1. Momentum, Impulse, and Collisions
Chapter 8
Momentum, Impulse, and Collisions

2. Goals for Chapter 8
To learn the meaning of the momentum of a particle and how an impulse causes it to change
To learn how to use the principle of conservation of momentum
To learn how to solve problems involving collisions

3. Goals for Chapter 8
To learn the definition of the center of mass of a system and what determines how it moves
To analyze situations, such as rocket propulsion, in which the mass of a moving body changes

4. Introduction
In many situations, such as a bullet hitting a carrot, we cannot use Newton’s second law to solve problems because we know very little about the complicated forces involved.
In this chapter, we shall introduce momentum and impulse, and the conservation of momentum, to solve such problems.

5. Momentum and Newton’s second law
The momentum of a particle is the product of its mass and its velocity:
Newton’s second law can be written in terms of momentum as
What is the momentum of a 1000kg car going 25 m/s west?

6. Momentum and Newton’s second law
The momentum of a particle is the product of its mass and its velocity:
Newton’s second law can be written in terms of momentum as
What is the momentum of a 1000kg car going 25 m/s west?
p = mv = (1000 kg)(25 m/s) = 25,000 kgm/s west

7. Impulse and momentum
Impulse of a force is product of force & time interval during which it acts.
Impulse is a vector!
On a graph of Fx versus time, impulse equals area under curve.

8. Impulse and momentum
Impulse-momentum theorem: Impulse = Change in momentum J of particle during time interval equals net force acting on particle during interval
J = Dp = pfinal – pinitial (note all are VECTORS!)
J = Net Force x time = (SF) x (Dt) so…
Dp = (SF) x (Dt)
SF = Dp/(Dt)

9. Compare momentum and kinetic energy
Changes in momentum depend on time over which net force acts But…
Changes in kinetic energy depend on the distance over which net force acts.

10. Ice boats again (example 8.1)
Two iceboats race on a frictionless lake; one with mass m & one with mass 2m.
Wind exerts same force on both.
Both boats start from rest, both travel same distance to finish line.
Questions!
Which crosses finish line with more KE?
Which crosses with more Momentum?

11. Ice boats again (example 8.1)
Two iceboats race on a frictionless lake; one with mass m and one with mass 2m.
The wind exerts the same force on both.
Both boats start from rest, and both travel the same distance to the finish line?
Which crosses the finish line with more KE?
In terms of work done by wind? W = DKE = Force x distance = same;
So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22

12. Ice boats again (example 8.1)
Two iceboats race on a frictionless lake; one with mass m and one with mass 2m.
The wind exerts the same force on both.
Both boats start from rest, and both travel the same distance to the finish line?
Which crosses the finish line with more KE?
In terms of work done by wind? W = DKE = Force x distance = same;
So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22
Since m2> m1, v2 < v1 so more massive boat loses

13. Ice boats again (example 8.1)
Two iceboats race on a frictionless lake; one with mass m and one with mass 2m.
The wind exerts the same force on both.
Both boats start from rest, and both travel the same distance to the finish line?
Which crosses the finish line with more p?
In terms of momentum? Force on the boats is the same for each, but TIME that force acts is different. The second boat accelerates slower, and takes a longer time.
Since t2> t1, p2 > p1 so second boat has more momentum

14. Ice boats again (example 8.1)
Check with equations!
Force of wind = same; distance = same
Work done on boats is the same, gain in KE same.
½ m1v12 = DKE1 = W = DKE2 = ½ m2v22
And m1v1 = p1; m2v2 = p2
½ m1v12 = ½ (m1v1) v1 = ½ p1 v1 & same for ½ p2 v2
½ p1 v1 = ½ p2 v2
Since V1 > V2, P2 must be greater than P1!
p2 > p1 √

15. A ball hits a wall (Example 8.2)
A 0.40 kg ball moves at 30 m/s to the left, then rebounds at 20 m/s to the right from a wall.
What is the impulse of net force during the collision, and if it is in contact for 0.01 s, what is the average force acting from the wall on the ball?

16. Kicking a soccer ball – Example 8.3
Soccer ball 0.40 kg moving left at 20 m/s, then kicked up & to the right at 30 m/s at 45 degrees. If collision time is 0.01 seconds, what is impulse?

17. An isolated system
The total momentum of a system of particles is the vector sum of the momenta of the individual particles.
No external forces act on the isolated system consisting of the two astronauts shown below, so the total momentum of this system is conserved.

18. Conservation of momentum
External forces (the normal force and gravity) act on the skaters, but their vector sum is zero. Therefore the total momentum of the skaters is conserved.
Conservation of momentum: If the vector sum of the external forces on a system is zero, the total momentum of the system is constant.

19. Remember that momentum is a vector!
When applying conservation of momentum, remember that momentum is a vector quantity!
Use vector addition to add momenta in COMPONENTS!

20. Recoil of a rifle – 1D conservation of momentum
A rifle fires a bullet, causing the rifle to recoil.

21. Objects colliding along a straight line (example 8.5)
Two gliders collide on an frictionless track.
What are changes in velocity and momenta?

22. A two-dimensional collision
Two robots collide and go off at different angles.
You must break momenta into x & y components and deal with each direction separately

23. Elastic collisions
In an elastic collision, the total momentum of the system in a direction is the same after the collision as before if no external forces act in that direction:
Pix = Pfx
mavai +mbvbi =mavaf +mbvbf
But wait – there’s more!

24. Elastic collisions
In an elastic collision, the total kinetic energy of the system is the same after the collision as before.
KEi = KEf
½ mavai2 + ½ mbvbi2 = ½ mavaf2 + ½ mbvbf2 =

25. Elastic collisions In an elastic collision,
Difference in velocities initially = (-) difference in velocities finally
(vai - vbi) = - (vaf – vbf)

26. Elastic collisions In an elastic collision,
Difference in velocities initially = (-) difference in velocities finally
(vai - vbi) = - (vaf – vbf)
Solve generally for final velocities:
vaf = (ma-mb)/(ma+mb)vai + 2mb/(ma+mb)vbi
vbf = (mb-ma)/(ma+mb)vbi + 2ma/(ma+mb)vai

27. Elastic collisions

28. Elastic collisions
Behavior of colliding objects is greatly affected by relative masses.

29. Elastic collisions

30. Elastic collisions
Behavior of colliding objects is greatly affected by relative masses.

31. Elastic collisions

32. Elastic collisions
Behavior of colliding objects is greatly affected by relative masses.
Stationary Bowling Ball!

33. Inelastic collisions
In any collision in which the external forces can be neglected, the total momentum is conserved.
A collision in which the bodies stick together is called a completely inelastic collision
In an inelastic collision, the total kinetic energy after the collision is less than before the collision.

34. Some inelastic collisions
Cars are intended to have inelastic collisions so the car absorbs as much energy as possible.
Example 8.7

35. The ballistic pendulum
Ballistic pendulums are used to measure bullet speeds

36. A 2-dimenstional automobile collision (Example 8.9)
Two cars traveling at right angles collide.

37. An elastic straight-line collision

38. Neutron collisions in a nuclear reactor

39. A two-dimensional elastic collision (page 258)

40. Center of mass of symmetrical objects
It is easy to find the center of mass of a homogeneous symmetric object

41. Motion of the center of mass
The total momentum of a system is equal to the total mass times the velocity of the center of mass.
The center of mass of the wrench at the right moves as though all the mass were concentrated there.

42. Tug-of-war on the ice Start 20 meters away; each 10 m from coffee!
Pull light rope; when James moved 6 m, how far is Ramon?

43. External forces and center-of-mass motion
When a body or collection of particles is acted upon by external forces, the center of mass moves as though all the mass were concentrated there.

44. Rocket propulsion
Conservation of momentum holds for rockets, too! F = dp/dt = d(mv)/dt = m(dv/dt) + v(dm/dt)
As a rocket burns fuel, its mass decreases (dm< 0!)

45. Rocket propulsion Initial Values m = initial mass of rocket
v = initial velocity of rocket in x direction
mv = initial x-momentum of rocket
vexh = exhaust velocity from rocket motor (This will be constant!)
(v - vexh) = relative velocity of exhaust gases

46. Rocket propulsion Changing values
dm = decrease in mass of rocket from fuel
dv = increase in velocity of rocket in x-direction
dt = time interval over which dm and dt change

47. Rocket propulsion Final values
m + dm = final mass of rocket less fuel ejected
v + dv = increase in velocity of rocket
(m + dm) (v + dv) = final x-momentum of rocket (+x direction)
[- dm] (v - vexh) = final x-momentum of fuel (+x direction)

48. final momentum of the ejected mass of gas
Rocket propulsion
Final values
mv = [(m + dm) (v + dv)] + [- dm] (v - vexh)
initial momentum
final momentum of the lighter rocket
final momentum of the ejected mass of gas

49. Rocket propulsion Final values
mv = [(m + dm) (v + dv)] + [- dm] (v - vexh)
mv = mv + mdv + vdm + dmdv –dmv +dmvexh

50. Rocket propulsion Final values
mv = [(m + dm) (v + dv)] + [- dm] (v - vexh)
mv = mv + mdv + vdm + dmdv –dmv +dmvexh

51. Rocket propulsion
Final values
mv = mv + mdv + vdm + dmdv –dmv +dmvexh

52. Rocket propulsion Final values mv = mv + mdv + vdm + dmdv –dmv +dmvexh
0 = mdv + dmdv + dmvexh

53. Rocket propulsion Final values mv = mv + mdv + vdm + dmdv –dmv +dmvexh
0 = mdv + dmdv + dmvexh
Neglect the assumed small term:
m(dv) = -dmdv – (dm)vexh

54. Rocket propulsion Final values m(dv) = – (dm)vexh
Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas!

55. Rocket propulsion Final values m(dv) = – (dm)vexh
Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas! Differentiate both sides with respect to time!

56. Rocket propulsion Final values m(dv) = – (dm)vexh
m(dv/dt) = – [(dm)/dt] vexh

57. Rocket propulsion Final values m(dv) = – (dm)vexh
m(dv/dt) = – [(dm)/dt] vexh
ma = Force (Thrust!) = – [(dm)/dt] (vexh)

58. Rocket propulsion Final values m(dv) = – (dm)vexh
m(dv/dt) = – [(dm)/dt] vexh
ma = Force (Thrust!) = – [(dm)/dt] (vexh)
rate of change of mass
Velocity of gas exhausted

59. Rocket propulsion Final values Thrust = – [(dm)/dt] (vexh)
Example: vexhaust = 1600 m/s
Mass loss rate = 50 grams/second
Thrust =?

60. Rocket propulsion Final values Thrust = – [(dm)/dt] (vexh)
Example Problem 8.62 vexhaust = 1600 m/s
Mass loss rate = 50 grams/second
Thrust = m/s (-0.05 kg/1 sec) = +80N

61. Rocket propulsion Gain in speed? m(dv) = – (dm)vexh
dv = – [(dm)/m] vexh integrate both sides

62. Rocket propulsion Gain in speed? m(dv) = – (dm)vexh
dv = – [(dm)/m] vexh (integrate both sides)
vf - vi = (vexh) ln(m0/m)
m0 = initial mass
m0 > m, so ln > 1
“mass ratio”
gain in velocity
Velocity of gas exhausted

63. Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m)
m0 = initial mass
m0 > m, so ln > 1
“mass ratio”
gain in velocity
Velocity of gas exhausted
Faster exhaust, and greater difference in mass, means a greater increase in speed!

64. Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m) Example:
Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s?

65. Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m) Example:
Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s? ln(mo/m) = (3000/2000) = 1.5 so m/mo = e-1.5 = .223

66. Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m)
m0 = initial mass
m0 > m, so ln > 1
“mass ratio”
gain in velocity
Velocity of gas exhausted
STAGE rockets to throw away mass as they use up fuel, so that m0/m is even higher!