1. Momentum and Its Conservation
Physics Chapter 6
Momentum and Its Conservation

2. Linear Momentum
The velocity and mass of an object determine what is needed to change its motion.
Linear Momentum (ρ) is the product of mass and velocity
ρ =mv
Unit is kgm/s

3. Example 1
Example 1. Find the magnitude of the momentum of each cart shown below, before the collision.

4. Example 1. Find the magnitude of the momentum of each cart shown below, before the collision.
Given:
mA = 450 kg mB = 550 kg
nA = 4.50 m/s nB = 3.70 m/s
Find:
ρ A and ρ B
Solution:
ρ A = mAnA = (450 kg)(4.50 m/s) = 2025 = 2.0 x 103 kg . m/s
ρ B = mBnB = (550 kg)(3.70 m/s) = 2035 = 2.0 x 103 kg . m/s

5. Example 2. The man in this picture has a mass of 85. 0 kg
Example 2. The man in this picture has a mass of 85.0 kg. If he strikes the ground at 7.7 m/s, what is his change in momentum (Dρ) Assume his mass remains the same after the collision.
Given:
m = 85.0 kg nf = 0.0 m/s
ni = -7.7 m/s (negative for down)
Find:
D ρ
D ρ = m(nf - ni)
= (85.0 kg)((0.0 m/s – (-7.7 m/s)) = 650 kg . m/s
Solution:

6. If the velocity is changed by an outside force, then the momentum is also changed.
F=ma = mv/t = ( ρ /t)
Ft = m v = mvf-mvi=  ρ Ft =  ρ = “impulse” in N  s
t is the time during which the force is applied.
This is called the “Impulse-Momentum Theorem”.

7. Example 3. A player exerts a force (assume constant) of 12
Example 3. A player exerts a force (assume constant) of 12.5 N on a ball over a period of 0.35 s. Find the total impulse .
Given:
F = 12.5 N Dt = 0.35 s
Find:
Impulse
Solution:
Impulse = FDt = (12.5 N)(0.35 s) = 4.4 N . s

8. Example 4:
Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it without splashing it back (kind of a fake problem, but that’s OK). What is the force exerted by the water on the car each second.
m = 1.5 kg vi=20m/s vf=0m/s t=1s

9. FC means force on water by car (Fw is force on car by water which is equal and opposite)
FC t = ρ
FC= ρ /t
FC = (ρ f – ρi) / t
FC = (mvf – mvi) / t
FC = m (vf – vi) / t
FC = 1.5 kg (0 m/s – 20 m/s) / 1 s
FWC,x = -30 N  FCW,x = 30 N
FC = -30 N  FW = 30 N

10. Stopping Distance
Stopping distance depends on the Impulse-Momentum Theorem.
A heavy object takes longer to stop…
Also, A change in momentum over a longer time requires less force.
Ex: nets, air mattresses, padding (see fig. 6-5)

11. Calculating time and distance
Stopping time can be calculated using Ft =  p, and solving for t.
Stopping distance can be calculated using x = ½(vi+vf)t

12. Example
A kg beryllium sphere traveling to the west slows down uniformly from m/s to m/s. How long does it take the sphere to decelerate if the force on the sphere is 840 N to the east? How far does the sphere travel during the deceleration?

13. First find change in time
m=720.0kg
vi= m/s vf=-12.00m/s
F= +840N
Ft=p
t=p/F=(mvf-mvi)/F
=[(720kgX-12.00m/s)-
(720.0kgX-230.0m/s)]/840N=186.9s

14. Then find change in x x=1/2 (vi+vf)t=
1/2(-230.0m/s-12.00m/s)(186.9s)
= m
= m to the West

15. Newton’s 3rd Law
As two objects collide, the force that one exerts on the other multiplied by the time (Ft) is the impulse for each.
Newton’s 3rd law tells us that the force on mass1 is equal and opposite to the force on mass2 when they collide, which leads to conservation of momentum.
Forces in real collisions are not constant (see fig. 6-9, pg. 220)

16. Conservation of Momentum in Collisions
When two objects collide, the momentum of the individual objects change, but the total momentum remains the same.
p1,i + p2,i = p1,f + p2,f
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Friction is disregarded

17. Example 1
A 76 kg person, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the person moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?

18. List Knowns and Unknowns and formula…
vi,boat=0 vi,person=0
vf,boat=? vf,person=2.5m/s
mperson=76kg mboat=45kg
mpvp,i + mbvb,i = mpvp,f + mbvb,f
0 + 0 = (76kg)(2.5m/s) + (45kg)(vb,f)
Vbf= - (76X2.5)/45 = -4.2m/s
(negative means the boat is moving in the opposite direction)

19. Example 2. A proton with an initial velocity of 235 m/s strikes a stationary alpha particle(He) and rebounds off with a velocity of –188 m/s. Find the final velocity of the alpha particle.
Mass of proton= 1.67 x kg
Mass of alpha particle (He)= 6.64 x kg

20. mp = 1.67 x kg mHe = 6.64 x kg
np,i = 235 m/s nHe,i = 0.0 m/s np,f = -188 m/s
Given:
Find:
nHe,f
Original Formula:
Now, the initial velocity of He (alpha particle) is zero,
so cross it out!

21. mp = 1.67 x kg mHe = 6.64 x kg
np,i = 235 m/s nHe,i = 0.0 m/s np,f = -188 m/s
Given:
Find:
nHe,f
Original Formula:
Solve for
nHe,f

22. Working Formula:

23. np = 235 m/s nHe = 0.0 m/s np,f = -188 m/s Given:
mp = 1.67 x kg mHe = 6.64 x kg
np = 235 m/s nHe = 0.0 m/s np,f = -188 m/s
Given:
Find:
nHe,f
Working Formula:
The mass of the proton doesn’t change, so

25. Perfectly Inelastic Collisions
A collision is perfectly inelastic when they collide and move together as one mass with a common velocity.
m1v1,i + m2v2,i = (m1 + m2)vf

26. Example 1. A moving railroad car, mass=M, speed=3m/s, collides with an identical car at rest. The cars lock together as a result of the collision. What is their common speed afterward?

27. Vi1
Vi2=0 M
before
Vf? M
after

28. x
Vi1
Vi2=0 M
before
Vf? M
after
p1i = p2f
M(+Vi1) + 0= (M+M)(Vf)
Vf = MVi1 / 2M = Vi1 / 2
Vf = (3m/s) / 2 = 1.5m/s

29. Kinetic Energy and Perfectly Inelastic Collisions
During perfectly inelastic collisions, some kinetic energy is lost because the objects are deformed.
Once mass and velocities are known (using conservation of momentum), solve for initial and final kinetic energies (1/2mv2) using the formula from Ch. 5 to determine energy lost.
KEi = KE1,i + KE2,i
KEf = KE1,f + KE2,f
KE = KEf - KEi

30. Example 1. A red clay ball with a mass of 1
Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.
Given:
mr = 1.5 kg mb = 2.8 kg nr,i = 4.5 m/s nb,i = 0 m/s
Find: nf
Original Formula:

31. Example 1. A red clay ball with a mass of 1
Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.
Given:
mr = 1.5 kg mb = 2.8 kg nr,i = 4.5 m/s nb,i = 0 m/s
Find: nf
Original Formula:
The black ball starts at rest, so

32. Example 1. A red clay ball with a mass of 1
Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.
Given:
mr = 1.5 kg mb = 2.8 kg nr,i = 4.5 m/s nb,i = 0 m/s
Find: nf
Original Formula:
The black ball starts at rest, so

33. Example 1. A red clay ball with a mass of 1
Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.
Given:
mr = 1.5 kg mb = 2.8 kg nr = 4.5 m/s nb = 0 m/s
Find: nf
Working Formula:

34. Example 1. A red clay ball with a mass of 1
Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.
Given:
mr = 1.5 kg mb = 2.8 kg nr,i = 4.5 m/s nb,i = 0 m/s
Find: nf
Solution:

35. Recall the formula for kinetic energy.
Was the kinetic energy conserved in example 1?

36. KEbefore = 15 J

38. KEbefore = 15 J
KEafter = 5.3 J
KE = 5.3 J-15 J = -9.7 J

39. Where does the missing kinetic energy go?
Some energy is converted to internal energy,
some is converted into sound.

40. Elastic Collisions
Two objects collide and return to their original shape with no loss of Kinetic Energy.
The two objects move separately after the collision.
Total momentum and KE remain constant. See page 230.
Most collisions lose energy as sound, internal elastic potential energy and friction…we assume perfect collisions.

41. Formula for elastic collisions
m1v1,i + m2v2,i = m1v1,f + m2v2,f
½ m1v1,i2 + ½ m2v2,i2 = ½ m1v1,f2 + ½ m2v2,f2

42. Example 1. A kg blue marble sliding to the right on a frictionless surface with a velocity of 17 m/s makes an elastic head-on collision with kg red marble moving to the left with a velocity of 15 m/s. After the collision, the
blue marble moves to the left with a velocity of 15.0 m/s. Find the velocity of the red marble.

43. Example 1.
Given:
mb = kg mr = kg nb,i = 17 m/s
nr,i, = -15 m/s nb,f = -15 m/s
Find:
nr,f
Original Formula:

44. =17m/s

45. Now…you can plug the numbers into the conservation of kinetic energy formula to check your work which they will ask you to do!
(Remember – there is NO LOSS OF KINETIC ENERGY with elastic collisions!!)
½ m1v1,i2 + ½ m2v2,i2 = ½ m1v1,f2 + ½ m2v2,f2

46. ½ m1v1,i2 + ½ m2v2,i2 = ½ m1v1,f2 + ½ m2v2,f2 6.425J = 6.425J
½ (.025)(17)2+ ½ (.025)(-15)2 = ½ (.025)(-15)2+ ½ (.025)(17)2
6.425J = J