A bullet of mass m and velocity v is fired into a wood block of
mass M intially at rest on a frictionless surface. The bullet buries
itself into the block and then the block is seen to have a final
velocity vf.
Was this an elastic collision? A) Yes
B) No
2 = ½ (M+m)v 2 True
or
False:
½
mv
A) True
B) False
f
This is an inelastic collision because
the
bullet sticks
to the
block.
True or False: mv = Mvf A) True B) False
L30 M 11/3/14 a*er lecture 1 A bullet of mass m and velocity v is fired into a wood block of
mass M intially at rest on a frictionless surface. The bullet buries
itself into the block and then the block is seen to have a final
velocity vf.
Was this an elastic collision? True or False: ½ mv2 = ½ (M+m)vf2
A) Yes
B) No
A)
B) False
True
True or False: mv = Mvf A) True B) False
As an inelastic collision, KE is not conserved.
L30 M 11/3/14 a*er lecture 2 A bullet of mass m and velocity v is fired into a wood block of
mass M intially at rest on a frictionless surface. The bullet buries
itself into the block and then the block is seen to have a final
velocity vf.
Was this an elastic collision? True or False: ½ mv2 = ½ (M+m)vf2
A) Yes
B) No
A)
B) False
True
True or False: mv = Mvf A) True B) False
Even
in an inelastic collision, momentum is conserved, but
we must consider the entire mass of the bullet + block.
L30 M 11/3/14 a*er lecture 3 Assignments
Announcements:
•  CAPA 10 and HW 10 are due this week.
•  Clicker questions from Prof. Radzihovsky’s and Prof. Munsat’s
lectures last week will appear at the end of today’s Lecture Notes.
•  You should read Ch. 10 now.
Today:
•  Finish up our discussion of orbits and gravity, found in Ch. 8, as
well as linear momentum and its conservation (including impulse),
found in Ch. 9. •  Profs Radzihovsky’s and Munsat’s clicker questions are placed
at the end of today’s lecture notes.
•  Will move on to rotational motion, found in Ch. 10. L30 M 11/3/14 a*er lecture 4 Linear Momentum



p = mv is conserved ⇒ ptotal = constant (derived from NII and NIII)
Two objects collide:




m1v1i + m2 v2i = m1v1 f + m2 v2 f (initial momentum = final momentum)
In 1D (scalar equation):
m1v1i + m2 v2i = m1v1 f + m2 v2 f
If it is an "elastic collision", KE is also conserved:
1
1
1
1
m1v1i2 + m2 v2i2 = m1v12f + m2 v22 f
2
2
2
2
If it is not elastic, it is inelastic. Total energy is conserved, KE is not.
If it is a totally inelastic collision, then the objects stick together after the
collision so that there is a single final speed v f :
m1v1i + m2 v2i = (m1 + m2 )v f
L30 M 11/3/14 a*er lecture 5 Linear Momentum
NII is written elegantly in terms of momentum:

d 
Fnet = p
dt
This can be productively applied to one particle.
If the net force = constant:


Fnet Δt = Δp (impulse)
(Units: [F][t] = N s)
The change of momentum of a particle during a collision
is referred to as the impulse.
L30 M 11/3/14 a*er lecture 6 A ball bounces off the floor as shown.
The direction
of the impulse of the

ball, Δ p , is
A) Straight up,
B) Straight down,

 
I = Δ p = F net Δt

v1

v2
θ
θ
C) To the right,
D) To the left,
E) Indeterminate.
Note: the angle of incidence is equal
to the angle of reflection.

 
p1 + Δ p = p 2
L30 M 11/3/14 a*er lecture 7 Comment on Elastic Collisions
L30 M 11/3/14 a*er lecture 8 Comment on Elastic Collisions
for elastic collisions
Derivation is in the book and in Prof. Dubson’s notes on
Ch. 9.
L30 M 11/3/14 a*er lecture 9 Consider a suspended blue ball
that collides elastically with a
suspended red ball.
(like Newton’s cradle)
m m

vB
L30 M 11/3/14 a*er lecture 
vA = 0
10 Consider a suspended blue ball
that collides elastically with a
suspended red ball.
m m

vB

vA = 0
m
'
vB = 0
m
' 
vA = vB
After impact is the situation at right possible: A) Yes B) No
vB − 0 = vB = −(0 − v ) = v
'
A
L30 M 11/3/14 a*er lecture '
A
11 Newton’s Law of Universal Gravitation
m1m2
F=G 2
r
G ≈ 6.67 x 10 −11 Nm 2 / kg 2
(1) Acceleration of gravity:
g(r) =
GM
r2
(4) Saw how the mass of the earth
and the sun can be computed from
knowledge of their satellites’ orbits.
(5) Also saw how planetary radii and
periods can be inferred from
knowledge of other planetary radii
and periods.
(2) Speed of a satellite in a circular orbit:
vcircular orbit =
GM
r
(3) Relation between period τ and radius r of a circular orbit:
⎛ 4π 2 ⎞ 3
τ =⎜
r
⎟
GM
⎝
⎠
L30 M 11/3/14 a*er lecture 2
12 Gravitational Potential energy
−GmM
U(r) =
r
dU GmM
F=−
=
dr
r2
Note: we chose the reference level for potential energy such that
U(r) < 0 everywhere, except at infinity where it is = 0.
L30 M 11/3/14 a*er lecture 13 Gravitational Potential energy
−GmM
U(r) =
r
dU GmM
F=−
=
dr
r2
1 2
mM
Etot = KE + PE = mv − G
2
r
Total energy of an object of mass m moving at speed v subject only to the
attractive force of gravity of an object of mass M a distance r away.
L30 M 11/3/14 a*er lecture 14 A projectile is fired with initial speed v0 at radius r0 from the center of an
airless planet of mass M (r0 is significantly greater than the radius of the
planet). The projectile rises to a maximum distance rmax and then falls back
down to the ground, and achieves a final speed vfinal just before it hits the
ground. Initial point
Final point,
v=0
A student wishes to compute rmax. What is the correct Conservation of
Energy equation, she should start with? 1 2
Mm
Mm
mv0 − G
= −G
2
r0
rmax
1 2
Mm
C) mv0 = −G
2
rmax
A)
L30 M 11/3/14 a*er lecture 1 2
Mm 1 2
Mm
mv0 − G
= mv final − G
2
r0
2
rmax
Mm 1 2
Mm
D) − G
= mv final − G
r0
2
rmax
B)
15 Gravitational Potential energy
−GmM
U(r) =
r
Etot = 0
dU GmM
F=−
=
dr
r2
Etot = 0
Etot < 0
Etot > 0
Unbound Orbit
Bound Orbit Parabola 1 Ellipse
Circle,
2
0 = Etot = KE + PE =
Unbound Orbit
L30 M 11/3/14 a*er lecture 2
mv − G
Hyperbola
v0 = vescape
= 2GM
RE
mM
v0 < vescape
r2
v0 > vescape
16 Does escape speed depend on launch angle? That is, if a
projectile is given an initial speed vo, is it more likely to
escape an (airless) planet, if fired straight up than if fired
at an angle?
A)  Yes
B)  No
C)  Impossible to tell from the information given.
The speed of escape depends only on the relationship between the
object’s KE and PE at launch, which are independent of angle. L30 M 11/3/14 a*er lecture 17 The following clicker questions were used by Profs Radzihovsky
and Munsat in their lectures on Wed and Friday, Oct 29 and 31.
These are Lectures 28 and 29.
L30 M 11/3/14 a*er lecture 18 In which situation is the magnitude of the total momentum the largest?
A)  Situation I.
B)  Situation II.
C)  Same in both.
Magnitudes are the same:
mv + 0 = mv
mv + (2m)(-v) = -mv
L30 M 11/3/14 a*er lecture 19 A car is sitting on the surface of the Earth and both the car and the
Earth are at rest. (Pretend the Earth is not rotating or revolving around
the Sun.) The car then accelerates to a final velocity. After the car reaches its final velocity, the magnitude of the Earth’s
momentum is ___________ the magnitude of the car’s momentum.
Total momentum of the car/Earth system
A)  more than
is zero. So, as the car acquires
B)  the same as
momentum to the right, the Earth
C)  less than
acquires the same momentum to the left.
D)  Cannot answer, need
⎛ Mc ⎞
| M E vE |=| M c vc | → | vE |= ⎜
more information.
⎟ | vc || vc |
⎝ ME ⎠
L30 M 11/3/14 a*er lecture 20 Two masses of size m and 3m are at rest on a frictionless table. A
compressed, massless spring between the masses is suddenly allowed
to uncompress, pushing the masses apart.
v2
v1
2
1
After the masses separate, the speed of m is _______ the speed of 3m.
A) the same as
B) twice C) 3 times
D) 4 times.
ptot = 0 = mv1 + 3mv2
v1 = −3v2 → | v1 | = 3 | v2 |
L30 M 11/3/14 a*er lecture 21 Two masses of size m and 3m are at rest on a frictionless table. A
compressed, massless spring between the masses is suddenly allowed
to uncompress, pushing the masses apart.
-3v
2
1
Let v2 = v, then v1 = −3v
After the masses separate, the KE of m is _______ the KE of 3m.
A) the same as
B) twice C) 3 times
D) 9 times.
v
1
3 2
2
KE2 = (3m)v = mv
2
2
1
9 2
2
KE1 = m(−3v) = mv = 3KE2
2
2
L30 M 11/3/14 a*er lecture 22 Two masses of size m and 3m are at rest on a frictionless table. A
compressed, massless spring between the masses is suddenly allowed
to uncompress, pushing the masses apart.
v
-3v
2
1
Let v2 = v, then v1 = −3v
While the spring is in contact with the masses, how does the magnitude
of the force of m on 3m compare to the magnitude of the force of 3m
on m.
A) the same as
B) less than C) greater than
D) unknown
By NIII
L30 M 11/3/14 a*er lecture 23 A ball bounces off the floor as shown.
The direction
of the impulse of the

ball, Δ p , is
A) Straight up,
B) Straight down,
C) To the right,

 
I = Δ p = F net Δt

v1
py

v2
θ
θ
px
D) To the left,
E) Indeterminate.

 
p1 + Δ p = p 2
-py

p1 = ( px , − py )

p 2 = ( px , py )
 
Δp = p 2 − p1 = ( px , py ) − ( px , − py )
= (0, 2 py )
L30 M 11/3/14 a*er lecture 24 Suppose a tennis ball and a bowling ball are
rolling toward you. Both have the same
momentum p and you exert the same force F to
stop each. How do the time intervals to stop them
compare?

 
I = Δ p = F net Δt
A)  It takes less time to stop the tennis ball.
B)  Both are the same.
C)  It takes less time to stop the bowling ball.
Δp
Δt =
F
L30 M 11/3/14 a*er lecture which is the same for both the tennis and
bowling balls. 25 
 
I = Δ p = F net Δt
A fast-ball thrown at a batter has a momentum
of magnitude |pi| = (0.3kg)(40m/s) = 12 kg m/s.
The batter hits the ball back in a line drive with
momentum of magnitude |pf| = (0.3kg)(80m/s)
= 24 kg m/s.
What is the magnitude of the impulse |Δp|?
= 24 - (-12) kg m/s = 36 kg m/s
A)  12 kg m/s
B)  24 kg m/s
= -12 kg m/s
= 24 kg m/s
C)  36 kg m/s
D)  0 kg m/s
E)  None of these.
L30 M 11/3/14 a*er lecture 26 A big ball of mass M = 10m and speed v strikes a small ball of
mass m at rest. p
i = (10m)v = 10mv
p f = m(10v) = 10mv
KEi = 1 (10m)v 2 = 5mv 2
2
f = 1 m(10v)2 = 50mv 2
KE
2
After the collision, could the big ball come to a complete stop
and the small ball take off with speed 10 v?
A) Yes this can occur.
B) No, because it’d violate conservation of momentum.
C) No, because it’d violate conservation of energy. L30 M 11/3/14 a*er lecture 27