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Impulse & Momentum
Physics 11
Problem

A baseball of mass 0.145kg is pitched
toward a batter with an initial velocity
of 35 m/s. If the batter hits the ball in
the opposite direction at 45 m/s,
determine the force that is applied by
the bat on the ball if the contact time
was 0.013s.
Impulse


When an object is accelerated,
typically the force will only be applied
for a given time
So instead of considering Newton’s
Second Law as we have previously
discussed it, we will rearrange the
equation
Impulse



Use Newton’s Second
Law and substitute the
definition for
acceleration
Rearrange the equation
so there are no terms in
the denominator
This expression is
known as impulse (J)


F  ma

 mv
F
t
 

J  Ft  mv
Example

A tennis ball is
struck by a racquet
with a force of
750N; if the time of
contact was
0.023s, what
impulse was
delivered to the
ball?
 

J  Ft  mv

J  750 N (0.023s)

J  17 Ns
Momentum



Momentum can also be
defined starting from
Newton’s Second Law
The rate of change of
momentum can also be
used to determine the force
Momentum is defined as the
product of an object’s
mass and velocity  1st
Law of Motion


F  ma
 mv
F
t


 mv f  mvi
F
t


p  mv
Example

A cyclist is
travelling at 32km/h
and the bike and
rider have a mass
of 85kg. What is
their momentum?


p  mv

p  85kg(8.89m / s )

p  760kgm / s
Impulse-Momentum


Impulse and
momentum can be
related in order to
solve dynamics
problems
Substitute the
definition for
momentum into the
impulse equation


Ft  mv



Ft  mv f  mvi



Ft  p f  pi


Ft  p
Example – again…

A baseball of mass 0.145kg is pitched
toward a batter with an initial velocity
of 35 m/s. If the batter hits the ball in
the opposite direction at 45 m/s,
determine the force that is applied by
the bat on the ball if the contact time
was 0.013 s.
Example


Ft  p

 mv
F
t
 .145kg(45m / s  (35m / s )
F
.013s

F  890 N
Collisions
Physics 11
Conservation of Momentum

The vector quantity momentum will be
conserved in any collision


That is, the sum of all momenta prior to
the collision will be equal to the sum of all
momenta following a collision
Every object that has mass and velocity
will have momentum and must be
included in the total momentum of the
system
Collisions



With any collision, it is imperative that you
diagram the system prior to and following
the collision and identify all objects involved
in the collision
This allows you to ensure that you calculate
the total momentum for the system to
properly analyze the situation
While this may seem onerous, generally we
will be looking at a maximum of two particles
Recall Momentum

Momentum

Impulse


p  mv
 


J  Ft  mv  p
Momentum Conservation

Momentum is conserved

This is an expression of Newton’s first law:


pI  pF


 
p  p'
“An object at rest or in uniform motion will remain at
rest or in uniform motion unless acted on by an
external force.”
External forces can change the momentum of a
system (Impulse)
 


J  Ft  mv  p
Momentum Conservation

In interactions between two bodies,
momentum of one object can change,
but the total momentum of the system
remains constant.
 
 
p1  p2  p1 ' p2 '




m1v1  m2v2  m1v1 'm2v2 '
Types of Momentum Problems

Elastic collisions
Initial

Inelastic collisions
Initial

Final
Final
Explosions
Initial
Final
Explosions: Recoil

A Barrett M82 is a high calibre sniper
rifle. Below are it’s specifications:





Barrel length: 73.7 cm
Weight: 14.0 kg
Muzzle Velocity: 853 m/s
Typical ammunition weight: 50.0g
Calculate the magnitude force exerted
on the riflemen.
Explosion: Recoil

The plan is this


Calculate the momentum of the rifle
knowing the momentum of the bullet
Calculate the impulse imparted to the
riflemen to stop the gun.


Impulse is change in momentum
Impulse is force multiplied by time

Need to know the time (how long) the explosion
takes.
The Momentum Problem




Barrel length: 73.7 cm
Weapon mass: 14.0 kg
Muzzle Velocity: 853 m/s
Typical ammunition mass: 50.0g




m1v1  m2 v2  m1v1 ' m2v2 '


0  m1v1 ' m2 v2 '


 m1v1 '  m2v2 '

 (14.0kg)(v1 ' )  (0.0500kg)(853 ms )

v1 '  3.05 ms
Impulse

We know the person stops the gun, so
to find the force, we need to know the
interaction time.
 


J  Ft  mv  p
 
p  Ft
Explosions: Recoil




Barrel length: 73.7 cm
Weapon mass: 14.0 kg
Muzzle Velocity: 853 m/s
Typical ammunition mass: 50.0g
M
L
m
vI
vF
1
d  vF  vI t
2
2d
2(0.737m)
t 

 0.00173s
m
m
vF  vI  853 s  0.00 s 
Bring it all together
pRifle  p f  pI

p  m1v1  0
p  (14.0kg)( 3.05 ms 1 ' )
p  42.65
kg m
s
t  0.00173s
 
p  Ft
 p
42.65
4
F

 2.46 x10 N
t 0.00173
For Comparison

How many people would have to sit on
your shoulder to get the same force?

Fperson  mg  (70kg)(9.81 sm2 )  6.9 x102 N

FGun  2.46 x10 4 N


FGun  36 FPerson
Collision

A billiard ball, mass 155g, is travelling
at 3.5m/s across the table. It strikes
another ball at rest, mass 150g and
comes to rest. What is the velocity of
the second ball after the collision?