1. Chapter 9B - Conservation of Momentum
A PowerPoint Presentation by
Paul E. Tippens, Emeritus Professor
Southern Polytechnic State University
© 2007

2. Momentum is conserved in this rocket launch
Momentum is conserved in this rocket launch. The velocity of the rocket and its payload is determined by the mass and velocity of the expelled gases. Photo: NASA
NASA

3. Objectives: After completing this module, you should be able to:
State the law of conservation of momentum and apply it to the solution of problems.
Distinguish by definition and example between elastic and inelastic collisions.
Predict the velocities of two colliding bodies when given the coefficients of restitution, masses, and initial velocities.

4. A Collision of Two Masses
When two masses m1 and m2 collide, we will use the symbol u to describe velocities before collision. m1 u1 m2 u2
Before
The symbol v will describe velocities after collision.
After m1 v1 m2 v2

5. A Collision of Two Blocksm1 u1 m2 u2
Before
Collision m1 B m2
“u”= Before
“v” = After m2 v2 m1 v1
After

6. Conservation of Energym1 m2 u1 u2
The kinetic energy before colliding is equal to the kinetic energy after colliding plus the energy lost in the collision.

7. Example 1. A 2-kg mass moving at 4 m/s collides with a 1-kg mass initially at rest. After the collision, the 2-kg mass moves at 1 m/s and the 1-kg mass moves at 3 m/s. What energy was lost in the collision?
It’s important to draw and label a sketch with appropriate symbols and given information. m2
u2 = 0 m1
u1 = 4 m/s
m1 = 2 kg
m1 = 1 kg
BEFORE m2
v2 = 2 m/s m1
v1 = 1 m/s
m1 = 2 kg
m1 = 1 kg
AFTER

8. Energy Conservation: K(Before) = K(After) + Loss
Example 1 (Continued). What energy was lost in the collision? Energy is conserved. m2
u2 = 0 m1
u1 = 4 m/s
m1 = 2 kg
m1 = 1 kg
v2 = 2 m/s
v1 = 1 m/s
BEFORE:
AFTER
Energy Conservation: K(Before) = K(After) + Loss
Loss = 16 J – 3 J
Energy Loss = 15 J

9. mBvB - mBuB = -(mAvA - mAuA)
Impulse and Momentum A B uA uB vA vB
Impulse = Dp
-FA Dt
FB Dt
FDt = mvf– mvo
Opposite but Equal F Dt
FBDt = -FADt
mBvB - mBuB = -(mAvA - mAuA)
Simplifying:
mAvA + mBvB = mAuA + mBuB

10. Conservation of Momentum
The total momentum AFTER a collision is equal to the total momentum BEFORE.
mAvA + mBvB = mAuA + mBuB A B uA uB vA vB
-FADt
FB Dt
Recall that the total energy is also conserved:
Kinetic Energy: K = ½mv2
KA0 + KB0 = KAf + KBf + Loss

11. mAvA + mBvB = mAuA + mBuB mBvB mA mAvA = - mBvB vA = -
Example 2: A 2-kg block A and a 1-kg block B are pushed together against a spring and tied with a cord. When the cord breaks, the 1-kg block moves to the right at 8 m/s. What is the velocity of the 2 kg block? A B
The initial velocities are zero, so that the total momentum before release is zero.
mAvA + mBvB = mAuA + mBuB
vA = -
mBvB mA
mAvA = - mBvB

12. Example 2 (Continued) vA2 mAvA+ mBvB = mAuA + mBuB mBvB mA
mAvA = - mBvB
vA = -
mBvB mA A B
2 kg
1 kg
8 m/s
vA2
vA = -
(1 kg)(8 m/s)
(2 kg)
vA = - 4 m/s

13. Example 2 (Cont.): Ignoring friction, how much energy was released by the spring?
2 kg
1 kg A B
8 m/s
4 m/s
Cons. of E: ½kx2 = ½ mAvA + ½mBvB2 2
½kx2 = ½(2 kg)(4 m/s)2 + ½(1 kg)(8 m/s)2
½kx2 = 48 J
½kx2 = 16 J + 32 J = 48 J

14. Elastic or Inelastic?
An elastic collision loses no energy. The deform-ation on collision is fully restored.
In an inelastic collision, energy is lost and the deformation may be permanent. (Click it.)

15. Completely Inelastic Collisions
Collisions where two objects stick together and have a common velocity after impact.
Before
After

16. Example 3: A 60-kg football player stands on a frictionless lake of ice. He catches a 2-kg football and then moves at 40 cm/s. What was the initial velocity of the football? A
Given: uB= 0; mA= 2 kg; mB= 60 kg; vA= vB= vC vC = 0.4 m/s B
Momentum:
mAvA + mBvB = mAuA + mBuB
Inelastic collision:
(mA + mB)vC = mAuA
(2 kg + 60 kg)(0.4 m/s) = (2 kg)uA
uA= 12.4 m/s

17. Example 3 (Cont.): How much energy was lost in catching the football?
½(2 kg)(12.4 m/s)2 = ½(62 kg)(0.4 m/s)2 + Loss
154 J = 4.96 J + Loss
Loss = 149 J
97% of the energy is lost in the collision!!

18. General: Completely Inelastic
Collisions where two objects stick together and have a common velocity vC after impact.
Conservation of Momentum:
Conservation of Energy:

19. Common speed after colliding: 2.4 m/s.
Example 4. An 87-kg skater B collides with a 22-kg skater A initially at rest on ice. They move together after the collision at 2.4 m/s. Find the velocity of the skater B before the collision. A B
uB = ?
uA = 0
Common speed after colliding: 2.4 m/s.
22 kg
87 kg
vB= vA = vC = 2.4 m/s
(87 kg)uB = (87 kg + 22 kg)(2.4 m/s)
(87 kg)uB =262 kg m/s
uB = 3.01 m/s

20. Example 5: A 50 g bullet strikes a 1-kg block, passes all the way through, then lodges into the 2 kg block. Afterward, the 1 kg block moves at 1 m/s and the 2 kg block moves at 2 m/s. What was the entrance velocity of the bullet?
1 kg
2 kg
uA= ?
2 kg
1 kg
1 m/s
2 m/s

21. mAuA + mBuB + mCuC = mBvB + (mA+mC) vAC
50 g
Find entrance velocity of bullet: mA= 0.05 kg; uA= ?
2 kg
1 kg
1 m/s
2 m/s
Momentum After = Momentum Before
mAuA + mBuB + mCuC = mBvB + (mA+mC) vAC
(0.05 kg)uA =(1 kg)(1 m/s)+(2.05 kg)(2 m/s)
(0.05 kg) uA =(5.1 kg m/s)
uA= 102 m/s

22. Completely Elastic Collisions
Collisions where two objects collide in such a way that zero energy is lost in the process.
APPROXIMATIONS!

23. Velocity in Elastic CollisionsuB uA
1. Zero energy lost. A B
2. Masses do not change. vA vB A B
3. Momentum conserved.
Equal but opposite impulses (F Dt) means that:
(Relative Dv After) = - (Relative Dv Before)
vA - vB = - (uA - uB)
For elastic collisions:

24. From conservation of energy (relative v):
Example 6: A 2-kg ball moving to the right at 1 m/s strikes a 4-kg ball moving left at 3 m/s. What are the velocities after impact, assuming complete elasticity? A B
3 m/s
1 m/s vA vB
1 kg
2 kg
vA - vB = - (uA - uB)
vA - vB = uB - uA
vA - vB = (-3 m/s) - (1 m/s)
From conservation of energy (relative v):
vA - vB = - 4 m/s

25. Example 6 (Continued) mAvA + mBvB = mAuA + mBuB vA + 2vB = -5 m/s
1 kg
2 kg
Energy: vA - vB = - 4 m/s
Momentum also conserved:
mAvA + mBvB = mAuA + mBuB
(1 kg)vA+(2 kg)vB=(1 kg)(1 m/s)+(2 kg)(-3 m/s)
vA + 2vB = -5 m/s
Two independent equations to solve:
vA - vB = - 4 m/s

26. Example 6 (Continued) vA + 2vB = -5 m/s vA - vB = - 4 m/s vA
1 kg
2 kg
vA + 2vB = -5 m/s
vA - vB = - 4 m/s
Subtract:
0 + 3vB2 = - 1 m/s
vB = m/s
vA2 - ( m/s) = - 4 m/s
Substitution:
vA= m/s
vA - vB = - 4 m/s

27. (0.150 kg)vA+ (2 kg)(40 m/s) = (0.150 kg)(715 m/s)
Example 7. A kg bullet is fired at 715 m/s into a 2-kg wooden block at rest. The velocity of block afterward is 40 m/s. The bullet passes through the block and emerges with what velocity? B A
uB = 0
(0.150 kg)vA+ (2 kg)(40 m/s) = (0.150 kg)(715 m/s)
0.150vA+ (80 m/s) = (107 m/s)
0.150vA = 27.2 m/s)
vA = 181 m/s

28. Example 8a: Inelastic collision: Find vC.B
5 kg
7.5 kg
uB=0
2 m/s
After hit: vB= vA= vC
(5 kg)(2 m/s) = (5 kg kg)vC A B
Common vC after vC
12.5 vC =10 m/s
vC = m/s
In an completely inelastic collision, the two balls stick together and move as one after colliding.

29. Example 8. (b) Elastic collision: Find vA2 and vB2
5 kg
7.5 kg
vB1=0
2 m/s
Conservation of Momentum:
(5 kg)(2 m/s) = (5 kg)vA2 + (7.5 kg) vB A B vA vB
5 vA vB = 10 m/s
For Elastic Collisions:
Continued . . .

30. Example 8b (Cont). Elastic collision: Find vA & vB
Solve simultaneously: A B
5 kg
7.5 kg
vB =0
2 m/s vA vB
x (-5)
5 vA v B = 10 m/s
5 vA vB = 10 m/s
-5 vA vB = +10 m/s
vA m/s = -2 m/s
vA = m/s
12.5 vB = 20 m/s
vB = 1.60 m/s

31. General: Completely Elastic
Collisions where zero energy is lost during a collision (an ideal case).
Conservation of Momentum:
Conservation of Energy:

32. The ballistic pendulum!
Example 9: A 50 g bullet lodges into a 2-kg block of clay hung by a string. The bullet and clay rise together to a height of 12 cm. What was the velocity of the 50-g mass just before entering? uA B A B A
12 cm
The ballistic pendulum!

33. Example (Continued): mAuA+0= (mA+mB)vC vC = 2gh
Collision and Momentum: B A
12 cm
50 g uA
2.05 kg
2 kg
mAuA+0= (mA+mB)vC
(0.05 kg)uA = (2.05 kg)vC
To find vA we need vC .
After collision, energy is conserved for masses.
vC = 2gh

34. Example (Continued): mAuA+0= (mA+mB)vC uA = 62.9 m/s
vC = 2gh = 2(9.8)(0.12) B A
12 cm
50 g uA
2.05 kg
2 kg
After Collision: vC = 1.53 m/s
Momentum Also Conserved:
mAuA+0= (mA+mB)vC
(0.05 kg)uA = (2.05 kg)(1.53 m/s)
uA = 62.9 m/s

35. Conservation of Momentum:
Summary of Formulas:
Conservation of Momentum:
Conservation of Energy:
For elastic only:

36. CONCLUSION: Chapter 9B Conservation of Momentum