CHAPTER 8

SYSTEMS OF PARTICLES,EXTENDED OBJECTS

&CONSERVATION OF MOMENTUM

• Center of Mass ! finding the center of mass

• Motion of the center of mass

• Conservation of momentum

• Collisions ! impulse ! inelastic collisions ! elastic collisions ! coefficient of restitution

Center of Mass

So far we have considered all objects as “point” masses but what happens to the equations of motion, etc., if we have a large or ‘extended’ system (like a collection of masses or an object with a complicated shape)?

Examples:

(1) 2-d molecule ( H2O):

(2) dumbell:

(3) hammer:

(4) pool balls:

Then the concept of a center of mass becomes important because we can relate the motion of the object (or objects) to the motion of the center of mass.

If we have a collection of masses, we choose some origin, then the center of mass (cm) relative to that origin is at:

! r cm = (xcm,ycm,zcm),

where:

xcm =

m1x1 + m2x2 + ...m1 + m2 + ...

=mixii∑mii∑

=mixii∑M

,

ycm =

m1y1 + m2y2 + ...m1 + m2 + ...

=miyii∑mii∑

=miyii∑M

,

and

zcm =

m1z1 + m2z2 + ...m1 + m2 + ...

=mizii∑mii∑

=mizii∑M

.

mi (xi ,yi, zi)

cm

x

y

! r cm

O

mi cm

x

y

! r cm

O ! r i

i.e.,

! r cm =mi! r ii∑

mii∑=

mi! r ii∑

M

or M! r cm = mii∑

! r i

where M (= mii∑ ) is the total mass of the ensemble.

1. Calculation (possibly using symmetry).2. Experimentally.

Question 8.1: Find the center of mass of the sheet of metal of uniform thickness, shown below, relative to the lower left hand corner. The mass of the sheet is 35 kg.

So, how can I find thecenter of mass?

1 m

3 m 2 m

1 m

1 m 3 m

3 m

Consider the sheet as 3 separate object (masses), then

mass 1 + mass 2 + mass 3 = 35 kg.

Since the sheet has constant thickness, mass ∝ area.

Area 1 + Area 2 + Area 3 = 3 +1+ 3( ) m2 = 7 m2.

∴mass

m2 =35 kg7 m2 = 5.00 kg/m2.

mass 1 = mass 3 = 3 m2 × 5.00 kg/m2 = 15.0 kg,

and mass 2 = 1 m2 × 5.00 kg/m2 = 5.00 kg.

Choose axes and an origin (0,0). By symmetry, the positions of the individual cm’s are:

1 (0.5 m,1.5 m): 2 (1.5 m,0.5 m): 3 (2.5 m,1.5 m)

1 m

3 m 2 m

1 m

1 m 3 m 1 3

2 origin(0,0)

x

y

1 m 1 m 1 m

The coordinates of the cm are given by:

xcm =

mixii∑M

and ycm =

miyii∑M

∴xcm =

(15.0 × 0.5) + (5.00 ×1.5) + (15.0 × 2.5)35.0

,

i.e., xcm =

52.5 kg ⋅m35.0 kg

= 1.50 m.

~ Surprise, surprise ??also

ycm =

(15.0 ×1.5) + (5.00 × 0.5) + (15.0 ×1.5)35.0

i.e., ycm =

47.5 kg ⋅m35.0 kg

= 1.36 m.

∴! r cm = (1.50 m,1.36 m),

from the origin chosen.

There is another way to find the center of mass of the object but with less calculations.

Consider a 3 m × 3 m sheet with a 2 m ×1 m region of negative mass (shaded). By symmetry, the cm of the whole sheet is at (1.5 m,1.5 m), and the cm of the shaded region is at (1.5 m,2.0 m).

The mass of the

3 m × 3 m sheet is

97× 35 kg = 45.0 kg,

and the mass of the shaded region is

27× 35 kg =10.0 kg.

∴xcm =

45.0 ×1.5 + (−10.0) ×1.545.0 + (−10.0)

= 1.50 m

and

ycm =

45.0 ×1.5 + (−10.0) × 245.0 + (−10.0)

= 1.36m,

the same as before!

origin(0,0)

x

y 1 m

3 m 2 m

1 m

1 m

3 m

3 m

DISCUSSION PROBLEM [8.1]:

A baseball bat is cut through its center of mass (cm) producing two parts. Which piece has the smaller mass?

A: The piece on the left.B: The piece on the right.C: They both have the same mass.

cm•

The gravitational potential energy of an ensemble of particles is easily found:

If the y-direction is vertical then:

UG = Uii∑ = migyii∑ = g miyii∑ .

But ycm =

miyii∑M

.

∴UG = Mgycm,

i.e., UG depends on the total mass and the vertical

coordinate of the cm only.

NOTE: We have assumed that the value of g is the same for all particles, i.e., g is constant.

m1

m2

m3

y1 y2

y3

The center of mass of an object can also be found by integration. The x- component for the object shown is

xcm =

1M

xii∑ (Δmi ),

where Δmi is the mass of an element located at (xi,yi ) and M is the total mass of the object. If the number

of elements → ∞, then Δmi → 0, and

xcm = Limit

Δmi→0

1M

xii∑ (Δmi )⎧ ⎨ ⎩

⎫ ⎬ ⎭ =

1M

x.dm∫ .

Similarly, for the y- and z-components:

ycm =

1M

y.dm∫ and zcm =

1M

z.dm∫ .

The limits of the integrals are determined by the x, y and z dimensions of the object. Also, generalizing an earlier expression, we can write:

! r cm =

1M! r .dm∫ .

y

x O

Δmi

xi

yi ! r i

Here is an example of using integration to locate the center of mass of an object of uniform thickness.

Question 8.2: Find the position of the center of mass of the triangular sign shown below, from the point O.

Assume the sign is of uniform thickness.

a

b

b

O

Define x and y axes. Let the thickness of the triangle be δ and its density be ρ. Then, from before,

! r cm =

1M! r ∫ .dm.

Clearly, by symmetry the center of mass lies on the x-axis, i.e., ycm = 0, so we only need to concern ourselves with the x-coordinates. Therefore, we have to evaluate

xcm =

1M

x∫ .dm,

where dm = 2yδρ.dx and M = abδρ. We need to express y as a function of x,

i.e., y =

ba

x.

O a

b

b

y

dm

dx x

Then, substituting for M and dm, we get

xcm =

1abδρ

2bδρa

⎛ ⎝

⎞ ⎠ 0

a∫ x2.dx =

2a2 x2

0

a∫ .dx

=

2a2

x3

3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ 0

a

=2a3

.

O a

b

b

y

dm

dx x

cm

The center of mass of an irregular object can be found by experiment ... by suspending the object from two different points.

The center of mass lies on the vertical line through each suspension point. Therefore, the center of mass is where two such lines intersect.

cm

Question 8.3: In question 8.1 we determined the coordinates of the center of mass of a sheet of metal relative to the origin (0,0). If the sheet is suspended from the origin, what will be the angle between the vertical and the original x-axis?

xy

(0,0)

cm

We found the coordinates of the center of mass of the sheet were (1.50 m, 1.36 m). So, since the center of mass lies on the vertical line through the suspension point, the angle, θ, between the vertical and the x-axis is

θ = tan−1 ycm

xcm

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = tan−1 1.36 m

1.50 m⎛ ⎝

⎞ ⎠

= 42.2!.

xy

cm

θ 1.36 m 1.50 m

Despite the complicated overall motion of the wrenches, the motion of the cm is simple if the gravitational force is the only external force, e.g., no friction, etc. Then the translational motion of the cm is described by the normal equations of motion, e.g.,

! F ext = M! a cm.

1-dimensionalmotion.

2-dimensionalmotion.

If a rocket explodes in flight, providing that the gravitational force is the only external force, i.e., neglecting air resistance, the cm will continue on a parabolic trajectory, as will the individual fragments.

A similar effect is seen with exploding fireworks:

cmcm

××

×

See the article I wrote ...

“Center of Mass, binary stars and newplanets”

on the web-site

Newton’s Laws for the center of mass:

First Law:If the center of mass of a system is at rest (or is moving with constant velocity) it will remain at rest (or moving with constant velocity) unless a net external force acts on the system.

Second Law:If a net external force acts on a system of particles, the center of mass is given an acceleration that is proportional to, and in the same direction as, the force,

i.e, ! F net = M! a cm.

DISCUSSION PROBLEM [8.2]:

A canoeist drops a paddle in the lake. Assuming both the canoe and paddle are stationary, can he retrieve the paddle by moving towards the bow? ... Why ... or why not?

What happens to the center of mass if the various parts of a system are moving relative to each other ... ?

Question 8.4: A 1500 kg car is moving west with a speed of 20 m/s, and a 3000 kg truck is traveling east with a speed of 16 m/s. What is the speed of the center of mass and it what direction is it moving?

What’s happening to the center of mass?Define East as the positive direction. By definition, the x-component of the cm is

xcm =

mixii∑mii∑

.

But vcm =

dxcmdt

=mi

dxidti∑

mii∑=

mivii∑mii∑

=mivii∑M

=(3000 kg ×16 m/s) + (1500 kg × (−20 m/s))

3000 kg +1500 kg

= +4 m/sEven though the car is traveling faster to the West than the truck, the greater product of mass and velocity of the truck means the center of mass is moving to the East.

In vector form

! v cm =mi! v ii∑

mii∑.

West East

20 m/s 16 m/s

1500 kg 3000 kg

East

In chapter 4, we stated Newton’s 2nd Law in the form:

! F net =

d! p dt

=d(m! v )

dt

where ! p = m! v is called the momentum of the particle of

mass m and velocity ! v . This is a more general

expression for the 2nd Law and is more like the way Newton stated it, i.e.,

The net applied force equals the rate of change of momentum.

But, as we saw earlier, ! F = m! a is applicable only when m

is constant.

Dimension: p⇒ [M][L]

[T](vector)

Units: kg.m/s

For a collection of objects, the momentum of the ensemble is the vector sum of the individual momenta,

i.e., ! P = !

p ii∑ = mii∑! v i.

Also ! r cm =

mii∑ ! r iM

so ! v cm =

mii∑ ! v iM

.

∴M! v cm = mii∑

! v i =! P .

Take the time derivative,

d! P

dt=

ddt

(mi! v i )i∑ =

d! p idt

=i∑! F ii∑ =! F net,

i.e., a net force acting on a collection of objects produces a change in the total momentum This is Newton’s 2nd Law for a collection of objects.

Note: if there is no net external force, i.e., ! F net = 0, then

! P

is constant in time. Also, since ! P is constant, then

! v cm

remains unchanged.

Conservation of momentum : If there is zero net external force on a system, the total momentum of the system remains constant (even when there’s a collision).

Question 8.5: A girl, of mass 55 kg, jumps from the bow of a boat, of mass 100 kg, onto a dock. If the boat is initially at rest and her velocity is 2.5 m/s to the right, what is the velocity of the boat after she jumps?

Take the girl and the boat as the system ... there are no external forces, e.g., wind or a push, so the total momentum is conserved, i.e., constant.

• Before she jumped, the initial momentum Pi = 0,

since the girl and the boat are both stationary.• After she jumped, the final momentum Pf = 0.

Taking the +x-direction as the +ve direction and the velocity of the boat = v, we have:

Pi + Pf = 0

i.e., (55 kg) × (2.5 m/s)[ ] + (100 kg × v)[ ] = 0

∴v = −1.375 m/s. Since v < 0 it is directed to the left. What about vcm?

x100 kg

55 kg

v

+2.5 m/s

• Before she jumped, vcm = 0, since the girl and the

boat are both stationary.

• After she jumped, vcm =

mivii∑M

=(55 kg)(2.5 m/s) − (100 kg)(1.375 m/s)

100 kg + 55 kg

= 0,i.e., vcm is unchanged.

DISCUSSION PROBLEM [8.3]:

A cheetah crouches, motionless, behind a bush (a). It spots an antelope and accelerates after its prey (b). Clearly, at (a) the cheetah’s momentum is zero whereas at (b) it has finite momentum.

I thought momentum was supposed to be conserved ... does this scenario violate the law of conservation of momentum? Explain.

(a) (b)

Let’s look at some “collisions” ... i.e., when a force is applied to change the motion.

Examples:Force ( ×105N)

Time (ms)

1

2

3

0 50 100“Crushing a Beetle”

“Soft”collision

F

t

“Hard”collision

From earlier we have:

! F =

d! p dt

, i.e., ! F dt = d

! p .

If the force is applied for a finite time Δt (= t2 − t1), we

define the impulse of the force as:

! I =

! F dt

t1

t2∫ = d! p

1

2∫ = ! p 2 − ! p 1 = Δ! p ,

i.e., the impulse is equal to the change in momentum. The average force for the time period Δt = t2 − t1( ) is:

! F av =

1Δt

! F dt

t1

t2∫ =

! I Δt

=Δ! p Δt

.

Note: for a given change in momentum, Δp:

Fav ∝

1Δt

, i.e., Fav .Δt = constant.

(Check web-site notes for “averages”.)

Fav

t1 t2 Δt ′ F av

Equal areas FavΔt

The impulse ! I = Δ! p =

! F av .Δt (vector)

Dimension: same as momentum ⇒ [M][L]

[T].

Units: N ⋅ s (alternatively kg ⋅m/s).

The impulse relation:

! F .Δt = Δ! p

is rather like the work-kinetic energy theorem:

! F • Δ! s = ΔK.

Whoops ... a mistake ?~ Can you spot it? ~

Let’s take a closer look at ...

Fav .Δt = Δp.

If an object (truck, ball, you, etc.) is brought to a stop, then the longer the time ( Δt ) it takes, the smaller the required force ( Fav) .... many examples in everyday life.

Catching a hard ball (same Δp):(a) Small Δt : large Fav (ouch!)

(b) Large Δt : small Fav (no problem!)

(a) (b)

Large Δt: small Fav⇒ little damage

v! v = 0

Small Δt: large Fav⇒ lots of damage!

Same Δp

v! v = 0

Question 8.6: A butterfly is fluttering down a road when it splatters on the windshield of a moving truck traveling in the opposite direction. Which object, the truck or the butterfly, experiences

(a) the greater change in the magnitude of momentum, or is the change the same for the truck and butterfly, and

(b) the greater force, or is the same for the truck and butterfly?

(a) Take the truck and butterfly as the system. With no external forces, momentum is conserved in the collision. So, if

! p 1 is the momentum of the truck and ! p 2 is the

momentum of the butterfly, before the collision,then

! p 1 +! p 2 = constant.

∴Δ! p 1 + Δ! p 2 = 0, i.e., Δ! p 1 = −Δ! p 2.

∴Δ! p 1 = Δ! p 2 ,

i.e., the magnitude of the change in momentum is the same for both!

(b) The force is related to the change in momentum via the impulse equation I = F.Δt = Δp. But Δp is the same

for the truck and butterfly so ! I TB =

! I BT . Since the

collision time Δt is the same for both then ! F TB =

! F BT .

(NB: We could also have used Newton’s 3rd Law.)

Question 8.7: A ball of mass 0.10 kg experiences the force shown in the plot. If the ball was at rest initially, what is its speed after 8 ms?

F (N)

t (ms) 0

2 4 6 8

10

20

30

From earlier, we know F.dt = Δp∫ , where F.dt∫ is the

area under the F − t plot, and Δp is the change in momentum. Since the ball was at rest to begin with,

Δp = mvf .

where vf is the (final) velocity after 8 ms. But

F.dt = Area =

12∫ (30 N)(0.008 s) = 0.12 N ⋅ s

= mv f .

∴vf =

0.12 N ⋅ s0.10 kg

= 1.2 m/s.

F (N)

t (ms) 0

2 4 6 8

10

20

30

Area = 0.12 N ⋅ s

Question 8.8: A 100 g ball is thrown horizontally against a wall with a speed of 5.0 m/s. It strikes the wall at an

angle of 40! and bounces off with the same speed at the

same angle. If it is in contact with the wall for 2.0 ms, what is the average force exerted by the ball on the wall?

The change in momentum of the ball is

Δ! p = ! p 2 −

! p 1 =! I WB

= (0.1 kg)(5 m/s) cos40")[ ]ˆ i − (0.1 kg)(−5 m/s) cos40")[ ]ˆ i = (0.766 kg ⋅m/s)ˆ i .

But, from the definition of impulse,

! I = Δ! p =

! F avΔt ,

∴! F av =

Δ! p Δt

=(0.766 kg ⋅m/s)ˆ i

2 ×10−3s= (383 N)ˆ i ,

i.e., in the positive i direction.

5m/s

5m/s

! I WB Δ

! p (= ! p 2 − ! p 1 =! I WB )

! p 1

! p 2 −

! p 1

40"

40"

i

vertical view

Question 8.9: A baseball of mass 150 g is thrown horizontally towards you with a speed of 10.0 m/s. In catching the ball, your hand moves back 15.0 cm. Assuming your hand moves back parallel to the flight of the ball,

(a) what is the impulse of the ball on your hand, and

(b) what is the average force exerted by the ball on your hand?

(a) When you catch the ball, by Newton’s 3rd Law

! F BH = −

! F HB.

Since Δt is the same for both forces,

! I BH = −

! I HB,

i.e., ! I BH =

! I HB = Δ! p ball

= 0 − (0.1 kg)(10 m/s)[ ] = 1.0 N ⋅ s ( kg ⋅m / s).

(b) If Δt is the time interval for the ball to stop, then IBH = FBH .Δt = Δp = 1.0 N ⋅ s.

But, what is Δt ?

• v2 = v"2 + 2aΔx.

∴a = −

v"2

2Δx=(10 m/s)2

2(0.15 m)= −333.3 m/s2.

• v = v" + aΔt .

∴Δt = −

v"a

= −(10 m/s)

(−333.3 m/s2 )= 0.030 s.

∴FBH =

ΔpΔt

=1.0 N ⋅ s0.030 s

= 33.3 N.

v" = 10 m/s v = 0

! I BH

! I HB

0.15 m

DISCUSSION PROBLEM [8.4]:

You fire two bullets at a log. The bullets have equal mass and travel with the same velocity. One bullet is made of steel and lodges in the log. The other bullet is made of rubber and bounces back off the log. Which bullet is more likely to tip the log over?

A: The steel bullet.B: The rubber bullet.C: Neither, they are both equally likely.

Closer look at collisions:~ Momentum is always conserved ~

But there are two types of collisions ...• non-elastic (inelastic) collisions

(Kinetic energy not conserved).• elastic collisions

(Kinetic energy conserved).

1. Non-elastic (inelastic) collisions.(e.g., when objects “stick” together)

! p 1 +! p 2 = ! p f

∴ ! v f (= ! v cm ) =

! p fm1 + m2

.

Examples:• bullet lodges in a target• book falling on a table

~ What happens to the “lost” KE ? ~

! p 1

! p 2

! p f m1

m2 (m1 + m2 )

Ballistic pendulum:

If the bullet lodges in the target, conservation of momentum tells us

initial momentum = final momentumi.e., m1v1 = (m1 + m2)vf

∴v1 =

m1 + m2m1

vf .

With conservation of mechanical energy, ΔK = −ΔU,

12(m1 + m2)vf

2 = (m1 + m2)gh

∴vf = 2gh .

i.e., v1 =

(m1 + m2)m1

2gh .

Can be used to determine the speed of a bullet.

m2 p1 = m1v1

p2 = 0 h

What about the kinetic energy?

Kinetic energy before collision:

Ki =

12

m1v12 =

12

m1v1( )2m1

=p1

2

2m1.

Kinetic energy immediately after collision:

Kf =

12(m1 + m2)vf

2 =p2

2

2(m1 + m2)=

p12

2(m1 + m2 ).

The change in kinetic energy:

ΔK = Kf − Ki =

p12

2(m1 + m2)−

p12

2m1< 0.

Since (m1 + m2) > m1, ΔK is always negative, i.e.,

kinetic energy is never conserved.What happens to the “lost” energy?

m2 p1 = m1v1

p2 = 0 h

2. Elastic collisions:momentum and kinetic energy are both conserved.

∴! p 1i +

! p 2i =! p 1f + ! p 2f

and

K1i + K2i = K1f + K2f

The analysis is more difficult in 2- and 3-dimensions because one needs to know three of the velocities involved in order to determine the fourth.

Elastic collisions in 1-dimension

! v 2f

! v 1f

! v 2i

! v 1i

! p 1i

! p 2i

! p 1f

! p 2f

Before collision After collision

v1i v2i v1f v2f

v2i < v1i v2f > v1f

m1 m2

m1 m2

Using conservation of momentum and kinetic energy:

m1v1i + m2v2i = m1v1f + m2v2f

and

12

m1v1i2 +

12

m2v2i2 =

12

m1v1f2 +

12

m2v2f2.

Re-arranging the two equations gives,

m1(v1i − v1f ) = m2(v2f − v2i ) ... ... (1)

and

m1(v1i2 − v1f

2) = m2(v2f2 − v2i

2) ... (2)

Dividing (2) by (1), we get

(v1i2 − v1f

2)(v1i − v1f )

=(v2f

2 − v2i2)

(v2f − v2i ),

i.e.,

(v1i + v1f )(v1i − v1f )(v1i − v1f )

=(v2f + v2i )(v2f − v2i )

(v2f − v2i ).

∴v1i + v1f = v2f + v2i,

i.e., −(v2i − v1i ) = v2f − v1f speed #1 approaches #2 = speed #2 moves away from #1

v1i v2i v1f v2f

v2i < v1i v2f > v1f

m1 m2

m1 m2

Note that this expression

−(v2i − v1i ) = v2f − v1f

is true only for a 1-dimensional perfectly elastic collision, i.e., one in which kinetic energy is conserved.

In the macroscopic world, collisions are somewhere between perfectly inelastic and perfectly elastic; at the microscopic level elastic collisions are common, e.g., between atoms.

We define a coefficient of restitution (e) to tell what type of a collision we have ...

e =

v2f − v1fv1i − v2i

=speed of recessionspeed of approach

.

Note that ... • if e = 1 the collision is perfectly elastic.• if e = 0 the collision is perfectly inelastic*.

* In a perfectly inelastic collision the two colliding objects stick together.

For a bouncing ball dropped vertically, if the speed at the instant before the collision is v1i and

the speed immediately after the collision is v1f , then

e =

v2f − v1fv1i − v2i

=v1fv1i

,

since v1i and v1f are measured with respect to the ground

(i.e., v2i = v2f = 0). Usually, a ball does not reach the

original height, so v1f < v1i , i.e., e <1.

Typical values (on concrete):ebaseball = 0.57.

ebasketball = 0.57.

eracquetball = 0.86.

egolfball = 0.89.

Show for yourselves: if a ball is dropped from a height h onto a solid floor, the height of the first bounce is

h1 = e2 h.

v1f v1i

Question 8.10: A 0.10 kg rubber ball strikes a solid wall horizontally with a speed of 5.00 m/s and rebounds in the opposite direction. Below is shown the force the wall exerts on the ball.

Neglecting the effect of gravity during the impact(a) What is the speed of the ball when it rebounds?(b) What is the coefficient of restitution of the impact?

225 N

6.5 ms

t

Force

225 N

6.5 ms

t

Force

(a) We need to find the speed of the ball after the rebound. During the impact, the change in momentum equals the impulse of the wall on the ball, i.e.,

Δp = p2 − p1 = F.dt∫ = area under the force-time curve.

∴mv2 − m −v1( ) = 1

2FmaxΔt =

12(225 N)(0.0065s),

i.e., v2 + v1 =

0.731 kg ⋅m/s0.10 kg

= 7.31 m/s.

∴v2 = 7.31 m/s − 5.00 m/s = 2.31 m/s

(b) The coefficient of restitution is

e =

v2v1

=2.31 m/s5.00 m/s

= 0.462.

v1

v2

Before

After

Question 8.11: A 2 kg object moving at 6 m/s collides with 4 kg object that is originally at rest. After the collision, the 2 kg object moves backward at 1 m/s. Find

(a) the velocity of the 4 kg object after the collision, (b) find the velocity of the center of mass before and

after the collision, (c) the coefficient of restitution, and (d) the kinetic energy “lost” in the collision.

We don’t know whether the collision is elastic or not!

(a) Since: p1f + p2f = p1i + p2i,

(2 kg)(−1 m/s) + (4 kg)v2f = 12 kg ⋅m/s.

∴v2f =

14 kg ⋅m/s4 kg

= +3.5 m/s.

(b) From earlier vcm =

mivii∑M

.

Before collision:

vcm =

(2 kg)(6 m/s) + (4 kg)(0)6 kg

= 2 m/s.

After collision:

vcm =

(2 kg)(−1 m/s) + (4 kg)(3.5 m/s)6 kg

= 2 m/s.

After collision ...Momentum after:

p1f + p2f = (2 kg)(−1 m/s) + (4 kg)(v 2f )

v1f = −1 m/s v2f = ?

Momentum before: p1i + p2i = 12 kg ⋅m/s

Before collision ...

v1i = +6 m/s v2i = 0 m1 = 2 kg m2 = 4 kg

The velocity of the center of mass is conserved in a collision.

(c) e =

v2f − v1fv1i − v2i

=(3.5 m/s) − (−1 m/s)

(6 m/s) − 0= 0.75.

Therefore, the collision is non-elastic; kinetic energy is not conserved.

(d) Change in kinetic energy ( ΔK = Kf − Ki):

ΔK =

12

m1v1f2 +

12

m2v2f2⎛

⎝ ⎜ ⎞ ⎠ ⎟ −

12

m1v1i2 +

12

m2v2i2⎛

⎝ ⎜ ⎞ ⎠ ⎟

=

12(2 kg)(−1 m/s)2 + (4 kg)(3.5 m/s)2( )

−

12(2 kg)(6 m/s)2( )

= 25.5 J − 36 J = −10.5 J. “lost”

Special case of an elastic collision in 2-dimensions

Consider the special case of an off-center elastic collision between two equal masses, with one initially at rest. Then we have:

m! v 1i = m! v 1f + m! v 2f , i.e.,

! v 1i =! v 1f + ! v 2f ,

and

12

mv1i2 =

12

mv1f2 +

12

mv2f2, i.e., v1i

2 = v1f2 + v2f

2.

But v1i2 = ! v 1i •

! v 1i = (! v 1f + ! v 2f )• (

! v 1f + ! v 2f )

= v1f2 + v2f

2 + 2! v 1f •! v 2f .

∴2! v 1f •! v 2f = 0,

i.e., ! v 1f and

! v 2f are perpendicular, so

θ + φ = 90".

θ

φ m

m ! v 1i

! v 1f

! v 2f

Question 8.12: A 170 g hockey puck moving at 5 m/s makes an off-center collision with another puck of equal mass that is initially at rest. The incoming puck is

deflected at an angle of 30! from its original direction of

motion while the second puck is deflected at an angle of

60!.

(a) What is the velocity of each puck after the collision?

(b) Show that the collision is elastic.

(a) Since momentum is conserved, we have:

mv1i( )ˆ i = mv1f cos30( )ˆ i + mv2f cos60( )ˆ i ,

i.e., 5 =

32

v1f +12

v2f .

∴ 3v1f + v2f = 10 ... ... ... (1)

and

0 = mv1f sin 30( )ˆ j − mv2f sin 60( )ˆ j ,

i.e., 0 =

12

v1f −3

2v2f .

∴v1f − 3v2f = 0 ... ... ... (2).

Multiply (2) by 3, then

3v1f − 3v2f = 0 ... ... ... (3).

Subtract (3) from (1) and we get 4v2f = 10,

i.e., v2f = 2.50 m/s.

60! 30!

" v 1i " v 1f

" v 2f

" v 2i

i j

Substituting in (2) we get

v1f = 3 × v2f = 3 × 2.50 m/s( ) = 4.33 m/s.

(b) To show the collision is elastic we must show that kinetic energy is conserved. (We cannot use the coefficient of restitution as the expression only works in 1-dimension.) The initial kinetic energy is

K1i =

12

mv1i2 =

12(0.170 kg)(5.0 m/s)2 = 2.125 J.

The final kinetic energy is

K1f + K2f =

12

mv1f2 +

12

mv2f2

=

12(0.170 kg)(4.33 m/s)2 +

12(0.170 kg)(2.50 m/s)2

= 1.594 J + 0.531 J = 2.125 J,

∴K1i = K1f + K2f .

Thus, kinetic energy is conserved so the collision is elastic.