1. Momentum and Collisions
Dr. Robert MacKay
Clark College, Physics

2. Introduction Review Newtons laws of motion Define Momentum
Define Impulse
Conservation of Momentum
Collisions
Explosions
Elastic Collisions

3. Introduction Newtons 3 laws of motion 1. Law of inertia
2. Net Force = mass x acceleration
( F = M A )
3. Action Reaction

4. Law of interia (1st Law)
Every object continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it.
acceleration = 0.0 unless the objected is acted on by an unbalanced force

5. Newton’s 2nd Law
Net Force = Mass x Acceleration
F = M A

6. Newton’s Law of Action Reaction (3rd Law)
You can not touch without being touched
For every action force there is
and equal and oppositely directed reaction force

7. Newton’s Law of Action Reaction (3rd Law)
Ball 1
Ball 2
F2,1
F1,2
F1,2 = - F2,1
For every action force there is
and equal and oppositely directed reaction force

8. Momentum , p Momentum = mass x velocity is a Vector
has units of kg m/s

9. Momentum , p (a vector) Momentum = mass x velocity p = m v p = ?
8.0 kg
6.0 m/s

10. Momentum , p Momentum = mass x velocity p = m v p = 160.0 kg m/s

11. Momentum , p Momentum is a Vector p = m v p1 = ? p2 = ? V= +8.0 m/s
7.5 kg
m2=
10.0 kg
V= -6.0 m/s

12. Momentum , p Momentum is a Vector p = m v
p1 = +60 kg m/s p2 = - 60 kg m/s
V= +8.0 m/s
m1=
7.5 kg
m2=
10.0 kg
V= -6.0 m/s

13. Momentum , p Momentum is a Vector p = m v
p1 = +60 kg m/s p2 = - 60 kg m/s
the system momentum is zero.,
V= +8.0 m/s
m1=
7.5 kg
m2=
10.0 kg
V= -6.0 m/s

14. Newton’s 2nd Law Net Force = Mass x Acceleration F = M a F = M (∆V/∆t)
F ∆t = M ∆V
F ∆t = M (VF-V0)
F ∆t = M VF- M V0
F ∆t = ∆p Impulse= F∆t
The Impulse = the change in momentum

15. Newton’s 2nd Law Net Force = Mass x Acceleration
F ∆t = ∆p Impulse= F ∆t
The Impulse = the change in momentum

16. Newton’s Law of Action Reaction (3rd Law)
Ball 1
Ball 2
F2,1
F1,2
F1,2 = - F2,1
For every action force there is
and equal and oppositely directed reaction force

17. Newton’s Law of Action Reaction (3rd Law)
Ball 1
Ball 2
F2,1
F1,2
F1,2 = - F2,1
F1,2∆t = - F2,1 ∆t
∆p2 = - ∆p1

18. Conservation of momentum
Ball 1
Ball 2
F2,1
F1,2
If there are no external forces acting on a system (i.e. only internal action reaction pairs), then the system’s total momentum is conserved.

19. “Explosions” 2 objects initially at rest
A 30 kg boy is standing on a stationary 100 kg raft in the middle of a lake. He then runs and jumps off the raft with a speed of 8.0 m/s. With what speed does the raft recoil?
V=8.0 m/s
after
V=?
M=100.0 kg
M=100.0 kg
before

20. “Explosions” 2 objects initially at rest
A 30 kg boy is standing on a stationary 100 kg raft in the middle of a lake. He then runs and jumps off the raft with a speed of 8.0 m/s. With what speed does the raft recoil?
V=8.0 m/s
after
V=?
M=100.0 kg
M=100.0 kg
before
p before = p after
= 30kg(8.0 m/s) kg V
100 kg V = 240 kg m/s
V = m/s

21. Explosions
If Vred=9.0 m/s
Vblue=?
9.0 m/s

22. Explosions
If Vred=9.0 m/s
Vblue=3.0 m/s
9.0 m/s
3.0 m/s

23. “Stick together” 2 objects have same speed after colliding
A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How fast does he and the raft move immediately after the collision?
V=8.0 m/s
V=?
M=100.0 kg
M=100.0 kg
before
after

24. “Stick together” 2 objects have same speed after colliding
A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How fast does he and the raft move immediately after the collision?
V=8.0 m/s
V=?
M=100.0 kg
M=100.0 kg
before
after
p before = p after
30kg(8.0 m/s) = 130 kg V
240 kg m/s = 130 kg V
V = m/s

25. “Stick together” 2 objects have same speed after colliding This is a perfectly inelastic collision
V=8.0 m/s
V=?
M=100.0 kg
M=100.0 kg
before
after
A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How fast does he and the raft move immediately after the collision?

26. A 20 g bullet lodges in a 300 g Pendulum
A 20 g bullet lodges in a 300 g Pendulum. The pendulum and bullet then swing up to a maximum height of 14 cm. What is the initial speed of the bullet?

27. Before and After Collision Before and After moving up
mv = (m+M) V
Before and After Collision
1/2(m+M)V2=(m+M)gh
After collision but
Before and After moving up

28. 2-D Stick together (Inelastic)
Momentum Before = Momentum After
P before= P after
For both the x & y components of P.
A 2000 kg truck traveling 50 mi/hr East on McLoughlin Blvd collides and sticks to a 1000 kg car traveling 30 mi/hr North on Main St. What is the final velocity of the wreck? Give both the magnitude and direction OR X and Y components.

29. 2-D Stick together (Inelastic)
A 2000 kg truck traveling 50 mi/hr East (V1) on McLoughlin Blvd collides and sticks to a 1000 kg car traveling 30 mi/hr North (V2) on Main St. What is the final velocity (V) of the wreck? Give both magnitude and direction OR X and Y components.
Pbefore=Pafter
PBx=PAy & PBy=PAy
2000Kg(50 mi/hr)=3000KgVx & 1000kg(30mi/hr)=3000kgVy
Vx=33.3 mi/h & Vy=10 mi/hr Or
V= 34.8mi/hr = (sqrt(Vx2+Vx2) & q =16.7° = tan-1(Vy/Vx) V V1
3000Kg
2000Kg V2
1000Kg

30. Elastic Collisions Bounce off without loss of energy
p before = p after
&
KE before = KE after m1 m2 m2 m1 v1
v1,f
v2,f

31. Elastic Collisions Bounce off without loss of energy
p before = p after & KE before = KE after
&
if m1 = m2 = m, then
v1,f = & v2,f = v1 m m m m v1
v2,f = v1
v1,f= 0.0

32. Elastic Collisions Bounce off without loss of energy
p before = p after & KE before = KE after
&
if m1 <<< m2 , then m1+m2 ≈m & m1-m2 ≈ -m2
v1,f = - v & v2,f ≈ 0.0 M M m m v1
v2,f ≈ 0.0
v1,f=- v1

33. Elastic Collisions Bounce off without loss of energy
if m1 <<< m2 and v 2 is NOT 0.0
Speed of Approach = Speed of separation
(True of all elastic collisions) M M m m v1 v2
v1,f=- (v1 +v2 +v2)
v2,f ≈ v2

34. M M Elastic Collisions Speed of Approach = Speed of separation m
if m1 <<< m2 and v 2 is NOT 0.0
Speed of Approach = Speed of separation
(True of all elastic collisions)
A space ship of mass 10,000 kg swings by Jupiter in a psuedo
elastic head-on collision. If the incoming speed of the ship
is 40 km/sec and that of Jupiter is 20 km/sec, with what speed
does the space ship exit the gravitational field of Jupiter? m
v1,f=?
v2=20 km/s M M m
v2,f ≈ ?
v1 = 40 km/s

35. M M Elastic Collisions Speed of Approach = Speed of seperation v1,f=?
if m1 <<< m2 and v 2 is NOT 0.0
Speed of Approach = Speed of seperation
(True of all elastic collisions)
A little boy throws a ball straight at an oncoming truck
with a speed of 20 m/s. If truck’s speed is 40 m/s and the
collision is an elastic head on collision, with what speed does
the ball bounce off the truck?
v1,f=? M m
v2=40 m/s M m
v1 = 20 m/s
v2,f ≈ ?

36. Elastic Collisions Bounce off without loss of energy
p before = p after & KE before = KE after
if m1 = m2 = m, then
v1,f = & v2,f = v1 m m m
90° m v1

37. Elastic Collisions Bounce off without loss of energy
p before = p after & KE before = KE after
if m1 = m2 = m, then
v1,f = & v2,f = v1
p1f
p2f
90° m m m
90° p1 m v1
p1 =p1f +p2f

38. The End