Lecture 6: Impulse and Momentum 1

PH1011 Physics
Week 3
Lecture 6: Impulse and Momentum 1

Impulse Linear Momentum Conservation

- Definition 1: In terms of force - Conservation Law:
)*
𝑚7 𝑣⃗70 + 𝑚9 𝑣⃗90 = 𝑚7 𝑣⃗72 + 𝑚9 𝑣⃗92
𝐼⃗ = $ 𝐹⃗ 𝑑𝑡 = 𝐹⃗avg (𝑡0 − 𝑡2 ) - Nice formula for 1D elastic
)+
- Definition 2: In terms of linear momentum 𝑢7 − 𝑢9 = 𝑣9 − 𝑣7
𝐼⃗ = 𝑝⃗0 − 𝑝⃗2 = 𝑚𝑣⃗0 − 𝑚𝑣⃗2

Examples
- Two Toy Cars Colliding (inelastic) on page 8
- Young & Freedman pg 254 example 8.9 on page 9
- Giancoli pg 228 example 9-13 on page 10

Examples
- Giancoli pg 216 example 9-1 on page 3
- Soccer on page 3
- Young (College Physics) pg 263 problem 8.41 on page 4
- Daily applications on page 4

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 Lecture 6: Impulse and Momentum 1

Impulse >> General Discussion:

When a constant force 𝐹 acts on an object, the impulse 𝐼 is the force multiplied by the time
interval 𝛥𝑡 = 𝑡0 − 𝑡2 during which the force acts.
𝐼 = 𝐹𝛥𝑡

1) Note that impulse 𝐼 is a vector quantity; its direction is the same as the force 𝐹.

2) Units: Force x time (N.s or kg.m/s)

<=
3) From Newton’s 2nd law, 𝐹 = , we can thus say 𝐼 = 𝛥𝑝 = 𝑝0 − 𝑝2 . This is called the
<)
“Impulse-Momentum” relation.

3) From the expression 𝐹𝛥𝑡, we deduce that the area under 𝐹 − 𝑡 graph gives the impulse.

Average force acting over time interval 𝛥𝑡. Same impulse, different 𝛥𝑡.

4) Using the area under 𝐹 − 𝑡 graph = impulse idea, we can deduce that for non-constant
forces, the impulse would be
)*
𝐼= 𝐹 𝑑𝑡
)+
5) Usually we can’t measure the force accurately because the force varies in time so rapidly.
Thus, for practical purposes, we want to define an average force 𝐹avg by Fig 9.10. 𝐹avg is defined
by the value of the force that gives the same area (or impulse) in the interval 𝛥𝑡.
)*
Area = 𝐹 𝑑𝑡 = 𝐼 = 𝐹avg 𝛥𝑡 = 𝐹avg (𝑡0 − 𝑡2 )
)+

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 Lecture 6: Impulse and Momentum 1

Impulse >> Example:

(Modified) Giancoli pg 216 example 9-1:

a) Calculate the momentum of a 57 g tennis ball moving

horizontally at 65 m/s.

b) The tennis ball is struck by a racquet and it moves in

the reverse direction at 70 m/s. Calculate the new
momentum.

c) Calculate the change in momentum of tennis ball.

Tennis racquet striking a ball.

This quantity, change in momentum, is also known as impulse. If we can control impulse, we
can control the final momentum.

d) If the ball is in contact with the racquet for 4 ms, estimate the average force on the ball. Will
this force be large enough to lift a 60-kg person?

Impulse >> Example:

Soccer:

You are the striker of a soccer team. A 450 g ball is crossed from the left to you (with horizontal
velocity 15 m/s) when you are just in front of the goal. If you give the soccer ball a gentle kick
(impulse = 5 kg.m/s) with the tip of your toe perpendicular to the ball’s initial velocity, which
direction will the ball eventually move? [Here, we simplify our analysis by regarding the
spherical ball as a point particle.]

Now you have another chance. With the same conditions as before and generating an impulse
of 8 kg.m/s, at what angle should you kick for the ball to go straight into the goal?

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 Lecture 6: Impulse and Momentum 1

Impulse >> Example:

Young (College Physics) pg 263 problem 8.41:

A 150 g baseball is hit towards the left by a bat. The magnitude of the force the bat exerts on
the ball as a function of time is shown in the figure below:

a) Find the magnitude and direction of the impulse that the bat imparts to the ball.
b) Find the ball's velocity just after it is hit by the bat if the ball is initially at rest.
c) Find the ball's velocity just after it is hit by the bat if the ball is initially moving to the right at
30.0 m/s.

Impulse >> Example:

Daily Applications:

Power shooting and free kicks in soccer:

Beckham on power shooting: ``... Kick through the back of the ball... Follow through with foot
and body...''; Relate this to impulse momentum relation…

Advice for the apprentice:

The young apprentice is challenged to catch a piece of tofu thrown at him without breaking it.
What would your advice to him be?

Cat landing:

Airbag:

Standing broad jump (landing):

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 Lecture 6: Impulse and Momentum 1

Conservation of Linear Momentum >> General Discussion:

What is a system?:

Impulse delivered on object Impulse delivered on object (car) by

(ball) by `external agent’ (bat). another object (car). These two objects
form a closed system.

Conclusion: We can choose any number of objects to be called a system. In a system, an

action-reaction pair of forces inside the system are called “internal forces”. If only the action
force or the reaction force is in the system then it is called an “external force”.

Newton’s Third Law + Impulse-Momentum relation = Conservation of Linear Momentum

(COLM):
Consider two particles colliding into each other and exerting a force on each other throughout
the duration 𝛥𝑡 = 𝑡0 − 𝑡2 of contact.

According to Newton’s third law, 𝐹79 = −𝐹97 . Applying the momentum-impulse relation, the
change in momentum of particle 1 during the time interval of collision is (we simplify the
analysis by assuming 𝐹 is constant – the result can be generalized to cases where 𝐹 is not
constant in time),
𝛥𝑝7 = 𝐹79 𝛥𝑡
Similarly for particle 2,
𝛥𝑝9 = 𝐹97 𝛥𝑡
Combining the two equations we get
𝛥𝑝7 + 𝛥𝑝9 = 𝐹79 𝛥𝑡 + 𝐹97 𝛥𝑡 = 0
since 𝐹79 = −𝐹97 . Thus, the change in momentum of the system of the two particles is zero
(when there are no other external forces acting).

Conservation of linear momentum states that the total linear momentum of an isolated
system of particles is constant at all times.

Remarks:
1. We can write 𝛥𝑝 = 𝑚𝑣0 − 𝑚𝑣2 . Thus for a two body collision,
𝑚7 𝑣70 − 𝑚7 𝑣72 + 𝑚9 𝑣90 − 𝑚9 𝑣92 = 0
𝑚7 𝑣70 + 𝑚9 𝑣90 = 𝑚7 𝑣72 + 𝑚9 𝑣92
Which essentially says that the total momentum before collision is the same as the total
momentum after collision.
2. This result applies to all collisions (even for many particle systems).
3. The same applies to a system of particles interacting at a distance (without touching).
4. It is important to identity the system correctly such that external forces are excluded before
the conservation of momentum is applied.

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 Lecture 6: Impulse and Momentum 1

Conservation of Linear Momentum >> Different Types of Collisions:

1. For an isolated system of particles, the conservation of linear momentum states that the total
linear momentum is constant at all times. Thus, in all types of collisions, the total linear
momentum (immediately) before the collision = total linear momentum (immediately) after the
collision.

2. In addition, the total energy of a closed system is also constant for all collisions. (Here, we
have to include energy such as heat, sound etc.)

3. For general (two body) collisions whereby kinetic energy of the system is lost to the
surrounding as heat and sound, it can be very difficult to predict the final velocities from the
masses and the initial velocities. We study some of these of collisions where we can solve
completely and we will study them in detail:
- Completely Inelastic Collision: objects moving off with common velocity after
collision.
- Elastic Collision: The total kinetic energy of the system is conserved. (This is
quite a special case.)
- Explosion: A single mass disintegrating into two (or several) pieces.

Conservation of Linear Momentum >> Nice Formula for 1D Elastic:

Consider a 1D elastic collision of 2 masses, 𝑚7 and 𝑚9 with velocities 𝑢7 and 𝑢9 to the right.
After the collision, the final velocities are 𝑣7 and 𝑣9 to the right respectively. Derive a
relationship 𝑢7 , 𝑢9 , 𝑣7 and 𝑣9 .

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 Collisions
Explosions
Completely
Perfectly Elastic Inelastic
Inelastic
Vector Sum of initial Vector Sum of initial Vector Sum of Vector Sum of
COLM momenta momenta initial momenta initial momenta
(holds for = = = =
all cases) Vector sum of final Vector sum of final Vector sum of final Vector sum of final
momenta momenta momenta momenta
Conservation of KE Need more information Final mass is the Vector sum of
Extra
equation is also from the question in sum of the 2 initial momenta = 0
conditions
needed order to solve the masses since they
problem stick together
Final KE change = 0 Intermediate KE change Maximum KE Since initial KE = 0,
velocities change change in KE = final
and KE
KE change
Conservation of Linear Momentum >> Bird’s Eye View of Collisions and Explosions:
Lecture 6: Impulse and Momentum 1

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 Lecture 6: Impulse and Momentum 1

Conservation of Linear Momentum >> Example:

Two Toy Cars Colliding (inelastic)

The diagram below shows two toy cars, X (2 kg, 7 m/s right) and Y (3 kg, 3 m/s left), about to
collide.

After the collision, the directions of motion of both toy cars are reversed and the magnitude of
the momentum of X is then 2 N.s .

Hints and Clues:

a) What is then the corresponding momentum of Y?

b) What is the velocity of Y?

c) What is the change in total kinetic energy of the system?

Note that we have some information about the toy cars after the collision. In general, we need
additional information on the nature of the collision in order to solve the problem.

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 Lecture 6: Impulse and Momentum 1

Conservation of Linear Momentum >> Example:

(Modified) Young & Freedman pg 254 example 8.9

A small compact car with a

mass of 1000 kg is travelling
north along Scotts road with a
speed of 15 m/s. At the
intersection of Scotts and
Orchard Road, it collides with
a truck with a mass of 2000
kg that is travelling east on
Orchard road at 10 m/s. The
two vehicles become
thoroughly tangled and moved
away from the point of contact
as one mass. Fortunately, all
passengers were wearing
seat belts and no one was
injured. What is the velocity of the wreckage immediate after the impact? [For simplicity, treat
each vehicle as a particle.]

Hints and Clues:

Calculate the impulse acting on the truck during the collision.

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 Lecture 6: Impulse and Momentum 1

Conservation of Linear Momentum >> Example:

Giancoli pg 228 example 9-13 proton-proton collisions

A proton travelling with speed 8.2 ×10G m/s collides elastically with a stationary proton. One of
the proton is observed to be scattered at a 60I angle. At what angle will the second proton be
observed, and what will be the velocities of the two protons after the collision?

Two dimensional elastic collision

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