1. AP Physics 1 Conroe High School
Impulse and Momentum
AP Physics 1
Conroe
High
School

2. The Big Mo’
Momentum
nfl football momentum

3. Momentum is a commonly used term in sports
Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop.

4. Momentum as a physics term; refers to the quantity of motion that an object has. A sports team which is "on the move" has the momentum. If an object is in motion (on the move) then it has momentum.

5. Momentum Defined v = 16 m/s p = mv
Momentum ρ is defined as the product of mass and velocity, mv. Units: kg m/s
p = mv
m = 1000 kg
v = 16 m/s
ρ = (1000 kg)(16 m/s)
ρ = 16,000 kg m/s

6. An object’s momentum will change if its mass and/or velocity changes.
Most common… a change in velocity.
What is a change in velocity called?
Acceleration a = Vf -Vo t
According to Newton’s laws,
a net force causes an object to accelerate,
or change its velocity.

7. Which is worse, 50mph car crash in a single crash with a wall, or 2 cars(of the same mass) in opposite directions going 50mph?
Lmh4
6min clip

8. IMPULSE
Impulse J is a force F acting for a small time interval Dt. F Dt
Impulse:
J = F Dt

9. The unit for impulse is the Newton-second (N·s)
Example 1: The face of a golf club exerts an average force of 4000 N for s. What is the impulse imparted to the ball? Dt F
J = F Dt
Impulse:
J = (4000 N)(0.002 s)
J = 8.00 Ns
The unit for impulse is the Newton-second (N·s)

10. Impulse from a Varying Force
Normally, a force acting for a short interval is not constant. It may be large initially and then play off to zero as shown in the graph. F
time, t
In the absence of calculus, we use the average force Favg.
Unless told otherwise, treat forces as average forces

11. Impulse Changes Velocity
Consider a mallet hitting a ball: F
Impulse = Change in “mv”

12. Impulse = Change in momentum
Impulse & Momentum
Impulse = Change in momentum
F Dt = mvf - mvo Dt F
A force F acting on a ball for a time Dt increases its momentum mv. mv

13. Impulse = Momentum Momentum is defined as “Inertia in Motion”
Consider Newton’s 2nd Law and the definition of acceleration Units of Impulse: Units of Momentum: Ns
Kg x m/s
Momentum is defined as “Inertia in Motion”

14. Racket and Bat Sports
The act of following through when hitting a ball increases the time of collision and contributes to an increase in the velocity change of the ball.
In tennis, baseball, racket ball, etc., giving the ball a high velocity often leads to greater success.

15. Choose right as positive.
Example 2: A 50-g golf ball leaves the face of the club at 20 m/s. If the club is in contact for s, what average force acted on the ball?
Given: m = 0.05 kg; vo = 0;
Dt = s; vf = 20 m/s Dt F mv +
Choose right as positive.
F Dt = mvf - mvo
F (0.002 s) = (0.05 kg)(20 m/s)
F = 500 N
Average Force:

16. Sample Problem
A 2200-kg sport utility vehicle traveling at 26 m/s can be stopped in 21 s by gently applying the brakes, in 5.5 s in a panic stop, or in 0.22 s if it hits a concrete wall. What average force is exerted on the SUV in each of these stops?
-2.7 x 103 N; -1.0 x 104 N; -2.6 x 105 N

17. Sample Problem
A hockey puck has a mass of kg and is at rest. A hockey player makes a shot, exerting a constant force of 30.0 N on the puck for 0.16 s. With what speed does it head toward the goal?
41.7 m/s

18. A 1000 kg car moving at 30 m/s (p = 30,000 kg m/s) can be stopped by
30,000 N of force acting for 1.0 s (a crash!) or
by 3000 N of force acting for 10.0 s (normal stop)

19. Car Crash
Inertia
Momentum
Impulse
air bag vs. windshield
d7iYZPp2zYY&feature=related

20. Airbag Deployment Real time
Slow motion 0

22. Another applications of impulse
Contact time is reduced if arm's deceleration is kept as small as possible.
This is done by using "follow-through", which means to continue to push during the contact period.

25. Bungee Jump

29. Bungee Jump
Take a look at how stopping time affects force.

30. Force – time graph
Force (N)
Time (sec)

31. Impulse is the Area
Since J=Ft, Impulse is the AREA of a Force vs. Time graph.

32. According to the graph, what impulse was given to the object over the three seconds?
(0.5)(3)(0.003) = NS
If this graph is showing the force acting on a 1 gram object, what is the object’s change in momentum over the first 9 seconds?
(0.5)(0.003)(3) + (0.003)(3) + (0.5)(0.003)(3) = NS
What is the object’s change in velocity over the first 9 seconds?
(0.018) / (0.001) = 18 m/s

33. So to summarize…
To minimize the effect of the force on an object involved in a collision, the time must be increased.
To maximize the effect of the force on an object involved in a collision, the time must be decreased.
BUT the change in momentum is the SAME either way!

34. Delta V V
p = m v
Watch the change in velocity when there is a change in direction.
V = Vf – Vi
Example; in a car crash, a car going 50mph west then 50mph east has a V of 100mph.

35. Stacked Ball Bounce https://www.youtube.com/watch?v=2UHS883_ P60
Click link

36. Vector Nature of Momentum
Consider the change in momentum of a ball that is dropped onto a rigid plate: + vo vf
A 2-kg ball strikes the plate with a speed of 20 m/s and rebounds with a speed of 15 m/s. What is the change in momentum?
Dp = mvf - mvo = (2 kg)(15 m/s) - (2 kg)(-20 m/s)
Dp = 30 kg m/s + 40 kg m/s
Dp = 70 kg m/s

37. + + - Dt F 40 m/s 20 m/s m = 0.5 kg F Dt = mvf - mvo
Example 3: A 500-g baseball moves to the left at 20 m/s striking a bat. The bat is in contact with the ball for s, and it leaves in the opposite direction at 40 m/s. What was average force on ball? + Dt F
40 m/s
20 m/s
m = 0.5 kg + -
F Dt = mvf - mvo
vo = -20 m/s; vf = 40 m/s
F(0.002 s) = (0.5 kg)(40 m/s) - (0.5 kg)(-20 m/s)
Continued . . .

38. + - Example Continued: m = 0.5 kg F 40 m/s 20 m/s Dt F Dt = mvf - mvo
F(0.002 s) = (0.5 kg)(40 m/s) - (0.5 kg)(-20 m/s)
F(0.002 s) = (20 kg m/s) + (10 kg m/s)
F = 15,000 N
F(0.002 s) = 30 kg m/s

39. 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.

41. Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.
Calculate the impulse suffered when a 70 kg person lands on firm ground after jumping from a height of 5 m. Then estimate the average force on the person’s leg if the landing is stiff legged where the body moves 1 cm during impact.
Hint:
Find the impact velocity using linear motion equations.
Find the Impulse
Find the average velocity during impact
Find the time of impact using the definition of average velocity
Find the average force acting during impact

42. A negative velocity since it is moving downward.
Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.
A negative velocity since it is moving downward.

43. Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.

44. Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.

45. Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.

46. Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.

47. A- What is his velocity as he hits the water?
Dr. Weatherly jumps off the high dive (7m) and splashes into the swimming pool. It was measured that it took the water .7 seconds to bring him to a stop underwater. His mass is 90kg.
A- What is his velocity as he hits the water?
B- What is his momentum as he strikes the water?
C- What is the average force the water puts on our Principal to bring him to rest underwater?

48. Impulse = Change in momentum
Summary of Formulas:
Impulse
J = FavgDt
Momentum
ρ = mv
Impulse = Change in momentum
F Dt = mvf - mvo

50. 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.
How many objects do I have before the collision?
How many objects do I have after the collision? 2 1
4 m/s

51. Explosion Example
A 3.8kg gun is fired. The 42 gram bullet leaves the muzzle at 470m/s. Find the guns recoil velocity.

52. Recoil – Explosion solution.
Mg x vg = Mb x vb
Vg = .042kg x 470m/s
3.8kg
Vg = 5.2 m/s

53. 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

54. How about a collision?
If the Impulses are equal then the MOMENTUMS are also equal!

55. 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.”

56. Several Types of collisions
Sometimes objects stick together or blow apart. In this case, momentum is ALWAYS conserved.
When 2 objects collide and DON’T stick
When 2 objects collide and stick together
When 1 object breaks into 2 objects
Elastic Collision = Kinetic Energy is Conserved
Inelastic Collision = Kinetic Energy is NOT Conserved

57. Prac Prob
In a hockey game, a 75kg goalie catches a 105g puck in his leather glove. The puck was traveling at 48m/s when he caught it. The goalie was at rest when he caught it. Calculate his velocity after the catch.

58. MPVP + MGVG = (MP + MG) VPG MPVP + MGVG = V’PG MPG
Given
MP =.105 kg
VP= 48m/s
MG= 75kg
VG= 0
V’P+G = ?
Inelastic
Formula
MPVP + MGVG = (MP + MG) VPG
MPVP + MGVG = V’PG
MPG
Solution
.105(48) + 0
75.105kg
V’PG = .067m/s

59. Prac. Prob
A bullet was fired into a block of wood. The bullet embedded into the wood and moved off with a velocity of 8.6m/s to the right. Find the bullets velocity before striking the wood. The bullets mass is 35 grams and the blocks mass is 5kg.

60. VB = 1237.2m/s Formula Given MBVB + MWVW = (MB + MW)V’BW MB= .035kg
VB = (MB + MW)V’BW – MWVW MB
Given
MB= .035kg
MW= 5kg
V’BW= 8.6m/s
VW= 0
VB=?
Inelastic
Solution
VB = 5.035(8.6) – 5(0)
.035
VB = m/s

61. Prac Prob
Another gun fires a bullet at a wooden block. This time the bullet goes through the block. Calculate the velocity of the wooden block after the bullet passes through. You know the bullets mass 35g, blocks mass 2.5kg, bullets initial speed 475m/s, and bullets final speed 275m/s.

62. MBVB + MWVW = MBV’B + MWV’W MBVB + MWVW – MBV’B = V’W MW
Given
MB = .035kg
VB = 475 m/s
MW= 2.5kg
V’B= 275 m/s
VW= 0m/s
V’W= ?
elastic
Formula
MBVB + MWVW = MBV’B + MWV’W
MBVB + MWVW – MBV’B = V’W MW
Solution
.035(475) + 2.5(0) kg(275m/s)
2.5
– 9.63
2.8 m/s

63. Conservation of Momentum
According to the law of conservation of linear momentum, when the vector sum of the external forces that act on a system of bodies equals zero, the total linear momentum of the system remains constant no matter what momentum changes occur within the system

64. For two objects interacting with one another, the conservation of momentum can be expressed as:
 v1 and v2 are initial velocities, and are final velocities

65. ELASTIC AND INELASTIC COLLISIONS
Elastic Collision: A collision in which objects collide and bounce apart with no energy loss.
Inelastic Collision: A collision in which objects collide and some mechanical energy is transformed into heat energy.

66. State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems.
A closed system is one in which objects can neither enter nor leave the system.
An isolated system is one in which there is no net external force acting on the system.
For example billiard balls have the force due to gravity acting on them but it is balanced by the table pushing up on them. So in that case there is no net external force.
There can be internal forces acting.

67. State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems.
The law of conservation of linear momentum only holds true in a closed and isolated system.
The net external force is zero
The total mass of the system does not change

68. State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems.
The Law of conservation of momentum states the total momentum before = the total momentum after.

69. State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems.
The law of conservation of momentum does NOT say that each particle within the system retains its same momentum.
The amount of momentum one particle in the system gains is equal to the amount of momentum lost by another.

70. State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems.
A 10,000 kg railroad car traveling at a speed of 24 m / s strikes an identical car at rest. If the cars lock together as a result of the collisions, what is their common speed afterward?
12 m / s

71. State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems.
Calculate the recoil velocity of a 4 kg rifle which shoots a kg bullet at a speed of 280 m / s.
-3.5 m / s

72. Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.
A collision is an interaction between two bodies if the interaction occurs over a short time interval and is so strong that other forces acting are insignificant compared to the forces the two objects are exerting on each other.

73. Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems.
Elastic collisions are collisions in which both momentum and kinetic energy are conserved.
Inelastic collisions are collisions in which only momentum is conserved.

74. Conservation of momentum equation
Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems.
Conservation of momentum equation
Conservation of Kinetic Energy Equation

75. The animation below portrays the elastic collision between a 1000-kg car and a 3000-kg truck. The before- and after-collision velocities and momentum are shown in the data tables.

76. The animation below portrays the elastic collision between a 3000-kg truck and a 1000-kg car.
The before- and after-collision velocities and momentum are shown in the data tables.

77. Before the collision, the momentum of the truck is 60 000 Ns and the momentum of the car is 0 Ns;
the total system momentum is Ns.
After the collision, the momentum of the truck is Ns and the momentum of the car is Ns; the total system momentum is
Ns.

78. The animation below portrays the inelastic collision between a very massive diesel and a less massive flatcar. The diesel has four times the mass of the freight car. After the collision, both the diesel and the flatcar move together with the same velocity.

79. In elastic collisions no permanent deformation occurs; objects elastically rebound from each other. In head-on elastic collisions between equal masses,
velocities are exchanged.

80. p before = p after m1V1 + m2V2 = m1Vf1 + m2V2f V2f = - 2 m/s
Example 4: A 0.50-kg ball traveling at 6.0 m/s collides head-on with a 1.00-kg ball moving in the opposite direction at a velocity of m/s. The 0.50-kg ball moves away at -14 m/s after the collision. Find the velocity of the second ball.
M1 = 0.50 kg M2 = 1.00 kg
V1 = 6.0 m/s V2 = m/s Vf1 = -14 m/s
p before = p after
m1V1 + m2V2 = m1Vf1 + m2V2f
(.5kg)(6m/s) + (1kg)(-12m/s) = (.5kg)(-14m/s) + (1kg)(V2f)
V2f = - 2 m/s

81. Inelastic collisions are characterized by objects sticking together and permanent deformation.

82. p before = p after m1V1 + m2V2 = (m1+ m2 )V V = 1.25 km/hr, right
Example 5: A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on with a 1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck.
p before = p after
m1V1 + m2V2 = (m1+ m2 )V
M1 = 3000 kg
V1 = 5.0 km/hr
M2 = 1000 kg
V2 = -10 km/hr
(3000kg)(5km/hr) + (1000kg)(-10km/hr)
(3000kg kg)
V = km/hr, right

83. Explosions
When an object separates suddenly, as in an explosion, all forces are internal.
Momentum is therefore conserved in an explosion.
There is also an increase in kinetic energy in an explosion. This comes from a potential energy decrease due to chemical combustion.

84. Recoil Guns and cannons “recoil” when fired.
This means the gun or cannon must move backward as it propels the projectile forward.
The recoil is the result of action-reaction force pairs, and is entirely due to internal forces. As the gases from the gunpowder explosion expand, they push the projectile forwards and the gun or cannon backwards.

85. Summary of Formulas: Impulse Momentum p = mv J = FavgDt
Impulse = Change in momentum
F Dt = mvf - mvo
Conservation of Momentum

86. Example 6 A 7500-kg truck traveling at 5 m/s east collides with a 1500-kg car moving at 20 m/s from a direction of 210. After the collision, the two vehicles remain tangled together. With what speed and in what direction does the wreckage begin to move?
m1 = 7500 kg v1 = 5 m/s, 0º
m2 = 1500 kg v2 = 20 m/s, 210º
m1 v1+ m2 v2 =( m1 +m2)V

87. 1500kg (20m/s cos 210º) 1500 kg (20m/s sin 210º)
m1 v1+ m2 v2 =( m1 +m2)V
x-comp y-comp
7500 kg (5 m/s)
1500kg (20m/s cos 210º) kg (20m/s sin 210º)
Σx = 11,519 kg m/s Σy = - 15,000 kg m/s
Initial Momentum
= 18,912.7 kg m/s

88. initial momentum = final momentum = (m1 + m2) V
= 2.1 m/s
= 52.5º
I quadrant
V = (2.1 m/s,52.5º)

89. Example 7: Suppose a 5.0-kg projectile launcher shoots a 209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher?

90. 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 kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of kg and is initially at rest. The collision is NOT head on. After the collision, the two pucks fly apart with angles shown in the drawing. Calculate the speeds of the pucks after the collision. vA
vAsinq
vAcosq
vBcosq
vBsinq vB

91. Collisions in 2 dimensionsvA
vAsinq
vAcosq
vBcosq
vBsinq vB

92. Collisions in 2 dimensions

93. 3 m/s
2 kg
8 kg
0 m/s
Before
2 m/s v
After
50o x y

94. Sample problem
Calculate velocity of 8-kg ball after the collision.
3 m/s
2 kg
8 kg
0 m/s
Before
2 m/s v
After
50o x y

95. 2D-Collisions Momentum in the x-direction is conserved.
SPx (before) = SPx (after)
Momentum in the y-direction is conserved.
SPy (before) = SPy (after)
Treat x and y coordinates independently.
Ignore x when calculating y
Ignore y when calculating x
Let’s look at a simulation:

96. Physics C Energy
4/28/2017
Center of Mass
Bertrand

97. Physics C Energy
4/28/2017
Center of Mass
The center of mass of a system is the point at which all the mass can be assumed to reside.
Sometimes the system is an assortment of particles and sometimes it is a solid object.
Mathematically, you can think of the center of mass as a “weighted average”.
Bertrand

98. Center of Mass – system of points
Physics C Energy
4/28/2017
Center of Mass – system of points
Analogous equations exist for velocity and acceleration, for example
Bertrand

99. y 2 x -2 -2 2 4 Determine the Center of Mass. 2 kg 3 kg 1 kg
Physics C Energy
4/28/2017
Determine the Center of Mass. y 2
2 kg x
3 kg
1 kg -2 -2 2 4
Bertrand

100. Using the conservation of momentum and Energy….
Certain problems involve the use of conservation of energy and the conservation of momentum
These problems require a special kind of Physics aptitude and are only attempted by the most savvy of Physics students…. You are those students.

101. The Ballistic Pendulum is used in the lab setting to find the velocity of fast moving projectiles

102. How fast was the bullet traveling before it struck the catcher
How fast was the bullet traveling before it struck the catcher? Is mechanical energy conserved before and after the collision?
821.3 m/s

103. Conservation of Momentum and Energy Example 2
A 4 kg tennis racket moving with a velocity of 21 m/s strikes a 50 g tennis ball initially at rest. Find the velocity of the tennis ball and tennis racket after the collision.