the Note

MOMENTUM AND IMPULSE (LIVE)
07 APRIL 2015
Lesson Description
In this lesson we revise:



Momentum
Collisions
Impulse
Section A: Summary Notes and Examples
1.1.
Momentum
Momentum is defined as the product of an object’s mass and velocity.
It is a vector which means that it has both magnitude and direction.
Momentum is calculated using the following equation:
p = momentum measured in 𝑘𝑔 ∙ 𝑚 ∙ 𝑠 −1
𝑝 = 𝑚𝑣
m = mass measured in kg
v = velocity measured in 𝑚 ∙ 𝑠 −1
When an object changes its velocity then it will experience a change in momentum. If the object
changes direction when changing velocity it will experience a much greater change in momentum
then if its direction doesn’t change.
Example 1:
-1
A tennis ball of mass 56 g is thrown towards a wall and hits the wall at 45 m.s . The ball then
-1
bounces off the wall with a velocity of 40 m.s as shown in the diagram. Calculate the change in
momentum of the ball.
Solution:
Hint 1: If there was no diagram given, then draw a sketch to help understand the question
Hint 2: Write a list of all the information given, including chosen positive direction. Also convert given
units to SI units.
Page 1
Right is positive
-1
vi = 45 m.s
-1
vf = −40 m.s
m = 56 g
= 0,056 kg
∆𝑝 = 𝑚∆𝑣
= 𝑚 𝑣𝑓 − 𝑣𝑖
= 0,056 −40 − 45
= −4,76
= 4,76 𝑚. 𝑠 −1 𝑙𝑒𝑓𝑡
In order for an object to change its velocity or change its direction a net force needs to be applied to
the object. This net force can be calculated as follows:
𝐹𝑛𝑒𝑡
Fnet = net force measured in N
∆𝑝
=
∆𝑡
Δp = change in momentum measured in kg.m.s
-1
Δt = time measured in s
This net force is the same net force that can be calculated using Newton’s second law. Therefore
Newton’s second law can state in terms of momentum:
Newton’s Second Law in terms of Momentum: The net force acting on an object is equal to the rate
of change of momentum.
Example 2:
A remote controlled toy rocket, with mass 0,5 kg, is launched so that is
-1
reaches a velocity of 35 m.s in 0,2 seconds. Calculate the net force
acting on the rocket.
Solution:
m = 0,5 kg
-1
vi = 0 m.s (the rocket started from rest)
vf = 35 m.s
-1
Δt = 0,2 s
𝐹𝑛𝑒𝑡 =
∆𝑝
∆𝑡
𝑚 𝑣𝑓 − 𝑣𝑖
∆𝑡
0,5 35 − 0
=
0,2
=
= 87,5 𝑁 𝑢𝑝
Collisions
When two objects collide with each other, their momentum will change. However the total momentum
of the system will not change if there are no external forces acting on the objects. External forces are
forces such as friction or the brakes on a car.
Law of Conservation of Momentum: The total linear momentum of an isolated system remains
constant (is conserved)
Page 2
As an equation:
Σ𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = Σ𝑝𝑓𝑖𝑛𝑎𝑙
This equation can be rewritten in several ways, depending on the type of the collision.
For example, if two objects collide with each other and then move apart after the collision.
𝑚1 𝑣𝑖1 + 𝑚2 𝑣𝑖2 = 𝑚1 𝑣𝑓1 + 𝑚2 𝑣𝑓2
m1 = mass of object 1
vi1 = initial velocity of object 1
m2 = mass of object 2
vi2 = initial velocity of object 2
vf1 = final velocity of object 1
vf2 = final velocity of object 2
Elastic and inelastic collisions
Although momentum is always conserved during collisions, kinetic energy may not always be
conserved. If the total kinetic energy of the system before collision isequal to the total kinetic energy
of the system after collision, then the collision is said to be elastic. However, if the total kinetic
energy before and after the collision is not the same, then the collision is said to be inelastic. The
kinetic energy is often transferred to sound, light and internal heat, due to friction.
Example 3:
-1
A ball of mass 100 g is moving at 10 m.s to the right when it collides with another ball of mass 150 g
-1
-1
moving at 10 m.s to the left. After the collision the 100 g ball moves with a velocity of 7 m.s to the
left.
1. Calculate the velocity of the 150 g ball after the collision.
2. Is the collision elastic or inelastic? Use appropriate calculations to prove your answer.
Solution
1.
Take to the right as positive.
Σ𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = Σ𝑝𝑓𝑖𝑛𝑎𝑙
m1 = 100 g
𝑚1 𝑣𝑖1 + 𝑚2 𝑣𝑖2 = 𝑚1 𝑣𝑓1 + 𝑚2 𝑣𝑓2
= 0,1 kg
vi1 = 10 m.s
-1
vf1 = - 7 m.s
0,1 10 + 0,15 −10 = 0,1 −7 + 0,15 𝑣𝑓2
-1
−0,5 = −0,7 + 0,15𝑣𝑓2
m2 = 150 g
0,2 = 0,15𝑣𝑓2
= 0,15 kg
vi2 = -10 m.s
∴ 𝑣𝑓2 = 1,33 𝑚 ∙ 𝑠 −1 𝑟𝑖𝑔𝑕𝑡
-1
vf2 = ?
Page 3
2.
Inelastic
1
1
2
2
Σ𝐸𝑘 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚1 𝑣𝑖1
+ 𝑚2 𝑣𝑖2
2
2
1
1
= (0,1)(10)2 + (0,15)(−10)2
2
2
1
1
2
2
𝑚1 𝑣𝑓1
+ 𝑚2 𝑣𝑓2
2
2
1
1
= (0,1)(−7)2 + (0,15)(1,33)2
2
2
Σ𝐸𝑘 𝑓𝑖𝑛𝑎𝑙 =
= 5 + 7,5
= 2,45 + 0,13
= 12,5 𝐽
= 2,58 𝐽
𝚺𝑬𝒌 𝒊𝒏𝒊𝒕𝒊𝒂𝒍 ≠ 𝚺𝑬𝒌 𝒇𝒊𝒏𝒂𝒍
1.2.
Impulse
Impulse is defined as the product of the net force acting on an object and the contact time that the
net force acts.
Impulse = FnetΔt.
Also, Fnet∆t =Δp (impulse-momentum theorem).
Therefore theimpulse equation can be written as follows:
FnetΔt = impulse measured in N.s
𝐹𝑛𝑒𝑡 ∆𝑡 = ∆𝑝
Fnet = net force measured in N
Δt = time measured in s
Δp = change in momentum measured in kg.m.s
-1
Impulse is a vector therefore it has both magnitude and direction.
Example
Consider the tennis ball in example 1.
1. If the tennis ball is in contact with the wall for 0,1s what is the impulse of the wall on the ball?
Solution:
𝛥𝑝 = 4,76 𝑚. 𝑠 −1 𝑙𝑒𝑓𝑡 (Calculated in Example 1)
FnetΔt = Δp
= 4,76 N.s left
2. Calculate the force exerted by the wall on the tennis ball.
Solution
FnetΔt = Δp
Fnet (0,1) = 4,76
 Fnet = 47,6 N left
If the ball experiences a force of 47,6 N left, then the wall experiences a force of 47,6 N right due to
the tennis ball. This is according to Newton’s third law. Both objects experience the same size force
but in opposite directions.
Page 4
Application
Air bags and seat belts are used in cars to increase the time it takes for passengers to stop during a
collision. The passengers will experience the same change in momentum but if the time to stop is
increased then the force experience decreases as F netΔt = Δp and Δp is a constant.
Section B: Practice Questions
Question 1
(Taken from NCS November Paper I 2012)
-1
The diagram below shows a car of mass m travelling at a velocity of 20 m.s east on a straight level
-1
road and a truck of mass 2m travelling at 20 m.s west on the same road. Ignore the effects of
friction.
The vehicles collide head-on and stick together during the collision.
1.1.
State the principle of conservation of linear momentum in words.
(2)
1.2.
Calculate the velocity of the truck-car system immediately after the collision.
(6)
1.3.
On impact the car exerts a force of magnitude F on the truck and experiences an
acceleration of magnitude a.
1.3.1.
Determine, in terms of F, the magnitude of the force that the truck exerts on the car
on impact. Give a reason for the answer.
(2)
1.3.2.
Determine, in terms of a, the acceleration that the truck experiences on impact.
Give a reason for the answer.
(2)
1.3.3.
Both drivers are wearing identical seat belts. Which driver is likely to be more
severely injured on impact? Explain the answer by referring to acceleration and
velocity.
(3)
[15]
Question 2
(Taken from NCS November Paper I 2011)
-1
A patrol car is moving on a straight horizontal road at a velocity of 10 m.s east. At the same time a
-1
thief in a car ahead of him is driving at a velocity of 40 m.s in the same direction.
-1
While travelling at 40 m.s , the thief’s car of mass 1000 kg, collides head-on with a truck of mass
-1
5000 kg moving at 20 m.s . After the collision, the car and the truck move together. Ignore the
effects of friction.
2.1.
State the law of conservation of linear momentum in words
(2)
2.2.
Calculate the velocity of the thief’s car immediately after the collision.
(6)
Page 5
2.3.
Research has shown that forces greater than 85 000N during collisions may cause fatal
injuries. The collision described above lasts for 0,5 s. Determine, be means of a calculation,
whether the collision above could result in a fatal injury.
(5)
[13]
Question 3
(Taken from Gauteng Prelim Paper I 2013)
A delivery van having a mass of 5 000 kg is moving towards the right at a constant velocity of
-1
10 m.s . The delivery van collides head on with a car of mass 2 000 kg moving at a velocity of 15
-1
-1
m.s towards the left. Immediately after the collision the car moves at a constant velocity of 5 m.s
towards the right.
3.1.
Calculate the magnitude of the velocity of the delivery van immediately after the collision.
(4)
3.2.
Calculate the change in the kinetic energy of the system.
(4)
3.3.
What type of collision is this? Write only ELASTIC or INELASTIC
(1)
3.4.
If the collision lasts for 0,4 s, calculate the magnitude of the force that the delivery can
exerts on the car.
(4)
[13]
Page 6
Section C: Solutions
Question 1
1.1.
The total linear momentum remains constant in an isolated system 
1.2.
East positive
1.3.1.
F
Newton’s Third Law of motion 
(2)
Σ𝑝𝑖 = Σ𝑝𝑖 
𝑚1 𝑣𝑖1 + 𝑚2 𝑣𝑖2 = 𝑚1 + 𝑚2 𝑣𝑓
𝑚 20  + 2𝑚 −20  = 𝑚 + 2𝑚 𝑣𝑓 
20𝑚 − 40𝑚 = 3𝑚 𝑣𝑓
−20𝑚 = 3𝑚 𝑣𝑓
∴ 𝑣𝑓 = −6,67
= 6,67 𝑚. 𝑠 −1 𝑤𝑒𝑠𝑡
(6)
(2)
1.3.2.
½a
1
The truck experiences the same Fnet but has twice the mass. Since 𝑎 ∝ then acceleration
𝑚
must be half 
(2)
1.3.3.
Car driver 
The car has a greater acceleration , due to the smaller mass but same Fnet. Therefore the car
experiences a greater change in velocity. 
(3)
Question 2
2.1.
The total linear momentum remains constant in an isolated system 
2.2.
To the right is positive
(2)
Σ𝑝𝑖 = Σ𝑝𝑖 
𝑚𝑐 𝑣𝑖𝑐 + 𝑚 𝑇 𝑣𝑖𝑇 = 𝑚𝑐 + 𝑚 𝑇 𝑣𝑓
1000 40  + 5000 −20  = (1000 + 5000)𝑣𝑓 
−60000 = 6000𝑣𝑓
∴ 𝑣𝑓 = −10
= 10 𝑚. 𝑠 −1 𝑙𝑒𝑓𝑡
(6)
2.3.
Force on car: (To the right is positive)
𝐹𝑛𝑒𝑡 ∆𝑡 = ∆𝑝 = 𝑚 𝑣𝑓 − 𝑣𝑖
𝐹𝑛𝑒𝑡 0,5  = 1000 −10 − 40 
∴ 𝐹𝑛𝑒𝑡 = −100 000
= 100 000 𝑁 𝑙𝑒𝑓𝑡
Fnet > 85 000 N, therefore the collision will be fatal 
(5)
Question 3
3.1.
Right Positive
Total p before = Total p after
m1 vi1 + m2 vi2 = m1 vf1 + m2 vf2
5000 10 + 2000 −15  = 5000 vf1 + 2000 5 
vf1 = 2 m. s −1 
OR
Page 7
(4)
Left Positive
Total p before = Total p after
m1 vi1 + m2 vi2 = m1 vf1 + m2 vf2
5000 −10 + 2000 15  = 5000 vf1 + 2000 −5 
vf1 = −2 m. s −1
= 2 m. s −1 
3.2.
(4)
∆K = Total K after − Total K before 
1
1
1
1
2
2
2
2
m1 vf1
+ m2 v2f
−
m1 vi1
+ m2 vi2
2
2
2
2
1
1
1
= ( (5000)(2)2 + 2000 5)2  − ( 5000 10
2
2
2
=
2
1
+ (2000)(−15)2 )
2
= −440 000 J
(4)
OR
Total K before =
1
1
m v2 + m v2
2 1 i1 2 2 i2
1
1
=
5000 10 2 + (2000)(−15)2 
2
2
= 475 000 J
Total K after =
1
1
2
2
m1 vf1
+ m2 vf2
2
2
1
=
5000 2
2
2
1
+ (2000)(5)2 
2
= 35 000 J
∆K = Total K after − Total K before 
= 35 000 − 475 000
= −440 000 J 
3.3.
Inelastic 
3.4.
Right positive
(4)
(1)
Fnet =
Left positive
∆p
m1 vf1 − m1 vi1
=

∆t
∆t
=
Fnet =
2000 5 − (2000)(−15)
0,4 
∆p
m1 vf1 − m1 vi1
=

∆t
∆t
=
= 100 000 N
2000 −5 − (2000)(15)
0,4 
= − 100 000 N
= 100 000 N
OR
Page 8
(4)
Right positive
Fnet =
Left positive
∆p
m1 vf1 − m1 vi1
=

∆t
∆t
=
Fnet =
∆p
m1 vf1 − m1 vi1
=

∆t
∆t
5000 2 − (2000)(10)
0,4
5000 −2 − (5000)(−10)
0,4 
=
= −100 000 N
= 100 000 N
= 100 000 N
(4)
OR
Right Positive
Left Positive
vf = vi + a∆t for both equations
2
vf = vi + a∆t for both equations
−15 = 5 + a(0,4) 
15 = −5 + a(0,4)2 
∴ a = −50 m. s −2
∴ a = 50 m. s −2
Fnet = ma
Fnet = ma
= 2000 −50 
= 2000 50 
= −100 000 N
= 100 000 N
= 100 000 N
(4)
Page 9