Collisions in One Dimension

(Topic)

Collisions in One Dimension

A collision is a short-duration interaction in which two bodies exert large forces on one another and exchange momentum and energy.

During a sufficiently short collision, the external impulse on the two-body system is often negligible compared with the internal collision impulse. Then total linear momentum is approximately conserved:

m--u--+-m--u--=-m--v-+--m-v-.-| --1-1-----2-2-----1-1-----2-2--

(1)

Here

u₁, u₂	= velocities before the collision,	(2)
v₁, v₂	= velocities after the collision.	(3)

Momentum conservation alone gives one equation for the two unknown final velocities. A second physical relation is needed. That second relation depends on the type of collision.

Figure 1. A one-dimensional collision is analyzed by comparing the system state immediately before and immediately after the short interaction interval.

1 Why momentum is usually conserved during a collision

For the two-object system,

Jext = ΔP.

(4)

During a short collision interval Δt, slowly varying external forces produce impulses of order

Jext ∼ FextΔt.

(5)

The internal collision forces can be much larger, but they occur in equal-and-opposite pairs and cancel from the total momentum balance.

|J | ≪ |Δp |, ext 1

(6)

then

|--------| P ≈ P .| --f----i--

(7)

This approximation applies to many carts, pucks, balls, and impact problems.

2 Sign convention in one dimension

Choose a positive direction before writing any momentum equation.

For example, let rightward be positive.

Then:

rightward velocity is positive,
leftward velocity is negative,
momentum automatically carries the correct sign.

The momentum equation is

m1u1 + m2u2 = m1v1 + m2v2.

(8)

No additional directional language is needed once the sign convention is established.

3 Classification by kinetic energy

The main collision classes are distinguished by what happens to total kinetic energy.

3.1 Elastic collision

An elastic collision conserves both total momentum and total kinetic energy:

m₁u₁ + m₂u₂	= m₁v₁ + m₂v₂,	(9)
m₁u₁² + m₂u₂²	= m₁v₁² + m₂v₂².	(10)

3.2 Inelastic collision

An inelastic collision conserves total momentum, but translational kinetic energy decreases:

|K---<-K--.| ---f-----i-|

(11)

The missing translational kinetic energy is transferred into other forms such as

deformation,
internal energy,
sound,
vibration,
heating.

3.3 Perfectly inelastic collision

A perfectly inelastic collision is the limiting inelastic case in which the objects stick together after impact:

|-------------| v1 = v2 = vf. | ---------------

(12)

Momentum is still conserved if external impulse is negligible.

Figure 2. Collision type is determined by the post-impact constraint and by the change in translational kinetic energy.

4 Perfectly inelastic collision

If two objects stick together,

v1 = v2 = vf.

(13)

Momentum conservation gives

m1u1 + m2u2 = (m1 + m2)vf.

(14)

Therefore

|--------------------| | m1u1 + m2u2 | |vf = ---m--+-m---- .| -----------1----2----

(15)

This is exactly the center of mass velocity before the collision:

|----------| |vf = VCM. | ------------

(16)

Figure 3. In a perfectly inelastic collision, the bodies stick together and move afterward with the pre-collision center of mass velocity.

5 Example 1: perfectly inelastic collision

A 2.0 kg cart moving right at

u = 4.0 m ∕s 1

(17)

collides with a 3.0 kg cart initially at rest.

The carts stick together.

Momentum conservation gives

(2.0)(4.0) + (3.0)(0) = (5.0)vf.

(18)

Thus

|-------------| vf = 1.6 m∕s. | ---------------

(19)

Initial kinetic energy:

1- 2 Ki = 2(2.0)(4.0) = 16J.

(20)

Final kinetic energy:

1 2 Kf = 2(5.0)(1.6) = 6.4 J.

(21)

Therefore the translational kinetic-energy decrease is

|--------------------------| -ΔK---=-Kf-−--Ki-=-−-9.6J.-|

(22)

Momentum is conserved, but kinetic energy is not.

6 Kinetic-energy loss in a perfectly inelastic collision

For a general perfectly inelastic collision,

m1u1--+-m2u2- vf = M ,

(23)

where

M = m1 + m2.

(24)

The kinetic-energy loss can be written compactly using the reduced mass

--m1m2--- μ = m1 + m2 .

(25)

The result is

|-------------------------| | 1 | Ki − Kf = -μ (u1 − u2 )2. | ------------2-------------

(26)

Thus the dissipated translational kinetic energy depends on the relative approach speed.

7 Derivation of the perfectly inelastic energy loss

The total kinetic energy can be decomposed into

K = KCM + Krel.

(27)

For two particles,

1- 2 Krel = 2μ(u1 − u2) .

(28)

In the perfectly inelastic final state, the two objects have no relative motion:

v1 − v2 = 0.

(29)

Therefore

Krel,f = 0.

(30)

Because momentum conservation keeps K_CM unchanged, the entire initial relative kinetic energy is removed from translational motion:

|--------------------------------| K − K = K = 1μ(u − u )2.| --i-----f-----rel,i---2----1----2---

(31)

8 Elastic collisions

For a one-dimensional elastic collision, both momentum and kinetic energy are conserved:

m₁u₁ + m₂u₂	= m₁v₁ + m₂v₂,	(32)
m₁u₁² + m₂u₂²	= m₁v₁² + m₂v₂².	(33)

A direct simultaneous solution is possible, but there is a cleaner route.

9 Relative-speed relation for elastic collisions

Start with momentum conservation:

m1 (u1 − v1) = m2 (v2 − u2).

(34)

Kinetic-energy conservation can be written

m1 (u2 − v2) = m2 (v2− u2 ). 1 1 2 2

(35)

Factor both sides:

m (u − v )(u + v ) = m (v − u )(v + u ). 1 1 1 1 1 2 2 2 2 2

(36)

Divide by the momentum relation, assuming a nontrivial collision:

u1 + v1 = v2 + u2.

(37)

Rearrange:

|----------------------| -u1-−-u2-=-−-(v1 −-v2).|

(38)

Equivalently,

|------------------| -v2 −-v1 =-u1 −-u2.|

(39)

Thus, in a one-dimensional elastic collision,

|--------------------------------------------------------| -relative-speed--of separation-=-relative-speed-of approach.-

(40)

Figure 4. In a one-dimensional elastic collision, the relative speed reverses sign while preserving its magnitude.

10 General one-dimensional elastic-collision formulas

The two equations

m₁u₁ + m₂u₂	= m₁v₁ + m₂v₂,	(41)
v₂ − v₁	= u₁ − u₂	(42)

can be solved for v₁ and v₂.

The results are

|-------------------------------| v = m1-−-m2-u + ---2m2---u , | |1 m1 + m2 1 m1 + m2 2 | --------------------------------

(43)

and

|-------------------------------| | 2m1 m2 − m1 | v2 = --------u1 + ---------u2. | ------m1-+-m2------m1--+-m2-----

(44)

These formulas are useful, but they should be understood rather than memorized blindly.

11 Derivation of the elastic formulas

From the relative-speed relation,

v = v + u − u . 2 1 1 2

(45)

Substitute into momentum conservation:

m₁u₁ + m₂u₂

= m₁v₁ + m₂(v₁ + u₁ − u₂).

(46)

Collect v₁:

(m1 + m2)v1 = (m1 − m2 )u1 + 2m2u2.

(47)

Thus

m1-−-m2-- ---2m2--- v1 = m1 + m2 u1 + m1 + m2 u2.

(48)

Substituting that result back into the relative-speed relation gives

--2m1---- m2--−-m1- v2 = m1 + m2 u1 + m1 + m2 u2.

(49)

12 Target initially at rest

A common special case has

u = 0. 2

(50)

Then

|----m---−-m------| v1 = --1-----2u1, | -----m1--+-m2------

(51)

and

|-----------------| | ---2m1--- | v2 = m1 + m2 u1. | -------------------

(52)

These formulas reveal several useful limiting cases.

13 Equal masses

m1 = m2 = m,

(53)

then

v₁	= u₂,	(54)
v₂	= u₁.	(55)

Thus equal masses exchange velocities in a one-dimensional elastic collision:

---------------------- v1 = u2, v2 = u1.| ----------------------

(56)

If object 2 starts at rest,

|--------------------| |v1 = 0, v2 = u1.| ---------------------

(57)

Figure 5. In a one-dimensional elastic collision of equal masses, the bodies exchange velocities.

14 Example 2: equal-mass elastic collision

A 1.0 kg cart moving right at

u = 5.0 m ∕s 1

(58)

collides elastically with an identical cart at rest.

Because the masses are equal,

v₁	= 0,	(59)
v₂	= 5.0 m∕s.	(60)

The first cart stops and the second cart carries away the original momentum and kinetic energy.

This is the idealized behavior seen in devices such as a Newton’s cradle.

15 Light projectile striking a heavy target

Suppose

m ≪ m 1 2

(61)

and the heavy target is initially at rest.

Then

v1 = m1-−--m2-u1 ≈ − u1. m1 + m2

(62)

The light projectile rebounds with nearly the same speed.

Meanwhile,

--2m1---- v2 = m + m u1 1 2

(63)

is small.

This resembles a ball elastically bouncing from a massive wall.

16 Heavy projectile striking a light target

m1 ≫ m2

(64)

and u₂ = 0, then

v1 ≈ u1

(65)

while

v2 ≈ 2u1.

(66)

A light target can leave with nearly twice the heavy projectile’s incoming speed.

This does not violate energy conservation because the light target has much less mass.

17 Coefficient of restitution

Real collisions are often neither perfectly elastic nor perfectly inelastic.

A useful empirical measure is the coefficient of restitution e, defined in one dimension by

|--------------------------------| | relative-speed-of-separation- | |e = relative speed of approach .| ----------------------------------

(67)

For the sign convention used here,

|------------| | v2 −-v1-| e = u1 − u2.| --------------

(68)

For ordinary passive impacts,

0 ≤ e ≤ 1.

(69)

Important limits are

e = 1	elastic collision,	(70)
e = 0	perfectly inelastic collision.	(71)

Figure 6. The coefficient of restitution compares relative separation speed after impact with relative approach speed before impact.

18 General collision formulas using restitution

Combine momentum conservation,

m1u1 + m2u2 = m1v1 + m2v2,

(72)

with

v2 − v1 = e(u1 − u2).

(73)

Solving gives

|---------------------------------| | m1 − em2 (1 + e)m2 | v1 = ----------u1 + ----------u2, | ------m1-+--m2-------m1-+--m2------

(74)

and

|---------------------------------| | (1 + e)m m − em | v2 = ---------1u1 + --2------1u2. | ------m1-+--m2-------m1-+--m2------

(75)

Setting e = 1 reproduces the elastic formulas.

Setting e = 0 gives the common final velocity of a perfectly inelastic collision.

19 Example 3: partially inelastic collision

Let

m₁	= 2.0 kg,	u₁	= 6.0 m∕s,	(76)
m₂	= 3.0 kg,	u₂	= 0,	(77)

with

e = 0.50.

(78)

Then

v₁	= (6)	(79)
	= 0.60 m∕s,	(80)

and

v₂	= (6)	(81)
	= 3.60 m∕s.	(82)

Thus

|--------------------------------| v1 = 0.60 m ∕s, v2 = 3.60m ∕s.| ----------------------------------

(83)

Momentum is conserved, but kinetic energy decreases because e < 1.

20 Kinetic-energy loss and restitution

For a two-body one-dimensional collision with conserved momentum, the center of mass kinetic energy does not change.

Only the relative kinetic energy can change.

Before collision,

1- 2 Krel,i = 2μ (u1 − u2) .

(84)

After collision,

1 Krel,f = --μ(v1 − v2)2. 2

(85)

Using

|v1 − v2| = e|u1 − u2|,

(86)

we obtain

|----------------| |Krel,f = e2Krel,i.| -----------------

(87)

Therefore the translational kinetic-energy loss is

|--------------------------------| | 1 2 2 | Ki − Kf = -μ (1 − e )(u1 − u2) . ------------2---------------------

(88)

Figure 8. For fixed initial relative motion, the retained relative kinetic energy fraction is e².

21 Center of mass frame

The center of mass velocity is

|----------------------| |V = m1u1--+-m2u2-. | --CM-------m1-+-m2-----|

(89)

Define velocities in the center of mass frame:

u₁′	= u₁ − V _CM,	(90)
u₂′	= u₂ − V _CM.	(91)

Because

′ ′ m1u 1 + m2u 2 = 0,

(92)

the momenta in the center of mass frame are equal and opposite.

For an elastic one-dimensional collision,

|-′------′------′------′-| -v1 =-− u-1,---v2-=-−-u2.-

(93)

Each object’s center of mass-frame velocity simply reverses direction.

Figure 7. In the center of mass frame, a one-dimensional elastic collision reverses both velocities while preserving their magnitudes.

22 Why the center of mass frame is useful

In the center of mass frame,

′ P = 0

(94)

both before and after the collision.

For an elastic collision, total kinetic energy in this frame is also unchanged.

Because the two momenta are opposite, the collision geometry becomes especially simple.

The laboratory-frame solution can be recovered by adding

VCM

(95)

to each final center of mass-frame velocity.

This viewpoint becomes even more useful in two-dimensional collisions and scattering.

23 Impulse during a collision

For object 1,

J1 = m1 (v1 − u1).

(96)

For object 2,

J2 = m2 (v2 − u2).

(97)

If external impulse is negligible,

J1 + J2 = 0.

(98)

Therefore

|----------| |J1 = − J2.| -----------

(99)

The objects receive equal-and-opposite collision impulses.

This is the time-integrated form of Newton’s third law.

24 Example 4: collision impulse

A 0.50 kg cart changes velocity from

u1 = 4.0 m ∕s

(100)

v1 = − 2.0m ∕s.

(101)

Its impulse is

J₁	= m₁(v₁ − u₁)	(102)
	= 0.50(−2.0 − 4.0)	(103)
	= −3.0 N s.	(104)

The other object receives

|--------------| -J2-=-+3.0-N-s-|

(105)

if external impulse is negligible.

25 Collision formulas versus physical reasoning

Closed-form equations are useful, but several qualitative checks should always be made.

Momentum must balance with signs.
In an elastic collision, kinetic energy must also balance.
In an ordinary inelastic collision, K_f < K_i.
If the objects stick, their final velocities must be equal.
For e = 1, relative speed reverses with equal magnitude.
For e = 0, relative separation speed is zero.
The center of mass velocity must remain constant if external impulse is negligible.

26 Common mistakes

Conserving kinetic energy in every collision.
Forgetting that momentum is signed in one dimension.
Setting the final velocities equal unless the objects actually stick.
Using the elastic-collision formulas for an inelastic impact.
Forgetting that e = 0 means no relative separation speed, not necessarily zero final speed.
Mixing velocities from different inertial frames.
Using speed instead of velocity in the momentum equation.
Assuming equal-and-opposite impulses imply equal velocity changes for unequal masses.
Forgetting to check whether external impulse is negligible during the collision.
Applying the restitution relation with the wrong order of relative velocities.
Forgetting that kinetic-energy loss appears in other energy forms rather than disappearing.
Memorizing formulas without checking limiting cases.

27 Practice exercises

A 2.0 kg cart moving at 5.0 m∕s sticks to a 3.0 kg cart at rest. Find the final velocity.
For the preceding collision, find the initial and final translational kinetic energies and the kinetic-energy loss.
Derive
$m1u1-+--m2u2- vf = m1 + m2$ (106)

for a perfectly inelastic collision.
Show that the kinetic-energy loss in a perfectly inelastic collision is
$1-μ(u − u )2. 2 1 2$ (107)
Derive the one-dimensional elastic relative-speed relation from momentum and kinetic-energy conservation.
Derive the general elastic-collision formulas for v₁ and v₂.
A 1.0 kg cart moving at 6.0 m∕s collides elastically with an identical stationary cart. Find both final velocities.
A 1.0 kg cart moving at 6.0 m∕s collides elastically with a 3.0 kg stationary cart. Find both final velocities.
A 3.0 kg cart moving at 6.0 m∕s collides elastically with a 1.0 kg stationary cart. Find both final velocities.
For a collision with e = 0.60, m₁ = 2.0 kg, m₂ = 4.0 kg, u₁ = 9.0 m∕s, and u₂ = 0, find the final velocities.
Show that K_rel,f = e²K_rel,i.
Explain physically why a light object can rebound from a very heavy stationary target with nearly its original speed.
Explain physically why equal masses exchange velocities in a one-dimensional elastic collision.
Transform an elastic collision into the center of mass frame and show that both velocities reverse.
A collision gives object 1 an impulse of −5 N s. What impulse does object 2 receive if external impulse is negligible?