1. 5.2 Conserving momentum 1 2 3 4 5 Snooker Collisions ‘Explosions’
Check-point 3
Conservation of momentum
Check-point 4
Check-point 5
Examples of conservation of momentum
Momentum not conserved?
Check-point 6 1 2 3 4 5

2. Then, he pots the black one.
Snooker
A player pots the pink ball and the white ball stops after the hitting (stop shot).
Then, he pots the black one.
Could he use a stop shot again?
No.
Why?
Video
5.3 Snooker

3. Two laws of conservation are related to collisions:
momentum
kinetic energy
Simulation
5.2 Inelastic collisions of trolleys
Expt 5b
Sticky crash

4. 1 Collisions
v-t graph of the trolley A in Expt 5b:

5. > KE after collision KE is not conserved.
1 Collisions
Typical result:
0.38 m s–1
0.51 m s–1
total KE = 0.14 J total p = 0.76 kg m s–1
total KE = 0.20 J total p = 0.77 kg m s–1
KE before collision
> KE after collision
KE is not conserved.
But total momentum is conserved.

6. 1 Collisions
Expt 5c
Hard crash

7. 1 Collisions
v-t graphs of the trolleys in Expt 5c:

8. > KE after collision KE is not conserved, but…
1 Collisions
0.21 m s–1
Typical result:
0.09 m s–1
0.67 m s–1
total KE = 0.11 J total p = 0.34 kg m s–1
total KE = 0.04 J total p = 0.36 kg m s–1
KE before collision
> KE after collision
KE is not conserved, but…
energy loss is less than that in sticky crash
Total momentum is conserved.

9. 1 Collisions Bouncy crash 5.3 Elastic collisions of trolleys
Simulation
5.3 Elastic collisions of trolleys
Expt 5d
Bouncy crash

10. v-t graph of the trolleys in Expt 5d:
1 Collisions
v-t graph of the trolleys in Expt 5d:
trolley A pushed to move from rest
vA after collision
collision
B stationary before collision
vB after collision

11. KE is conserved in elastic collision
1 Collisions
0.90 m s–1
0.40 m s–1
Typical result:
0.70 m s–1
total KE = 0.32 J total p = 1.05 kg m s–1
total KE = 0.37 J total p = 1.05 kg m s–1
KE before collision
= KE after collision
KE is conserved in elastic collision
Total momentum is also conserved.

12. 1 Collisions
In Expt 5b, 5c and 5d:
Total momentum are conserved.
Total KE unchanged
elastic collision (Expt 5d)
Total KE changed
inelastic collision (Expt 5b, 5c)
Colliding objects stick together
completely inelastic collision (Expt 5b)

13. A ball bounces back to the same height h.
1 Collisions
ball
A ball bounces back to the same height h.
No energy loss during collision h
completely elastic collision

14. Wet towel sticks to the ground (not bounces back)
1 Collisions
Wet towel sticks to the ground (not bounces back)
wet towel
KE of the towel is lost during collision (max KE loss)
completely inelastic collision

15. a sudden separation of an object into pieces
2 ‘Explosions’
Explosion:
a sudden separation of an object into pieces
Simulation
5.4 ‘Explosion’ of trolleys
Expt 5e
‘Explosion’ of trolleys

16. v-t graph of the trolleys in Expt 5e:
2 ‘Explosions’
v-t graph of the trolleys in Expt 5e: vA
explosion
Graphs of both trolleys are -ve,
since each moves towards its motion sensor. vB

17. < KE after ‘explosion’
2 ‘Explosions’
–0.22 m s–1
Typical result:
0.68 m s–1
total KE = 0 total p = 0
total KE = J total p = 0.01 kg m s–1
KE before ‘explosion’
< KE after ‘explosion’
KE is not conserved in ‘explosion’
Total momentum is conserved.

18. Total momentum conserved?
2 ‘Explosions’
Conservation of total momentum and total KE in different types of interaction:
Total momentum conserved?
Total KE conserved?
Completely inelastic collision
Inelastic collision
Elastic collision
explosion        
Simulation
5.5 Exploring collisions

19. Identify the type of the following collisions.
Check-point 3 – Q1
Identify the type of the following collisions.
Total KE before collision
before
5 m s–1
4 m s–1
= (2   42) 1 2
ball B
ball A
2 kg
1 kg
= 33 J
after
1 m s–1
4 m s–1
Total KE after collision
ball B
ball A
2 kg
 (2   42) 1 2 =
1 kg
= 9 J
______ (All/Some/No) KE is lost in the collision. The collision is ________.
Some
inelastic

20. Total KE before collision
Check-point 3 – Q1
Total KE before collision
before
2 m s–1
at rest 2 1
= (2   02)
ball A
2 kg
ball B
1 kg
= 4 J
after
Total KE after collision
2/3 m s–1
8/3 m s–1
 [2  ( )2 + 1  ( )2] 2 3 1 8
ball A
2 kg =
1 kg
ball B
= 4 J
_____(All/Some/No) KE is lost in the collision. No
The collision is ________.
elastic

21. 3 Conservation of momentum
a Collisions along a straight line
Total momentum is conserved in
Expt 5b: Inelastic collision
Expt 5c, 5d: Elastic collision
Expt 5e: ‘Explosion’ of trolleys
Total momentum is conserved in all cases.

22. a Collisions along a straight line
Law of conservation of momentum:
The total momentum of a system is conserved, provided that there is no external net force acting on the system.
It can be expressed mathematically:
mAuA + mBuB = mAvA + mBvB

23. a Collisions along a straight line
Law of conservation of momentum is closely related to Newton’s third law of motion.
Consider 2 objects in a head-on collision:
during collision (in time interval t )
after collision
before collision
FAB (a force on A by B)
FBA (a force on B by A)

24. a Collisions along a straight line
FAB and FBA is an action-and-reaction pair.
By Newton’s third law,
FAB = –FBA
If no external force acting on the objects,
mA (vA – uA) t
= –
mB (vB – uB)
mAuA + mBuB = mAvA + mBvB
law of conservation of momentum
Example 7
Ice hockey

25. a Collisions along a straight line
Example 8
Skating

26. a Collisions along a straight line
Example 9
A stop shot

27. Check-point 4 – Q1
If no ________ force acts on colliding objects, the total __________ of the colliding objects is conserved.
external
momentum
The _______ energy of colliding objects in elastic collisions is conserved. But it is not in ________ collisions.
kinetic
inelastic

28. Mike and Roger collide when they are skating.
Check-point 4 – Q2
Mike and Roger collide when they are skating.
If they ‘stick’ together after collision, their velocity after collision = ?

29. Take the direction towards the right as +ve.
Check-point 4 – Q2
Take the direction towards the right as +ve.
Let v be the velocity after collision.
Since no __________ force acts on the boys,
external
momentum
by conservation of __________ ,
total p before collision = total p after collision
40   (−4) = ( )  v
v = 0.57 m s–1

30. A 1500-kg car accelerated uniformly at 1 m s–2.
Check-point 4 – Q3
A 1500-kg car accelerated uniformly at 1 m s–2.
After travelling 50 m, it collided with a stationary lorry of mass 2500 kg.
The vehicles were stuck together after the collision.

31. (a) Velocity of the car just before collision = ?
Check-point 4 – Q3
(a) Velocity of the car just before collision = ?
By v 2 – u 2 = 2as,
v 2 – 0 = 2  1  50
v = 10 m s–1

32. (b) Velocity of both vehicles after collision = ?
Check-point 4 – Q3
(b) Velocity of both vehicles after collision = ?
total p before collision = total p after collision
mCuC + mLuL = (mC + mL)  v
1500  = ( )  v
v = 3.75 m s–1

33. 3 Conservation of momentum
b collisions on a plane
Oblique collisions (collisions on a plane) are more common.
In solving problems about oblique collisions,
momenta of the colliding objects are resolved into 2  components.
Example 10
Bowling and a pin

34. b Collisions on a plane
Example 11
Pot the black ball

35. P, Q and R are identical balls.
Check-point 5 – Q1
P, Q and R are identical balls.
P : 3 m s–1
Q : 4 m s–1
R : stationary
P and Q collide with R at the same time and stick together.
Speed of the combined mass = ? 5 3 7 3
A m s–1
B m s–1
C m s–1
D m s–1

36. List the equations for the conservation of momentum:
Check-point 5 – Q2
List the equations for the conservation of momentum:
x -direction:
m1u1cos 1 – m2u2cos  = –m1v1cos 1 + m2v2cos 2
y -direction:
m1u1sin 1 + m2u2sin  = m1v1sin 1 + m2v2sin 2

37. 4 Examples of conservation of momentum
a 'Newton’s cradle'
Newton’s cradle:
A toy consisting of a line of 5 steel balls.
can demonstrate the law of conservation of momentum
Simulation
5.6 ‘Newton’s cradle’
Expt 5f
‘Newton’s cradle’

38. When 1 to 4 balls are released on one side of the Newton’s cradle,
a 'Newton’s cradle'
When 1 to 4 balls are released on one side of the Newton’s cradle,
same number of balls rise to the same height on the other side after collision  
momentum is conserved

39. 4 Examples of conservation of momentum
b Recoil of cannons and guns
When a cannonball is fired,
recoil velocity v
by conservation of momentum, the cannon recoils.
forward + backward momentum = 0
The massive base reduces recoil velocity of the cannon.

40. b Recoil of cannons and guns
5.7 Adding mass while moving
Simulation
Simulation
5.8 Shedding mass while moving
Video
5.9 Conservation of momentum

41. b Recoil of cannons and guns
When a bullet is fired, the rifle recoils.
recoil velocity v
forward + backward momentum = 0
total momentum = 0 
Conservation of momentum
Example 12
Recoil velocity of an air-rifle

42. 4 Examples of conservation of momentum
c Spacewalk
Gaseous nitrogen is ejected from the manned maneuvering unit (MMU) of an astronaut.
 The astronaut moves in opposite direction.
Astronauts can control their moving direction using nozzle thrusters at different locations on the MMU.

43. 4 Examples of conservation of momentum
d Bombs
When a bomb explodes,
by conservation of momentum,
the debris would fly such that its total horizontal momentum remains zero.

44. 5 Momentum not conserved?
running
stop
mass = m
mass = m
momentum = 0
momentum = mv
Total momentum of you and the earth is not conserved!
Why?

45. 5 Momentum not conserved?
This is because mass of the earth is huge.
m = 6  1024 kg
The motion of the earth is not noticeable.

46. 5 Momentum not conserved?
A ball hits a wall fixed to the ground and rebounds with a change in momentum.
∵ Mass of the earth is too huge.
 The wall and the earth do not have noticeable movement.
Example 13
High diving

47. This is because ____________ force (i.e. weight) acts on the book.
Check-point 6 – Q1
A book is dropped from a height. Why is the momentum of the book not conserved?
gravitational
This is because ____________ force (i.e. weight) acts on the book.
This force accelerates the book downwards and changes its momentum.

48. A One ball on side B rises up to 2h.
Check-point 6 – Q2
Which statement is true when two balls are pulled to side A and released?
A One ball on side B rises up to 2h.
B Two balls on side B rise up to h.
C One ball on side A rises up 2h.
D Two balls on side A rise up to . h 2

49. The End