Impulse and Momentum
Unit - 4
HPFP - I
1
Linear Momentum
Definition: The linear momentum (“momentum” for short) of an object is the product of its
mass and its velocity.
Plural: Momenta
Symbol: p ⃗
p⃗=mv⃗
Momentum is a vector quantity whose direction is the same as that of velocity.
| p ⃗| = m | v ⃗|
p = mv
SI unit: kg.m/s. There is no special name for this unit.
2
Linear Momentum
•
The everyday usage of the term momentum is in accord with the definition on the previous
slide.
•
If two cars have the same mass and are moving with different speeds, the fast moving car is
said to have more momentum.
•
A heavier car has more momentum than a lighter car if they are moving with the same
speed.
•
The more the momentum of the object, the harder it is to stop it.
•
The more the momentum of the object, the greater its effect on a collision.
3
Linear Momentum
The momentum of an object can be changed by application of a force.
The net force acting on a particle is equal to the time rate of change of its linear momentum.
Δp
ΣF ⃗=
Δt
Newton’s second law of motion.
v
−
v
Δv
Δp
0
Σ F ⃗ = ma = m
=m
=
Δt
Δt
Δt
4
Example 1
A tennis ball leaves the racket on a serve with a speed of 55 m/s. If the ball has a mass of
0.060 kg and is in contact with the racket for 4 ms, estimate the average force on the ball.
5
Example 1
The force that acts on the tennis ball, a baseball or a football being hit is not constant.
Δp mv − mv0
F̄ =
=
Δt
Δt
The tennis ball is hit when the initial velocity v0 = 0 at the top
of the throw, and we assume v = 55 m/s is in the horizontal
direction.
6
Example 1
Δp mv − mv0
F̄ =
=
Δt
Δt
(0.060 × 55) − (0.060 × 0)
F̄ =
= 820 N
0.004
High speed photography and radar can give us an estimate of the contact time and the velocity
of the ball leaving the racket. But a direct measurement of the force is not practical. This
calculation shows a handy technique for determining the unknown force in the real world.
7
Example 2
A baseball (m = 0.14 kg) has an initial velocity of v0 ⃗ = − 38 m /s as it approaches a bat. We
have chosen the direction of approach as the negative direction. The bat applies an average
force F ⃗ that is much larger than the weight of the ball, and the ball departs from the bat with a
final velocity of v ⃗ = + 58 m /s. Assuming that the time of contact is Δt = 1.6 × 10 s, find
the average force exerted on the ball by the bat.
−3
Δp mv − mv0
F̄ =
=
Δt
Δt
(0.14 × 58) − (0.14 × −38)
N
F̄ =
=
8400
1.6 × 10−3
8
Collisions and Impulse
A tennis racket or baseball bat striking a ball, a hammer hitting a nail, etc, are examples of
collisions that occur in everyday life.
When a collision occurs the interaction between the objects involved is usually far stronger
than any external forces. We can then ignore the effect of other forces during the brief time
interval (typically milliseconds for macroscopic collisions) of the collision.
9
Collisions and Impulse
From Newton’s second law of motion
Δp
ΣF ⃗=
Δt
⃗ = Δp
Σ F Δt
The product of the force F ⃗ and the time interval over Δt which the force acts is called the the
impulse J .⃗
⃗
⃗
J = Σ F Δt = Δp
SI Unit: N.s
Impulse is a vector quantity and has the same direction as the average force.
10
Example 3
In Example 2, determine the impulse applied to the ball by the bat.
⃗ = Δp
J ⃗ = Σ F Δt
J ⃗ = 0.14 × (58 − (−38)) = + 13.4 N.s (kg.m/s)
11
Conservation of momentum
Consider a head-on collision of two billiard balls.
If no net external forces act on the system, the total momentum
of the system is a conserved quantity.
Although the momentum of each ball changes as a result of the
collision, the SUM of the MOMENTA before and after the
collision is the same.
12
Conservation of momentum
The total momentum is the vector sum of the momenta of the balls.
Ball 1
Ball 2
Ball 1 + Ball 2
Before Collision
⃗ = m1i v 1i
⃗
p 1i
⃗ = m2i v 2i
⃗
p 2i
⃗ + m2i v 2i
⃗
p i⃗ = m1i v 1i
After Collision
⃗ = m1f v 1f
⃗
p 1f
⃗ = m2f v 2f
⃗
p 2f
⃗ + m2f v 2f
⃗
p f⃗ = m1f v 1f
If the net force on the system is zero; i.e., ΣF = 0
pi ⃗ = pf ⃗
⃗ + m2i v 2i
⃗ = m1f v 1f
⃗ + m2f v 2f
⃗
m1i v 1i
13
Conservation of momentum
No matter what the velocities and the masses are, experiments
show that the total momentum before the collision is the same
as after the collision, whether the collision is head-on or not as
long as no net external force acts.
The law of conservation of momentum states that
THE TOTAL MOMENTUM OF AN ISOLATED SYSTEM OF
OBJECTS REMAINS CONSTANT.
14
Conservation of momentum
The law of conservation of momentum was discovered from
experiment but can also be derived from Newton's laws of motion.
During collision, suppose the force exerted by object 1 on object 2
⃗ = F ,⃗ the force exerted on object 2 by object 1 is
is F 1,2
⃗ =− F⃗
F 2,1
⃗
F
⃗
⃗
⃗
1,2 + F 2,1 = F − F = 0
15
Conservation of momentum
⃗ − p 1i
⃗
p 1f
⃗1
Δ
p
⃗
F 1,2 =
=
Δt
Δt
⃗ − p 1i
⃗
p 1f
Δt
+
⃗ − p 2i
⃗
p 2f
⃗2
Δ
p
⃗
F 2,1 =
=
Δt
Δt
⃗ − p 2i
⃗
p 2f
Δt
=0
⃗ − p 1i
⃗ + p 2f
⃗ − p 2i
⃗ =0
p 1f
⃗ + p 2f
⃗ = p 1i
⃗ + p 2i
⃗
p 1f
⃗ + m2f v 2f
⃗ = m1i v 1i
⃗ + m2i v 2i
⃗
m1f v 1f
16
Conservation of momentum
The derivation in the previous slide can be extended to include
any number of interacting objects.
⃗ + m2f v 2f
⃗ + m3f v 3f
⃗ + ⋯ + mnf v nf
⃗ = m1i v 1i
⃗ + m2i v 2i
⃗ + m3i v 3i
⃗ + ⋯ + mni v ni
⃗
m1f v 1f
17
Example 4
Starting from rest, two skaters (a woman and a man) push each other on smooth level ice
where friction is negligible. The woman has a mass of 54 kg and the man has a mass of 88 kg.
The woman moves away with a velocity of +2.5 m/s. Find the recoil velocity of the man.
⃗ + mmf v mf
⃗ = mwi v wi
⃗ + mmi v mi
⃗
mwf v wf
⃗ ) = (54 × 0) + (88 × 0)
(54 × 2.4) + (88 × v mf
⃗ )=0
129.6 + (88 × v mf
⃗
v mf
−129.6
=
= − 1.5 m/s
88
18
Example 5
A 10,000 kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If the
cars lock together as a result of the collision, what is their common speed just afterwards.
⃗ + mBf v Bf
⃗ = mAi v Ai
⃗ + mBi v Bi
⃗
mAf v Af
⃗ ) + (10000 × v Bf
⃗ ) = (10000 × 24.0) + (10000 × 0)
(10000 × v Af
(10000 × v )⃗ + (10000 × v )⃗ = 240000
20000 × v ⃗ = 240000
v ⃗ = 12 m/s
19
Conservation of momentum
Assumption
No external forces, i.e., ΣFext = 0
Let's be real
In the real world there are external forces acting of the systems.
1. Frictional forces
2. Gravitational forces
3. etc.
That is why we consider a very small observation time (before and after).
20
Conservation of momentum and Kinetic energy
In an elastic collision,
total kinetic energy after = total kinetic energy before
1
1
1
1
2
2
2
2
m1f v1f + m2f v2f = m1iv1i + m2iv2i
2
2
2
2
21
Conservation of momentum and Kinetic energy
Collisions in which kinetic energy is not conserved are said to
be inelastic collisions.
The kinetic energy that is lost is changed into other forms of
energy, often thermal energy, so that the total energy (as always)
is conserved.
In an inelastic collision
1
1
1
1
2
2
2
2
m1f v1f + m2f v2f + thermal energy and other forms of energy = m1iv1i + m2iv2i
2
2
2
2
22
Conservation of momentum and Kinetic energy
We now apply the conservation laws for momentum and kinetic
energy to an elastic collision between two small objects that
collide head-on, so all the motion is along a line.
⃗ + m2f v 2f
⃗ = m1i v 1i
⃗ + m2i v 2i
⃗
m1f v 1f
1
1
1
1
2
2
2
2
m1f v1f + m2f v2f = m1iv1i + m2iv2i
2
2
2
2
If we know the masses and velocities before the collision, then we can solve these two
equations for the velocities after the collision because we will have two equations with two
unknowns.
23
Conservation of momentum and Kinetic energy
⃗ + m2f v 2f
⃗ = m1i v 1i
⃗ + m2i v 2i
⃗
m1f v 1f
⃗ − m1i v 1i
⃗ = m2i v 2i
⃗ − m2f v 2f
⃗
m1f v 1f
⃗ − v 1i
⃗ ) = m2( v 2i
⃗ − v 2f
⃗ )
m1( v 1f
24
…. Eq 1
Conservation of momentum and Kinetic energy
1
1
1
1
2
2
2
2
m1f v1f + m2f v2f = m1iv1i + m2iv2i
2
2
2
2
2
m1f v1f
2
m1(v1f
+
−
2
m2f v2f
2
v1i)
=
=
2
m1iv1i
2
m2(v2i
−
+
2
m2iv2i
2
v2f )
m1(v1f − v1i)(v1f + v1i) = m2(v2i − v2f ) + (v2i + v2f )
25
… . Eq 2
Conservation of momentum and Kinetic energy
⃗ − v 1i
⃗ ) = m2( v 2i
⃗ − v 2f
⃗ )
m1( v 1f
m1(v1f − v1i)(v1f + v1i) = m2(v2i − v2f ) + (v2i + v2f )
… Eq 1
… Eq 2
Dividing Eq 1 into Eq 2
⃗ + v 1i
⃗ = v 2i
⃗ + v 2f
⃗
v 1f
which can be rewritten as
⃗ − v 2i
⃗ = − ( v 1f
⃗ − v 2f
⃗ )
v 1i
[head-on 1-D elastic collision]
26
Example 6
−27
4
A proton of mass 1.68 × 10
kg traveling with a speed of 3.60 × 10 m/s has an elastic
−27
head-on collision with a helium (He) nucleus (mHe = 6.64 × 10
kg) initially at rest. What
are the velocities of the proton and helium nucleus after the collision?
27
Example 6
⃗ + mHe,f v He,f
⃗
⃗ + mHe,i v He,i
⃗
mp,f v p,f
= mp,i v p,i
⃗ + mHe,f v He,f
⃗
⃗ (because initial speed of Helium nucleus is zero)
mp,f v p,f
= mp,i v p,i
⃗ − v He,i
⃗ = − ( v p,f
⃗ − v He,f
⃗ )
v p,i
⃗ = − ( v p,f
⃗ − v He,f
⃗ )
v p,i
⃗
⃗ + v p,f
⃗
v He,f
= v p,i
28
Example 6
⃗ + mHe,f v He,f
⃗
⃗
mp,f v p,f
= mp,i v p,i
⃗
⃗ + v p,f
⃗ (substitute in above equation)
v He,f
= v p,i
⃗ + mHe,f ( v p,i
⃗ + v p,f
⃗ ) = mp,i v p,i
⃗
mp,f v p,f
⃗ + mHe,f v p,i
⃗ + mHe,f v p,f
⃗ = mp,i v p,i
⃗
mp,f v p,f
⃗ (mp,f + mHe,f ) + mHe,f v p,i
⃗ = mp,i v p,i
⃗
v p,f
29
Example 6
⃗ (mp,f + mHe,f ) + mHe,f v p,i
⃗ = mp,i v p,i
⃗
v p,f
⃗ (mp,f + mHe,f ) = mp,i v p,i
⃗ − mHe,f v p,i
⃗
v p,f
⃗ (mp,f + mHe,f ) = v p,i
⃗ (mp,i − mHe,f )
v p,f
⃗ =
v p,f
⃗ (mp,i − mHe,f )
v p,i
(mp,f + mHe,f )
4
−27
3.6 × 10 × (1.68 × 10 − 6.64 × 10
=
−27
−27
(1.68 × 10 + 6.64 × 10 )
−27
⃗
⃗ + v p,f
⃗ = 3.6 × 10 + (−2.16 × 10 ) = 1.45 × 10 m/s
v He,f
= v p,i
4
4
30
4
)
4
= − 2.16 × 10 m/s
Example 7 - Completely inelastic collision
Example 5 is an example of a completely inelastic collision. Calculate how much of the initial
kinetic energy is transformed to thermal or other forms of energy.
1
1
1
1
2
2
2
2
mA,f vA,f + mB,f vB,f + Eth = mA,ivA,i + mB,ivB,i
2
2
2
2
1
1
1
1
2
2
2
2
ETh = mA,ivA,i + mB,ivB,i − mA,f vA,f − mB,f vB,f
2
2
2
2
1
1
2
2
2
2
ETh = mA,i(vA,i − vA,f ) + mB,i(vB,i − vB,f )
2
2
31
Example 7 - Completely inelastic collision
1
1
2
2
2
2
ETh = mA,i(vA,i − vA,f ) + mB,i(vB,i − vB,f )
2
2
1
1
2
2
2
2
ETh = × 10000(24 − 12 ) + × 10000(0 − 12 )
2
2
6
5
6
ETh = 2.16 × 10 − 7.20 × 10 = 1.4 × 10 J
32
Collision in 2-D
We now apply the law of conservation of momentum to collisions in 2D where the vector
nature of momentum is especially important.
33
Collision in 2-D
Momentum is a vector, and because the total momentum is conserved, its components in the x
and y directions also are conserved.
P1ix + P2ix = P1fx + P2fx
P1iy + P2iy = P1fy + P2fy
34
Example 8
Billiard ball 1 moving with speed v1i = 3.0 m/s in the ±x direction strikes an equal-mass ball
2 initially at rest. The two balls are observed to move off at 45° to the x axis, ball 1 above the
x axis and ball 2 below. That is, θ1 = 45° and θ2 = - 45°. What are the speeds of the two balls
after the collision?
35
Example 8
P1iy + P2iy = P1fy + P2fy
mv1iy + mv2iy = mv1fy + mv2fy
∘
∘
0 + 0 = mv1f sin(45 ) + mv2f sin(−45 )
∘
∘
mv1f sin(45 ) = − mv2f sin(−45 )
v1f =
∘
−mv2f sin(−45 )
m sin(45∘)
v1f = v2f
36
Example 8
P1ix + P2ix = P1fx + P2fx
mv1ix + mv2ix = mv1fx + mv2fx
∘
∘
(m × 3.0) + 0 = mv1f cos(45 ) + mv2f cos(−45 )
∘
∘
∘
∘
3.0 = v1f cos(45 ) + v2f cos(−45 )
3.0 = v2f cos(45 ) + v2f cos(−45 )
∘
∘
3.0 = v2f (cos(45 ) + cos(−45 ))
3.0
v2f =
=
2.1
m/s
(cos(45∘) + cos(−45∘))
37
Center of Mass
General Motion
•
•
motion that is not pure translation
motion of real extended objects (i.e., objects that have size)
So far we have approximated point particles or assumed that our bodies undergo only
translational motion.
Real extended objects, however, can undergo rotational and other types of motion as well.
38
Center of Mass
The motion of the diver is pure translation.
39
Center of Mass
The motion of the diver is translation plus rotation.
40
Center of Mass
Observations indicate that even if an object rotates, or several parts of a system of objects
move relative to one another, there is one point that moves in the same path that a particle
would move if subjected to the same net force.
This point is called the center of mass (abbreviated CM).
41
Center of Mass
The black dot represents the diver’s CM at each moment.
42
Center of Mass
43
Center of Mass
We define the CM of an extended object (a system of particles) in order to predict its motion.
The center of mass of a system of particles is the point that moves as though (1) all of the
system’s mass were concentrated there and (2) all external forces were applied there.
44
Center of Mass
Consider a system made up of only two particles (or small objects) of masses mA and mB. We
choose a coordinate system so that both particles lie on the x axis at positions XA and XB.
45
Center of Mass
The center of mass of this system is defined to be at the position XCM , given by
mAxA + mBxB
XCM =
mA + mB
The center of mass lies on the line joining mA and mB.
If one mass is greater than the other, then the CM is closer to the larger mass.
If there are more than two particles along a line, there will be additional terms:
mAxA + mBxB + mC xC + ⋯
XCM =
mA + mB + mC + ⋯
46
Center of Mass
If the particles are distributed in three dimensions, the center of mass must be identified by
three coordinates.
1
xcm =
mi xi
M∑
1
ycm =
mi yi
M∑
1
zcm =
mizi
M∑
47
Center of Gravity
A concept similar to center of mass is center of gravity (CG).
An object’s CG is that point at which the force of gravity can be considered to act.
The center go gravity of a system of a system of particles is determined as follow:
Xcg =
FgAxA + FgBxB + FgC xC + ⋯
FgA + FgB + FgC + ⋯
48
Center of Gravity
The force of gravity actually acts on all the different parts or particles of an object, but for
purposes of determining the translational motion of an object as a whole, we can assume that
the entire weight of the object (which is the sum of the weights of all its parts) acts at the CG.
There is a conceptual difference between the center of gravity and the center of mass, but for
nearly all practical purposes, they are at the same point.
49
References
• Giancoli, D.C., 2016. Physics: principles with applications. Pearson.
• Cutnell, J.D., 2009. Physics . John Wiley & Sons.
• Serway, R.A. and Faughn, J.S., 8th Edition. College physics. Saunders Publishers.
• Halliday, D. and Resnick, R., 2014. Fundamentals of physics. John Wiley & Sons.
• Harris, B., 1991. University Physics. John Wiley & Sons, Inc.
50