Performing a similar type of calculation as for the generalised motion of an

object on a frictionless surface (see Additional Materials 53), we may derive

the expression

1

2mv2

f +1

2k (∆sf )

2 =1

2mv2

i +1

2k (∆si)

2 , (8.13)

where ∆si and ∆sf are the equilibrium displacements before and after release.

Equation (8.13) has a similar form to equation (8.2), where in addition to the

kinetic energy mv2/2, we define the elastic potential energy, Us:

Us =1

2k (∆s)2 . (8.14)

Us is the energy stored in a spring.

Equation (8.13) is the conservation of mechanical energy for our idealised

ball-spring system.

Note: a spring moving vertically has both gravitational and elastic potential

energy, and a total mechanical energy

EM = K + Ug + Us . (8.15)

8.5 Elastic and Inelastic Collisions

Let us recall the concept of a collision. An example of a microscopic model

of a typical collision is shown in figure 62 54.

• Before collision: objects approach each other with a kinetic energy

due to their movement.

53Derivation of the Kinetic and Potential Energies of Some Simple Systems,

http://www.nanotech.uwaterloo.ca/∼ne131/54Knight, Figure 9.1, page 240

112

Figure 62: Microscopic model of a collision.

• During collision: objects interact – a very large number of molecular

bonds are compressed. The kinetic energy is transformed into elastic

potential energy stored in the spring-like molecular bonds.

• After collision: the stored elastic potential energy is converted into

kinetic energy as the objects move away from each other.

Although in reality, real collisions lie somewhere in between, we may consider

two limiting cases of a collision:

8.5.1 Perfectly Elastic Collisions

The situation discussed above – in which all of the initial kinetic energy is

stored as elastic potential energy, and then all of the stored elastic potential

energy is transformed into kinetic energy – characterises a perfectly elastic

collision.

113

• The colliding pair approach each other (both particles make up the iso-

lated system).

• The pair collide and move away (figure 63 55).

• Linear momentum is conserved:

m1 (vix)1 + m2 (vix)2 = m1 (vfx)1+ m2 (vfx)2

, (8.16)

where (vix)2(vfx)2 = 0 in figure 63.

• Mechanical energy is conserved:

1

2m1 (vix)

2

1 +1

2m2 (vix)

2

2 =1

2m1 (vfx)

2

1+

1

2m2 (vfx)

2

2, (8.17)

where (vix)2 = (vfx)2 = 0 in figure 63.

Figure 63: A perfectly elastic collision.

55Knight, Figure 10.24, page 287

114

8.5.2 Perfectly Inelastic Collisions

A collision in which the two objects stick together and move away with a

common final velocity is called a perfectly inelastic collision.

• The colliding pair approach each other and interact (both particles make

up the isolated system).

• Instead of moving away from each other, the two objects stick together

(figure 64 56)..

• Some of the kinetic energy is dissipated inside the object – not all of the

kinetic energy is recovered.

• The form of such dissipation may be, for example,

– A change in the molecular structure of either or both masses (e.g.

bonds broken, etc.).

– Sound.

– Heat.

• Linear momentum is conserved:

m1 (vix)1 + m2 (vix)2 = m1 (vfx)1+ m2 (vfx)2

, (8.18)

where (vix)2(vfx)2 = 0 in figure 63.

• Mechanical energy is not conserved (but the total energy is).

8.5.3 Special Cases

Let us consider elastic collisions only. In obeying both linear-momentum

and mechanical-energy conservation laws, given by (8.16) and (8.17), respec-

tively, they can be solved simultaneously, yielding the results

56Knight, Figure 9.20, page 255

115

Figure 64: A perfectly inelastic collision.

(vfx)1=

m1 − m2

m1 + m2

(vix)1 , (8.19)

and

(vfx)2=

2m1

m1 + m2

(vix)2 , (8.20)

for the final velocities of each particle.

Consider the following special cases (see figure 65 57):

• m1 = m2: equations (8.19) and (8.20) give

(vfx)1= 0 , (vfx)2

; (8.21)

– e.g. one billiard ball striking another of equal mass.

57Knight, Figure 10.25, page 288

116

� �

Figure 65: Special cases of elastic collisions.

• m1 � m2: in the limiting case of m1 → ∞, equations (8.19) and (8.20)

give

(vfx)1≈ (vix)1 , (vfx)2

≈ 2 (vix)1 ; (8.22)

– e.g. a bowling ball hitting a ping-pong ball.

• m1 � m2: in the limiting case of m1 → 0, equations (8.19) and (8.20)

give

(vfx)1≈ − (vix)1 , (vfx)2

≈ 0 ; (8.23)

– e.g. ping-pong ball hitting a bowling ball.

117

Example: 58 a spring with k = 2000 Nm−1 is sandwiched between a 1.0 kg

block and a 2.0 kg block on a frictionless table, as shown in figure 66. 59 The

blocks are pushed together to compress the spring by 10 cm, then released.

What are the velocities of the blocks as they fly apart?

Figure 66: Mass-spring system.

Solution: consider the total mechanical energy of the mass-spring system

before and after the spring is released:

• Before release: the total mechanical energy (EM)i is entirely made up of

elastic potential energy, since (vix) = 0:

(EM)i = Ki + Usi =1

2k (∆xi)

2 . (8.24)

• After release: the total mechanical energy (EM)f is now entirely kinetic:

(EM)f = Kf + Usf =1

2m1 (vfx)

2

1+

1

2m2 (vfx)

2

2. (8.25)

Using the conservation of mechanical energy, we have

58Knight, Example 10.8, page 28659Knight, Figure 10.23, page 286

118

1

2k (∆xi)

2 =1

2m1 (vfx)

2

1+

1

2m2 (vfx)

2

2. (8.26)

This gives one equation with two unknowns (the final velocities). In order

to find both, we must therefore find another expression involving them both.

The problem may, of course, be viewed as a collision/explosion problem; if

the table is frictionless, the system is isolated and we may use the conservation

of linear momentum:

• Before release: the total linear momentum Pi is zero, since both blocks

are initially at rest:

Pi = m1 (vix)1 + m2 (vix)2 = 0 . (8.27)

• After release: the total linear momentum Pf is

Pf = m1 (vfx)1+ m2 (vfx)2

. (8.28)

Using equations (8.27) and (8.28), conservation of linear momentum gives us

m1 (vfx)1+ m2 (vfx)2

= 0 (vfx)1= −m2

m1

(vfx)2, (8.29)

which provides us with an additional equation linking the final velocities. In-

deed, substitution of equation (8.29) in (8.26) yields the expression

(vfx)2=

√

k (∆xi)2

(1 + m2/m1),

and so substitution of the known values k = 2000 Nm−1, ∆xi = 0.1 m, m1 = 1

kg and m2 = 2 kg gives

(vfx)2=

√

2000 × (0.1)2

3= 1.826 m s−1 ,

and so in (8.29),

(vfx)1= −2 × 1.826 = −3.65 m s−1 .

119

8.6 Basic Energy Model

Summarising the last few sections:

• Kinetic energy, K: energy due to an object’s motion.

• Potential energy, U : energy due to an object’s position. Dependent

upon the interaction between particles. Examples are gravitational and

elastic.

• Mechanical energy, EM: energy due to mechanical effects.

– EM = K + U .

– In an isolated system (including when elastic collisions are present),

∆EM = 0 – i.e. EM is conserved.

So far, we have idealised the situation somewhat. In realistic isolated systems,

we consider two types of energy:

• Mechanical energy is an energy associated with the macroscopic

scale, given by EM = K + U .

• Internal energy is an energy associated with the microscopic scale,

given by

Eint = Eth + Ech + Enuc + · · · . (8.30)

where

– Thermal energy, Eth: energy of atoms/molecules due to their

microscopic motions – i.e. atoms inside a solid object vibrating

within the molecular-bond structure. Eth is associated with an

object’s temperature: Eth ∝ T 2.

– Chemical energy, Ech: energy involved in chemical reactions be-

tween the molecules in a system.

120

– Nuclear energy, Enuc: energy stored in the atomic nuclei, which

can be released during processes such as radioactive decay.

Let us define the total energy, ES, as

ES = K + U + Eint

= EM + Eint . (8.31)

Note: here, we will consider the internal energy to be entirely thermal, so

equation (8.31) becomes

ES = EM + Eth . (8.32)

• Isolated system: energy is exchanged/transformed within the system

(figure 67):

– ∆ES = 0 (i.e. ES is conserved).

– Generally, ∆EM 6= 0 (∆EM = 0 only if Eth = 0).

Figure 67: Isolated system.

• Non-isolated system: system is allowed to interact with the outside

environment (i.e. no longer isolated).

121