Linear Momentum and Impulse

(Definition)

Linear Momentum and Impulse

A moving object carries a quantity of motion that depends on both its mass and its velocity. In Newtonian mechanics this quantity is the linear momentum

|--------| p-=--mv.--

(1)

Momentum is a vector. Its direction is the direction of the velocity.

Newton’s second law can be written in momentum form as

|-----------| | dp- | Fnet = dt .| ------------

(2)

This form makes clear that a force changes momentum.

Integrating through a finite time interval leads to the impulse momentum theorem:

|----∫-tf----------------| J = F (t)dt = Δp. | | ti net | --------------------------

(3)

Impulse is therefore the accumulated effect of force over time.

Figure 1. Linear momentum has magnitude p = mv and points in the same direction as the particle velocity.

1 Definition of linear momentum

For a particle of constant mass m,

|--------| p-=--mv.--

(4)

In Cartesian components,

|------------------------| -p-=-pxex-+--pyey +-pzez,|

(5)

with

p_x	= mv_x,	(6)
p_y	= mv_y,	(7)
p_z	= mv_z.	(8)

The magnitude is

p = m |v|.

(9)

The SI unit is

|--------| |kg m ∕s.| ---------

(10)

2 Momentum is a vector

Momentum must be added vectorially.

If one particle moves in the positive x direction,

px > 0.

(11)

If it moves in the negative x direction,

px < 0.

(12)

In one-dimensional motion, the sign of momentum contains the direction information.

For example, if

m = 2.0 kg

(13)

and

vx = − 3.0m ∕s,

(14)

then

|------------------| |px = − 6.0kg m ∕s.| -------------------

(15)

3 Newton’s second law in momentum form

Newton’s second law is most generally expressed for a particle of fixed identity as

|-----------| | dp- | Fnet = dt .| ------------

(16)

For constant mass,

p = mv.

(17)

Differentiate:

	= m	(18)
	= ma.	(19)

Thus

|----------| -Fnet =-ma--

(20)

is recovered.

The momentum form emphasizes that force is the rate of change of momentum.

Figure 2. Newton’s second law connects force to the time rate of change of momentum. Integrating in time produces the impulse momentum theorem.

4 Impulse

Integrate Newton’s second law from t_i to t_f:

∫ tf ∫ tf Fnet(t) dt = dp-dt. ti ti dt

(21)

The right side is

p − p . f i

(22)

Define the net impulse

|--------------------| | ∫ tf | Jnet = Fnet(t)dt.| | ti | ----------------------

(23)

Therefore

|----------------------| -Jnet =-Δp--=-pf-−-pi.-|

(24)

For constant mass,

|------------------| |Jnet = m (vf − vi).| --------------------

(25)

5 Units of impulse

Impulse has units

N s.

(26)

Since

2 1N = 1kg m ∕s ,

(27)

we have

1N s = 1kg m ∕s.

(28)

Thus impulse and momentum have the same physical dimensions:

|----------------| |1N s = 1kg m ∕s.| ------------------

(29)

They are not the same type of quantity conceptually. Momentum is a state quantity of the particle, while impulse describes momentum transferred during a time interval.

6 Force time graphs

Because

∫ J = F dt,

(30)

the signed area under a force-versus-time graph equals impulse.

In one dimension,

|------------------| | ∫ tf | |Jx = Fx (t)dt.| -------ti----------|

(31)

Area above the time axis contributes positive impulse.

Area below the time axis contributes negative impulse.

Figure 3. The signed area under a force-versus-time curve is impulse. The impulse equals the resulting change in momentum.

7 Average force

Define the average force over the interval Δt by requiring it to produce the same impulse:

|----------------------| | ∫ tf | |FavgΔt = F (t) dt.| ------------ti---------|

(32)

Therefore

|------------------| | -J- Δp-- | |Favg = Δt = Δt .| -------------------

(33)

This does not mean the actual force is constant.

The average force is the constant force that would produce the same impulse over the same time interval.

Figure 4. A varying force and its average-force rectangle have the same area and therefore the same impulse.

8 Example 1: constant-force impulse

A 0.50 kg cart initially moves at

vi = 2.0m ∕s.

(34)

A constant force

F = 3.0 N

(35)

acts in the direction of motion for

Δt = 4.0s.

(36)

The impulse is

J = F Δt = (3.0)(4.0) = 12N s.

(37)

Initial momentum:

pi = mvi = (0.50)(2.0) = 1.0kg m ∕s.

(38)

Thus

p = p + J = 13kg m ∕s. f i

(39)

The final velocity is

pf 13 vf = m--= 0.50 = 26 m ∕s.

(40)

9 Example 2: triangular force pulse

A force pulse rises linearly from zero to

F = 800N max

(41)

and falls linearly back to zero over a total duration

Δt = 0.020s.

(42)

The impulse is the area of the triangle:

1 J = --FmaxΔt. 2

(43)

Thus

J	= (800)(0.020)	(44)
	= 8.0 N s.	(45)

The average force is

Favg = -J- = -8.0--= 400 N. Δt 0.020

(46)

For this symmetric triangular pulse,

|--------------| | 1- | |Favg = 2Fmax. | ----------------

(47)

10 Changing direction produces a large impulse

Momentum depends on velocity, including its direction.

Suppose a ball approaches a wall with velocity

vi = +v

(48)

and rebounds with the same speed:

vf = − v.

(49)

Then

Δp	= m(−v) − m(+v)	(50)
	= −2mv.	(51)

Therefore the impulse magnitude is

|----------| |J| = 2mv. | ------------

(52)

A rebound can require twice the momentum change of simply bringing the object to rest.

Figure 5. Reversing the velocity changes momentum by more than merely stopping the object. For equal incoming and outgoing speeds, |Δp| = 2mv.

11 Example 3: bouncing ball

A ball has

m = 0.20 kg.

(53)

Take upward as positive.

Just before striking the floor,

v = − 6.0m ∕s. i

(54)

Just after leaving the floor,

vf = +4.0 m ∕s.

(55)

The momentum change is

Δp	= m(v_f − v_i)	(56)
	= 0.20[4.0 − (−6.0)]	(57)
	= 2.0 kg m∕s.	(58)

Therefore the net impulse is

|----------------| |Jnet = +2.0 N s.| -----------------

(59)

If the contact lasts

Δt = 0.010s,

(60)

then the average net force is

F = --2.0--= 200N. net,avg 0.010

(61)

If one wants the average contact force from the floor rather than the average net force, gravity must also be included in the force balance.

12 Contact force versus net force during impact

During a collision with the floor,

Fnet = N + mg.

(62)

Therefore

J = J + J . net N g

(63)

The impulse momentum theorem gives

Δp = JN + Jg.

(64)

Hence the contact impulse is

|---------------| JN--=-Δp--−-Jg.--

(65)

For a very short impact,

|Jg| = mg Δt

(66)

may be much smaller than the contact impulse, but it should be neglected only after checking the scale.

13 Why increasing collision time reduces average force

For a specified momentum change,

Δp = F Δt. avg

(67)

Thus

|------------| | Δp | |Favg = ----.| ---------Δt--

(68)

If the same momentum change occurs over a longer time, the required average force magnitude is smaller.

This is the physics behind

airbags,
crumple zones,
padded landing surfaces,
bending the knees when landing,
padded gloves and helmets.

These devices do not necessarily reduce the required momentum change. They increase the time over which the change occurs.

14 Impulsive forces

An impulsive force is a force that becomes very large over a short time interval and produces a finite impulse.

During a sufficiently short collision, one often approximates

Jexternal,slow ≈ 0

(69)

for forces such as gravity, because their impulse over the short collision time is small compared with the collision impulse.

For example,

Jg = mg Δt.

(70)

Δt = 10 −3s,

(71)

the gravitational impulse can be tiny relative to an impact impulse of several newton-seconds.

This approximation becomes central in collision analysis.

15 Two-dimensional impulse

Impulse is a vector and can change different momentum components independently.

In two dimensions,

|----------------| J = Jxex + Jyey.| ------------------

(72)

The component equations are

J_x	= m(v_fx − v_ix),	(73)
J_y	= m(v_fy − v_iy).	(74)

The impulse magnitude is

|----∘----------| | 2 2 | J-=----Jx-+-Jy.--

(75)

Figure 7. In two dimensions, impulse is the vector difference p_f − p_i. The change can alter both the magnitude and direction of momentum.

16 Example 4: two-dimensional impulse

A 0.25 kg puck has

vi = (4.0ex)m ∕s.

(76)

After being struck,

vf = (1.0ex + 3.0ey) m∕s.

(77)

Then

J	= m(v_f − v_i)	(78)
	= 0.25[(−3.0)e_x + 3.0e_y].	(79)

Thus

|---------------------------| J = (− 0.75ex + 0.75ey)N s. | -----------------------------

(80)

Its magnitude is

√ ------------- J = 0.752 + 0.752 = 1.06 N s.

(81)

17 Impulse versus work

Impulse and work describe different effects of force.

Impulse is

|----∫-------------| | | |J = F dt = Δp. | --------------------

(82)

Work is

|------∫---------------| | | |W = F ⋅ dr = ΔK. | -----------------------

(83)

Impulse is a vector and changes momentum.

Work is a scalar and changes kinetic energy.

Figure 6. The same force can be integrated over time to obtain impulse or over displacement to obtain work. Impulse changes momentum; work changes kinetic energy.

18 Momentum and kinetic energy

For a particle of mass m,

p = mv.

(84)

Therefore

v = -p . m

(85)

Substitute into

1- 2 K = 2mv :

(86)

----------- | 2 | |K = -p--.| ------2m---|

(87)

Thus momentum and kinetic energy are related but not interchangeable.

For fixed momentum magnitude, a larger mass corresponds to smaller kinetic energy.

For fixed kinetic energy, momentum magnitude depends on mass:

√ ------ p = 2mK.

(88)

19 Same impulse does not mean same energy change

Suppose the same impulse J is applied in one dimension to two identical particles with different initial momenta.

Since

pf = pi + J,

(89)

the kinetic-energy change is

ΔK	=	(90)
	= .	(91)

Therefore

|---------------2--| | piJ- -J-- | -ΔK--=---m--+-2m--.|

(92)

The same impulse can produce different changes in kinetic energy depending on the initial momentum.

This is another reason impulse and work must not be confused.

20 Momentum is frame dependent

Velocity depends on the inertial reference frame, so momentum does also.

If frame S′ moves with constant velocity V relative to frame S, then

v′ = v − V.

(93)

Therefore

|-′------------------------| p--=-m-(v-−-V-)-=-p-−-mV.---

(94)

Momentum is not an absolute quantity independent of inertial frame.

The impulse momentum theorem remains valid when all quantities are evaluated consistently in the same inertial frame.

21 External impulse on a system of particles

For a system of particles, internal forces cancel in the total momentum balance under the usual Newton’s-third-law assumptions.

The total momentum is

|------------| | ∑ | |P = pi. | -------i-----

(95)

The net external force satisfies

|--------------| | dP | |Fext,net = ---.| -----------dt--

(96)

Integrating,

|-----------| Jext =-ΔP.---

(97)

This result is the bridge to conservation of linear momentum.

Jext = 0,

(98)

then

Pf = Pi.

(99)

The next article develops this result in detail.

22 Variable-mass caution

The relation

p = mv

(100)

is always the Newtonian definition of particle momentum.

However, applying

d Fext = --(mv ) dt

(101)

to an open system whose mass changes by material flowing across the system boundary requires care.

Rockets are the classic example.

The simple constant-mass result

F = ma

(102)

cannot be obtained by differentiating mv and treating ṁv as an ordinary external force term.

Variable-mass mechanics requires an explicit momentum-flux analysis and will be treated separately.

23 Common mistakes

Treating momentum as a scalar instead of a vector.
Dropping the sign of velocity in one-dimensional momentum problems.
Using speed instead of velocity when computing momentum change.
Forgetting that rebound momentum change can be larger than stopping momentum change.
Confusing impulse with force.
Confusing impulse with work.
Using peak force instead of average force in J = F_avgΔt.
Forgetting that the area under a force time graph is impulse.
Forgetting other forces when converting net impulse into a particular contact impulse.
Neglecting gravity during impact without first checking whether its impulse is small.
Treating N s as a unit of energy.
Assuming equal impulses imply equal kinetic-energy changes.

24 Practice exercises

A 4.0 kg object moves at 3.0 m∕s in the positive x direction. Find its momentum.
A 0.50 kg ball moves at −8.0 m∕s. Find its one-dimensional momentum.
A constant 12 N force acts for 0.25 s. Find the impulse.
A 2.0 kg particle changes velocity from 3.0 to −5.0 m∕s. Find the net impulse.
A force time graph is triangular with base 0.040 s and peak 600 N. Find the impulse and average force.
A 0.15 kg baseball approaches a bat at 40 m∕s and leaves in the opposite direction at 50 m∕s. Find the magnitude of the impulse.
If the contact time in the previous problem is 1.5 ms, find the average force magnitude.
Derive the impulse momentum theorem from F = dp∕dt.
Show that K = p²∕(2m) for a Newtonian particle.
Explain why doubling the collision time halves the average force for a fixed momentum change.
A 1.0 kg puck has initial velocity (3e_x + 2e_y) m∕s and final velocity (−1e_x + 5e_y) m∕s. Find the impulse vector.
A force F(t) = F₀t∕T acts from 0 to T. Find the impulse.
A ball bounces vertically from a floor. Write an equation separating the floor-contact impulse from the gravitational impulse.
Give an example where a force does zero work but produces a nonzero impulse.
Give an example where a force produces both nonzero work and nonzero impulse.