution
mC = 5.0x10-3kg
vCi = 0.20m/s
mD = 1.0x10-2kg
vDi = 0.10m/s
vCf = 0.08m/s
3
vDf
p Df  mD vDf
pDf
vDf 
mD
1.6 x10 kg * m / s

 0.16m / s
2
1.0 x10 kg
Momentum and Collisions
Problem
A 2575kg van runs into the back of an 825kg
compact car at rest. They move off together at
8.5m/s. Assuming that the friction with the road
is negligible, calculate the initial speed of the
van.
Momentum and Collisions
p  mv
Solution
mV = 2575kg
mC = 825kg
vVf = 8.5m/s
vCf = 8.5m/s
pVi  pCi  pVf  pCf
mV vVi  mC vCi  mV vVf  mC vCf
vVi 
(mV  mC )v f
mV
(2575kg  825kg)8.5m / s
vVi 
 11m / s
2575kg
Momentum and Collisions
Homework
Pages 234 - 235
Problems
32 (3.00m/s)
35 (a, 0.81m/s east b, 1.4x103J)
37 (4.0m/s)
48 (14.5m/s north)