Transcript dynamics - WordPress.com

Types of force
Direction
of force
Friction
Weight
Tension in string
Thrust in rod
Friction
Resolving forces into components
Why do some cricket bowls
bounce higher than others?


A A  CH means v  v cos 

C H
v cos
v 
v
vv 
sin 
O
O  SH means v  v sin 
S H
The velocity of the ball has a horizontal and a vertical component
Resolving (separating) these components helps you understand the trajectory of the ball
Resolving forces into components
Consider a block on a slope:
The weight of the block has:
 mg cos 
a component acting perpendicular to the slope
a component acting parallel to the slope
T

T cos
T sin 

mg
mg sin 
Consider a block being pulled along a surface:
The tension of the wire exerts:
a component acting vertically on the block
a component acting horizontally on the block
When you intend to resolve forces, you can use the following notation:
R
 resolve parallel to the slope
R  resolve perpendicular to the slope
R
 resolve horizontally
R  resolve vertically
Resultant forces
The combined effect of any number of forces are called the resultant force
5N
R
16 N
8N
R
 16  5  11N
 8  3  5N
3N
6N
8N
45
What direction would the resultant force act in?
30

2.685...N
R
 8 cos 30  6 cos 45  2.685...N
R  8 sin 30  6 sin 45  8.242...N
8.242...N
tan  
8.242...
2.685...
   71.953...
Now try Ex3C, Q2
Newton’s laws of motion
Newton’s 2nd law states that the force needed to accelerate a particle is
equal to the product of the particle’s mass and the acceleration produced.
F  ma
Eg 6kg block is being pulled by a
cable with tension 12N.
The surface exerts a resistance of 4N.
Find the acceleration of the block
a m s-2
4N
6 kg
If the resultant force is zero,
the object will not move
Eg a 70kg skydiver is falling at
terminal velocity. Write down the force
caused by air resistance
FN
12 N
0 m s-2
R
R
 12  4  6a
a
4
m s-2
3
 70 g  F  70  0
 F  70 g
70 g N
Now try Ex3A, Q7-10
Particles on slopes
One of the common scenarios you
encounter is a particle on a slope.
mg
By resolving the weight into components
acting parallel and perpendicular to the slope
before applying Newton’s law F = ma, the
acceleration of the particle can be determined.
Eg a block released on a smooth slope

 mg cos 
mg sin 
Eg a block released on a
rough slope with friction F N
F
mg sin 
mg sin 

R


mg sin   ma
 a  g sin 
R
 mg sin   F  ma
a
mg sin   F
m
Normal reaction
The normal reaction is the force which acts perpendicular to a surface when
an object is contact with the surface. This must be equal to the resultant
force an object is applying to the surface, as the object is not accelerating.
Eg a block at rest on a slope
Eg a block at rest on a surface,
despite being pulled by a string
R
R

mg

T

T sin 
mg cos 
mg
R

R  mg cos   0
 R  mg cos 
R

R  T sin   mg  0
 R  mg  T sin 
Friction
The coefficient of friction, μ measures the roughness of 2 surfaces
in contact. The larger the value of μ, the greater the friction.
As the applied force increases, so does the force of friction resisting movement,
up to a limiting value. If the applied force exceeds this limit, the block will move.
The limiting value depends on the normal reaction and
coefficient of friction between 2 surfaces, so that Fmax  R
Hence F  R
Will it move?
R
  0.4
A block of mass 5kg lies at rest on a rough
surface with a coefficient of friction of 0.4
A horizontal force of P N is applied to the block
P
5kg
Find the size of the frictional force and where
applicable, acceleration of the block when P is:
a) 10N
b) 19.6N
c) 30N
a)
R
 R  5 g  Fmax  2g  19.6N
R  P is less than F , so the block will not move
F
5g
max
b)
P is equal to Fmax, so the block will not move but is in limiting equilibrium
c)
P is greater than Fmax, so the block will move
R

30 19.6  5a  a 
10.4
 2.08 ms -2
5
Now try Ex3D
The typical scenario
For a block that is about to slide down a slope:
The weight of the block has components acting parallel & perpendicular to the slope
The slope in turn applies a normal reaction equal to the perpendicular component
Proportional to this reaction, friction acts up the slope up to a
limit equal to the component of weight acting down the slope
R
Weight
R
Reaction
Friction
 mg cos 
Acceleration
mg

mg sin 
Eg A particle P of mass 3 kg is projected up a line of greatest slope of a rough plane
inclined at an angle of 30° to the horizontal. The coefficient of friction between P and
the plane is 0.4. The initial speed of P is 6 m s-1. Find
(a) the frictional force acting on P as it moves up the plane,
6 ms -1
R
R
F
 R  3 g cos 30  3 23 g
F  R  3 5 3 g  10 N (2sf)
30
3 g cos 30
3g
30
3 g sin 30
(b) the distance moved by P up the plane before P comes to instantaneous rest.
R

F  3 g sin 30  3a
R
g  3 g sin 30
a
3
 8.294...
 0  62  2  8.294...  s
62
s
 2.170.. m
2  8.294...
3 3
5

v 2  u 2  2as
WB10 The diagram above shows a boat B of mass 400 kg held at rest on a slipway
by a rope. The boat is modelled as a particle and the slipway as a rough plane
inclined at 15° to the horizontal. The coefficient of friction between B and the slipway
is 0.2. The rope is modelled as a light, inextensible string, parallel to a line of
greatest slope of the plane. The boat is in equilibrium and on the point of sliding
down the slipway.
(a) Calculate the tension in the rope.
R
R
 R  400 g cos 15
F  R  80 g cos 15
R
 T  F  400 g sin 15
 T  400 g sin 15  80 g cos 15
B
F
50 m
15°
T
400 g
 260 N (2sf)
The boat is 50 m from the bottom of the slipway.
The rope is detached from the boat and the boat slides down the slipway.
(b) Calculate the time taken for the boat to slide to the bottom of the slipway.
R

400 g sin 15  F  400a
 a  0.643...ms 2
s  ut  21 at 2  50  21  0.643...  t 2
t 
250
0.643 ...
 12 s (2sf)
WB11 A small parcel of mass 2 kg moves on a rough plane inclined at an angle of
30° to the horizontal. The parcel is pulled up a line of greatest slope of the plane by
means of a light rope which is attached to it.
The rope makes an angle of 30° with the plane, as shown in the diagram above.
The coefficient of friction between the parcel and the plane is 0.4.
Given that the tension in the rope is 24 N,
(a) find, to 2 significant figures,
the acceleration of the parcel.
24
R
30°
F
30°
R
2g
 R  24 sin 30  2 g cos 30
 R  3 g  12
F  R  52
R
 3g 12
 24 cos 30  2 g sin 30  F  2  a

 a  21 12 3  g  52
 3g 12  4.5 ms
-2
(2sf)
The rope now breaks. The parcel slows down and comes to rest.
(b) Show that, when the parcel comes to this position of rest, it
immediately starts to move down the plane again.
R

R  2 g cos 30  3 g
F  R  Fmax  R 
R
2 3
5
 W  2 g sin 30  g
R
g  6.78...N
 9.8N
 W  Fmax
F30°
W
30°
2g
 net force down plane
 parcel moves
(c) Find, to 2 significant figures, the acceleration of the parcel as it moves
down the plane after it has come to this position of instantaneous rest.
R

W  F  2 a


 a  21 g  2 53 g  1.5 ms -2 (2sf)
Now try Ex3E
Ex 3E
Q1)
R

1
2
g sin 20  21 a
 a  g sin 20
 3.4 ms -2 (2sf)
Q2)
R
20
 R  2 g cos   20 sin   0
20 N
 R  2g  54  20  35

 28 N (2sf)
R

5
20 cos   2 g sin   2  a
 a  21 20  54  2 g  35 
 2.1 ms (2sf)
-2

4
tan  
3
3
4
 sin   35
 cos  
4
5
Q5)
R

R  2 g cos 20
 Fmax  2g cos 20 (1)
R

2 g sin 20  Fmax  2 1.5 (2)
 ?
20
2 g sin 20  2 g cos 2  3
(1) + (2)
2 g sin 20  3
 0.20 (2sf)

2 g cos 20
Q6)
R
R

R  4 g cos 25
30 N
 Fmax  4 g cos 25 (1)
 30  4 g sin 25  Fmax
 4  2 (2)
(1) + (2) 30  4 g sin 25  4 g cos 25  8
22  4 g sin 25
 0.15 (2sf)

4 g cos 25
25
Q3)
R

R  40 g cos 20
 Fmax  4 g cos 20
R
 40g sin 20  Fmax  40  a
 a  401 40 g sin 20  4 g cos 20
  0.1
20
 2.430... ms -2
R
 v 2  u 2  2as
 02  2  2.430...  5
 v  4.9 ms -1 (2sf)
Q4)
R
v  u  at  21  0  6a  a  3.5
 20 g sin 30  R  20  3.5  28N
30
Q7)
s  ut  21 at 2
 4  0  21  a  22
a 2
R

25
R  10 g cos 25
 Fmax  10 g cos 25 (1)
R
(1) + (2)
 10 g sin 25  Fmax
 10  2 (2)
10 g sin 25  10 g cos 25  20

10 g sin 25  20
 0.24 (2sf)
10 g cos 25
Q8)
R

R  mg cos   54 mg
20 ms 1
 Fmax  154 mg
R

B
mg sin   Fmax  ma
 a   35  154 g  13
15 g
A

 35 mg  154 mg  ma
5

4
v 2  u 2  2as
 0 2  20 2  2  13
15 g  s
202
s
 23.547... m
13
2  15 g
v  u  at  0  20 
13
15
  31
20
gt  t  13  2.4 m (2sf)
15 g
tan  
3
3
4
 sin   35
 cos  
4
5
Finding the value of μ
If no other forces but friction and weight are acting on a particle, the coefficient
of friction can be determined by raising a slope until the block slides
R
R
 R  mg cos   0 1
 mg sin   R  0 2
R
R
1  R  mg cos 
 mg cos 
mg sin 
 tan 
2   
R
mg

Eg a block can be raised to an
angle of 45o before sliding, what
is the coefficient of friction?
  tan 45  1
mg sin 
WB12 A particle P of mass 0.4 kg is moving
under the action of a constant force F newtons.
Initially the velocity of P is (6i – 27j) m s–1 and 4 s
later the velocity of P is (-14i + 21j) m s–1.
(a) Find, in terms of i and j, the acceleration of P.
a
v  u  14i  21j  6i  27 j  20i  48 j


  5i  12j ms -2
4
t
4
(b) Calculate the magnitude of F.
F  ma  0.4   5i  12j   2i  4.8 j N
F  22  4.82  5.2 N
Connected particles moving in the same direction
Problems concerning connected particles moving in the same direction
should usually be solved by considering the particles separately
Eg Two particles P and Q, of masses
5kg and 3kg respectively, are
connected by a light, inextensible
string. Particle P is pulled by a
horizontal force of 40N along a rough
horizontal plane. The coefficient of
friction is 0.2 and the string is taut.
5g
3g
Q
T
3 kg
3
5
P
T
5 kg
40
g
g
5g
3g
Find the acceleration of each particle and the tension in the string
(1) + (2)
For Q:
R

T  35 g  3a (1)

 a  81 40  85 g  3.0 ms -2 (2sf)
Sub in (1) T  3a  35 g  15 N (2sf)
For P:
R
 40  85 g  8a
40  g  T  5a (2)
Now try Ex3F, Q1-3
0.4 m s -2
Ex 3F Q1
R
R
For P:
For Q:
 F T  8  0.4 (1)
 T  2 0.4 (2)
Q
T
P
T
2 kg
8 kg
(2)  T  0.8N
Sub T = 0.8 in (1)
Q2
For P:
For Q:
 F  3.2  0.8  4 N
 200  5 g  R  T  5 1.2 (1)
R  T  5 g  R  5  1.2 (2)
200
R
(1)  (2)  200  10 g  2 R  12
R
5g
 R  21 200  10 g  12  45N
Sub R = 45 in (2)  T  6  5 g  45
P 5 kg
 100N
T
T
Q 5 kg
5g
R
1.2 m s-2
F
a m s-2
Q3
T
T
500 kg
For the trailer:
10000
3000
1000
For the car:
1500 kg
R
 10000  3000  T  1500a (1)
R  T 1000  500a (2)
(1)  (2)  6000  2000a
 a  3 ms -2
Sub a = 3 in (2)
T  1500  1000  2500 N
NB: if the tensions are acting in opposite directions then the whole system
can be considered as one particle, sometimes making the algebra simpler
a m s-2
For the whole system:
T
500 kg
1000
4000
T
2000 kg
3000
1500 kg
10000
R
 10000  4000  2000a
 a  3 ms -2
WB16 This figure shows a lorry of mass 1600 kg towing a car of mass 900 kg along
a straight horizontal road. The two vehicles are joined by a light towbar which is at
an angle of 15° to the road. The lorry and the car experience constant resistances
to motion of magnitude 600 N and 300 N respectively. The lorry’s engine produces
a constant horizontal force on the lorry of magnitude 1500 N.
T
T
1500N
15°
300N
900 g
600N
Find
(a) the acceleration of the lorry and the car,
R
 1500  300  600  900  1600a
 600  2500a
 a  0.24 ms 2
1600 g
(b) the tension in the towbar.
For the car,
R
 T cos15  300  900 0.24
516
 534N (3sf)
 T  cos
15
When the speed of the vehicles is 6 m s–1, the towbar breaks. Assuming that the
resistance to the motion of the car remains of constant magnitude 300 N,
(c) find the distance moved by the car from the moment the towbar breaks to the
moment when the car comes to rest.
After towbar brakes, for the car:
 31 m s-2
T
6ms -1
R
T
 a   31 ms -2
15°
300N
  300  900 a
v 2  u 2  2as  02  62  2   31  s
900 g
 s  362 3  54m
(d) State whether, when the towbar breaks, the normal reaction of the road on the
car is increased, decreased or remains constant. Give a reason for your answer.
R
Before
After
T
15°
900 g
The vertical component of T
is no longer present, so the
normal reaction increases
Now try Ex3F
Connected particles moving in different directions
Problems concerning connected particles moving in different
directions must be solved by considering the particles separately
Eg two particles connected by a light, inextensible string over a smooth pulley
Find the acceleration of the particles
For P:
For Q:
R
 T  2mg  2ma 1
R  3mg  T  3ma 2
F
T
1  2  mg  5ma
T
 a  51 g
a
Find the tension in the string
For P, using (1): T  2ma  2mg 
12
5
2mg
mg
Find the force exerted by the string on the pulley
2T  245 mg
2m
P
T
T
3m
Q
3 mg
a
WB13 A particle A of mass 0.8 kg rests on a horizontal table and is attached to one
end of a light inextensible string. The string passes over a small smooth pulley P
fixed at the edge of the table. The other end of the string is attached to a particle B of
mass 1.2 kg which hangs freely below the pulley, as shown in the diagram above.
The system is released from rest with the string taut and with B at a height of 0.6 m
above the ground. In the subsequent motion A does not reach P before B reaches
the ground. In an initial model of the situation, the table is assumed to be smooth.
Using this model, find
(a) the tension in the string before B reaches the ground,
For A:
For B:
R
R


T  54 a
6
5
1
g  T  56 a
1  2  56 g  2a
 a  35 g
Sub in (1)  T  4.7N (2sf)
2
a m s-2
T
P
A
4
5
T
g
a m s-2
B
6
5
g
0.6 m
(b) the time taken by B to reach the ground.
s  ut  21 at 2  0.6  21  35 gt 2
0.6  2
 0.45 s 2sf 
t 
3
5 g
In a refinement of the model, it is
assumed that the table is rough
and that the coefficient of friction
between A and the table is 15
Using this refined model,
(c) find the time taken by B to
reach the ground.
 R  54 g  Fmax  254 g
R  T  254 g  54 a 1
For B: R  56 g  T  56 a 2
For A:
R
a m s-2
R
T
P
A
4
5
T
g
a m s-2
B
6
5
R
1  2  56 g  254 g  54 a  56 a
 2a 
26
25
g  a  13
25 g
g
0.6 m
2
gt
s  ut  21 at 2  35  21  13
25
t 
1
2

3
5
13
25
 0.49 s (2sf)
g
WB14 A particle A of mass 4 kg moves on the inclined face of a smooth wedge.
This face is inclined at 30° to the horizontal. The wedge is fixed on horizontal
ground. Particle A is connected to a particle B, of mass 3 kg, by a light inextensible
string. The string passes over a small light smooth pulley which is fixed at the top of
the plane. The section of the string from A to the pulley lies in a line of greatest slope
of the wedge. The particle B hangs freely below the pulley, as shown in the diagram
above. The system is released from rest with the string taut. For the motion before A
reaches the pulley and before B hits the ground, find
(a) the tension in the string,
a m s-2
T
T  4 g sin 30  4a 1
For A: R
For B:
 
R 
1  2
3 g  T  3a
A (4 kg)
 3 g  4 g sin 30  7a
a
Sub in (1)
2

a m s-2
4g
g
7
T  74 g  4 g sin 30
 25 N (2sf)
T
30
B (3 kg)
3g
(b) the magnitude of the resultant force exerted by the string on the pulley.
T
30
30
T
T
A (4 kg)
2T cos 30
2T cos 30  44 N 2sf 
T
B (3 kg)
30
(c) The string in this question is described as being ‘light’.
(i) Write down what you understand by this description.
(ii) State how you have used the fact that the string is light in your answer to part (a).
(i) The string has no weight/mass
(ii) The tension in the string is constant (same at A as B)
WB15 The diagram above shows two particles A and B, of mass m kg and 0.4 kg
respectively, connected by a light inextensible string.
Initially A is held at rest on a fixed smooth plane inclined at 30° to the horizontal.
The string passes over a small light smooth pulley P fixed at the top of the plane.
The section of the string from A to P is parallel to a line of greatest slope of the plane.
The particle B hangs freely below P. The system is released from rest with the string
1
taut and B descends with acceleration 5 g.
(a) Write down an equation of motion for B.
R

2
5
g  T  252 g
1
5
gms
-2
P
1.4 m
(b) Find the tension in the string.
g m s-2
A (m kg)
T  52 g  252 g 
8
25
(c) Prove that m =
16
35
R
B (0.4kg)
1
5
Using (a)
For A:
T
T
g  3.1N 2sf 
 T  mg sin 30  m  51 g
 T  21 mg  51 mg
 T  107 mg
mg
2
5
g1 m
30
80
 m  107 gT  710g  825g  175

16
35
(d) State where in the calculations you have used the information that P is a light
smooth pulley.
The tension in the string is constant (same at A as B)
On release, B is at a height of one metre above the ground and AP = 1.4 m.
The particle B strikes the ground and does not rebound.
(e) Calculate the speed of B as it reaches the ground.
v 2  u 2  2as
P
 v  0  2  g 1
2
v
2
2
5
2
5
 21 g m s-2
1
5
2
5
m
g ms -1
g  2.0 ms -1 (2sf)
16
35
g
(f) Show that A comes to rest as it reaches P.
When B hits the floor, A is moving at
2
5
g ms
-1
Between then and A stopping, only gravity is acting:
R
  1635 g sin 30  1635  a  a   21 g ms 2
v 2  u 2  2as
 v 2  52 g  2   21 g  52  0
 v  0 at P
Ex 3F Q4
For A:
For B:
1  2
Sub a 
1
7
R

R 
4 g  T  4a 1
T  3 g  3a 2
 g  7a  a  1.4 ms -2
T
g in (1):  4 g  T  74 g
T 
24
7
g  34 N (2sf)
For A after falling for 2 metres:
v 2  u 2  2as  v 2  02  2  1.4  2  v  5.6
For B after A hits the ground:
v 2  u 2  2as  02  5.6  2  g  s  s  72 m
Total height gained by B = 2 72 m
T
T
T
a
3kg
4 kg
B
A
4g
3g
a
Q5
For A:
For B:
R

R 
1  2 
1
2
T  g  5a 1
a
3 g  T  3a 2
5g
5
2
g  8a  a 
1
16
g
 0.61ms -2 (2sf)
Sub
a  0.6125 in (1):  T  52 g  5  0.6125
 T  28 N (2sf)
T
F  2T 2  39 N (2sf)
T
F
T
A
5 kg
5
2
T
g
5g
3kg B
3g
a
Q6 For P:
For Q:
R
 T  mg sin   ma 1
R  mg  T  ma 2
a
2m
1  2  52 mg  2ma
2
2
 v 2  02  2  51 g  2
 v  2.8 ms -1 (2sf)
Q
m
mg
mg
-2
For P: v  u  2as
T
m
 a  g  2.0 ms (2sf)
1
5
T
P
a
3m

5

4
tan  
3
3
4
 sin   35
 cos  
4
5
a
Q7
T
R
T
a
Q
m
mg
30
For P:
For Q:
 R  mg cos 60
R  mg sin 60  T  21 R  ma 1
R  T  mg sin 30  ma 2
1
2
R
m
P
mg
60
R
1  2  mg sin 60  mg sin 30  21 mg cos 60  2ma

3
2
g  21 g  41 g  2a
a
Sub in (2):  T

3
4

 38 g  0.57 ms -2 (2sf)

m
3
4

g  38 g  mg sin 30
 128 3 mg
sin 30 
1
2
sin 60 
3
2
cos 30 
3
2
cos 60 
1
2
Q8
a
For the whole system:
R

12000
12000  2200  1400 g sin   1400a
 1568  1400a
 a  1.12 ms
T
T
-2
900 g
1600
For the trailer:
R
 T  600  500 g sin   500a
T  4100N
600 500 g

5

4
tan  
3
3
4
 sin   35
 cos  
4
5
Q9
R
For Q:
 3 g  T  3  2 .5
2.5
 T  3 g  7.5
For P:
R

R 


R
P
R  2 g cos 30
T  R  2 g sin 30  2  2.5
T  2 g sin 30  5
R
3 g  2 g sin 30  12.5
2 g cos 30
 0.42 (2sf)
T
30
30
2T cos 30
T
2T cos 30  38 N (2sf)
T
2kg
3kg Q
2g
3g
R
30
T
2.5
a
Q10 For whole system:
R

 2400  1200a
 a  2 ms -2
T
300 kg
T
900 kg
2400
For the trailer:
R
 T  300 2
T  600N
v 2  u 2  2as
 02  202  2  2  s
20 2
s
 100 m
4
When decelerating, the tow bar
exerts a thrust on the car and trailer
Ex 3F Q4
3kg
4 kg
A
B
Q5
A
5 kg
3kg B
Q6
P
m

m
Q
Q7
Q
m
30
m
P
60
12000
Q8

Q9
P
2kg
30
3kg Q
Q10
300 kg
900 kg
2400
Review Exercise 1, Q41
R

R 
7
2
R  mg cos  1
R
R
mg sin   R  72 m 2
Sub (1) in (2)  mg sin   mg cos  
 
g sin  72
cos
g  72
 4
5 g
3
5
7
2
m
mg

 0.30 (2sf)
5

4
tan  
3
3
4
 sin   35
 cos  
4
5
0
Review Exercise 1, Q45
R
R

R  50 g  P sin 30  0
R
PN
 R  50 g  P sin 30 1

P cos 30  35 R  0
 R  53 P cos 30 2
1  2  50 g  P sin 30  53 P cos 30
 50 g  P53 cos 30  sin 30
P
50 g
 520 N (2sf)
5
cos
30

sin
30
3
3
5
R
30
50 kg
50 g
Review Exercise 1, Q47
a
A
v  u  2as
2
2
5m
B
 82  102  2  a  5
20
a
R

82 10 2
10
 3.6 ms -2
R  10 g cos 20  0
 R  10 g cos 20
R
 10 g sin 20  10g cos 20  36
20 36
 0.75 (2sf)
   1010g sin
g cos 20
v 2  u 2  2as
 02  102  2  3.6 s
s
10 2
23.6
.
 13. 8 m
C
Momentum
The momentum of a body of mass m kg which is moving with velocity v ms-1 is mv N
By Newton’s 3rd law, when two bodies collide, each exerts an equal and
opposite force on the other. By the impulse – momentum principle, these forces
are equal & opposite, hence the total momentum of the system is unchanged.
This is called the principle of conservation of momentum
In words,
Total momentum before impact = Total momentum after impact
Eg A railway truck P of mass 2000 kg is moving along a straight horizontal track with
speed 10 m s-1. The truck P collides with a truck Q of mass 3000 kg, which is at rest
on the same track. Immediately after the collision Q moves with speed 5 m s-1.
Calculate the speed of P immediately after the collision,
Before
I
After
10 ms -1
0 ms -1
P
Q
2000kg
3000kg
v ms -1
5 ms -1
Conservation of momentum
 
2000 10  3000  0  2000  v  3000  5
I
 20000  2000v  15000
 5000  2000v
 v  2.5 ms -1
Impulse
If a constant force F acts for time t then we define the impulse of the force to be Ft
vu
Substituting a 
in F  ma gives Ft  mv  u 
t
This is called the impulse - momentum principle
In words,
Impulse = change in momentum
Eg Calculate the magnitude of the impulse exerted by P on Q during the collision.
Before
I
After
10 ms -1
I  mv  u 
0 ms -1
P
Q
2000kg
3000kg
For Q:
  I  3000  5  0
 15000 Ns
I
NB: the impulse Q exerts on P
must be the equal and opposite
2.5 ms -1
5 ms -1
For P:
 I  2000  2.5 10
 15000 Ns
Eg A railway truck P of mass 2000 kg is moving along a straight horizontal track with
speed 10 m s-1. The truck P collides with a truck Q of mass 3000 kg, which is at rest
on the same track. Immediately after the collision Q moves with speed 5 m s-1.
Calculate the speed of P immediately after the collision,
Before
I
After
10 ms -1
0 ms -1
P
Q
2000kg
3000kg
v ms -1
Conservation of momentum
 
2000 10  3000  0  2000  v  3000  5
I
 20000  2000v  15000
 5000  2000v
 v  2.5 ms -1
5 ms -1
Eg Calculate the magnitude of the impulse exerted by P on Q during the collision.
I  mv  u 
For Q:
 
I  3000  5  0
 15000 Ns
NB: the impulse Q exerts on P
must be the equal and opposite
For P:
 I  2000  2.5 10
 15000 Ns
WB17 Two particles A and B have mass 0.12 kg and 0.08 kg respectively.
They are initially at rest on a smooth horizontal table. Particle A is then given an
impulse in the direction AB so that it moves with speed 3 m s-1 directly towards B.
(a) Find the magnitude of this impulse, stating clearly the units in which your answer
is given.
Before
I
After
3 ms -1
0 ms -1
A
B
0.12kg
0.08kg
1.2 ms -1
v ms -1
I  0.12  3  0  0.36 Ns
I
Immediately after the particles collide,
the speed of A is 1.2 m s-1, its
direction of motion being unchanged.
(b) Find the speed of B immediately
after the collision.
Conservation of momentum
(c) Find the magnitude of the impulse
exerted on A in the collision.
I  0.12  3  1.2  0.216 Ns
 
0.12 3  0.08  0  0.12 1.2  0.08  v
 0.36  0.144  0.08v
 0.216  0.08v
 v  2.7 ms -1
WB18 A particle P of mass 2 kg is moving with speed u m s–1 in a straight line on a
smooth horizontal plane. The particle P collides directly with a particle Q of mass 4 kg
which is at rest on the same horizontal plane. Immediately after the collision, P and Q
are moving in opposite directions and the speed of P is one-third the speed of Q.
(a) Show that the speed of P immediately after the collision is 51 u ms -1
Before
After
u ms -1
0 ms -1
P
Q
2kg
4 kg
v ms -1
3v ms -1
Conservation of momentum
 
2  u  4  0  2   v  4  3v
 2u  10v
 v  51 u ms -1
After the collision P continues to move in the same straight line and is
brought to rest by a constant resistive force of magnitude 10 N. The distance
between the point of collision and the point where P comes to rest is 1.6 m.
(b) Calculate the value of u.
For P:
R

 10  2a
 a  5 ms 2
v 2  u 2  2as
 0  51 u   2   51.6
-1
 251 u 2  16  u  2516  20 ms
2
WB19 Two trucks A and B, moving in opposite directions on the same horizontal
railway track, collide. The mass of A is 600 kg. The mass of B is m kg. Immediately
before the collision, the speed of A is 4 m s–1 and the speed of B is 2 m s–1.
Immediately after the collision, the trucks are joined together and move with the
same speed 0.5 m s–1. The direction of motion of A is unchanged by the collision.
Find
(a) the value of m,
Before
I
4 ms -1
2 ms -1
A
B
600kg
m kg
 
600  4  m   2  600  m 0.5
Conservation of momentum
 2400  2m  300  21 m
 2100  2 21 m
After
I
A&B
600  m kg
0.5 ms -1
(b) the magnitude of the impulse
exerted on A in the collision.
I  600  4  0.5  2100 Ns
 m  840 kg
Now do Ex 3I, Q1,3,7,14,15
Before
0 ms -1
0 ms -1
 
0  m  200  4  m  5
Conservation of momentum
Q1
Gun
4  m kg
Bullet
m kg
 0  200m  20  5m
After
Before
Q3
After
5 ms -1
2u ms -1
20
 394 kg
195
 4  m  4 394 kg
200 ms -1
m
u ms -1
Conservation of momentum
6mu  4mu  4mv
P
Q
3m kg
4m kg
0 ms -1
v ms
 
-1
 2u  4v
I  mv  u 
For P:
 
 v  21 u ms -1
I  3m  2u  0  6mu Ns
Q7
For the driver:
R

v 2  u 2  2as
 v 2  02  2  g 10
Before
14 ms -1
0 ms -1
 v  14
After
Driver
1000kg
Conservation of momentumR
Pile driver
1200 kg
Pile
1000 14  1200v
v ms -1
 v  353
200 kg
R
120000
 1200 g  120000  1200a
 a  90.2
Pile driver
a
1200 kg
35
3
ms -1
v 2  u 2  2as
 02  353   2   90.2 s
2
1200 g
353 2
 s  180.4  0.75m (2sf)
 
Before
Q14
After
3 ms -1
4 ms -1
P
Q
1.5 kg
2.5 kg
2.5 ms -1
v ms -1
Conservation of momentum
 
1.5  3  2.5   4  1.5   2.5  2.5  v
 5.5  3.75  2.5v
 5.5  3.75
v
 0.7 ms -1
2.5
 Speed  0.7 ms -1
As v <0, direction is unchanged
For P:
Before
Q15
After
2u ms
-1
 
I  1.5  3  2.5 8.25 Ns
u ms -1
A
B
m kg
km kg
A&B
m  km kg
Conservation of momentum
2um  kum  m  km32 u 
 2um  kum  32 um  32 kum
 34  53 k
 k  54
2
3
u ms -1
 
Assignment 4, Q2
Before
I  mv  u 
For A:
    7mu2  2mvA  2u 
 7mu  4mvA  8mu
 u  4vA
 v A  u4
For B:
   7mu2  mvB   3u 
 7mu  2mvB  6mu
 u  2vB
 vB   u2
 speed  u2
I
After
2u
3u
A
B
2m
m
vA
vB
I
4
9
Assignment 4, Q3
For B:
For A:
R
R
 2mg  T  89 mg
R
 T  109 mg
 R  mg  Fmax  mg
R  T  Fmax  94 mg

10
9
g
T
A
P
m
T
F
mg  mg  mg
mg
2m B
4
9
   109  94  32
2mg
For B when it hits the ground:
a  94 g , u  0, s  h
v  u  2as  v  gh
2
2
2
8
9
For A after B hits the ground:
R
  32 mg  ma  a   32 g
Giving Information : u 
2
8
9
8
9
a
A
m
F
gh, a   32 g , s  31 h
2
v 2  u 2  2as  v  89 gh  94 gh  94 gh  v  32 gh
gh
1
3
h
4
9
g
Assignment 4, Q4
R

R
49 cos   6 g sin 30
 cos  
R

6 g sin 30
49
49 N
P

 35
30
R  6 g cos 30  49 sin   90 N (2sf)
5

a
4
P
49 N
30
R
30
2kg
6g
 49 cos 30  6 g sin 30  6a
 a  2.2 ms -2 2sf 
2kg
6g
tan   34
3
 sin   35
 cos   54