Lecture 9

Last Lecture
• Momentum: p
 m r v
• Newton’s 2 nd : F


 t p
• Conserved for isolated objects
•

F
 t
  p
•

• Collisions (p always conserved)
• Elastic (E conserved too)
• Inelastic (E not conserved)

Example 6.7
A proton (m p
=1.67x10
-27 kg) elastically collides with a target proton which then moves straight forward. If the initial velocity of the projectile proton is 3.0x10
6 m/s, and the target proton bounces forward, what are a) the final velocity of the projectile proton?
b) the final velocity of the target proton?
0.0
3.0x10
6 m/s

Elastic collision in 1-dimension

1. Conservation of Energy:
1
2 m
1 v
1 i
2 
1
2
2 m
2 v
2 i

1
2 m
1
2 v
1 f

1
2
2 m
2 v
2 f
2. Conservation of Momentum: m
1 v
1 i
 m
2 v
2 i
 m
1 v
1 f
 m
2 v
2 f
(1)
(2)
• Rearrange both equations and divide:
 m
1
 v
1 i
2  v
2
1 f

 m
2
 v
2
2 f
 v
2
2 i
 m
1
 v
1 i
 v
1 f
  v
1 i
 v
1 f

 m m
1
 v
1 i
 v
1 f

 m
2
2
 v
2 f
 v
2 i
  v
2 f
 v
2 i

 v
2 f
 v
2 i

(1)
(2) v
1 i
 v
1 i
 v
1 f
 v
2 i
 v
2 f
 v
2 i
 
 v
1 f
 v
2 f



Elastic collision in 1-dimension
Final equations for head-on elastic collision: m
1 v
1 i v
1 i
 m
2 v
2 i
 v
2 i
 m
1 v
1 f
 
 v
1 f
 v
2 f

 m
2 v
2 f
• Relative velocity changes sign

• Equivalent to Conservation of Energy

Example 6.8
An proton (m p
=1.67x10
-27 kg) elastically collides with a target deuteron (m
D
=2m p
) which then moves straight forward. If the initial velocity of the projectile proton is 3.0x10
6 m/s, and the target deuteron bounces forward, what are a) the final velocity of the projectile proton?
b) the final velocity of the target deuteron?
v p v d
=-1.0x10
6 m/s
= 2.0x10
6 m/s
Head-on collisions with heavier objects always lead to reflections

Example 6.9a
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
=M
1 and which is initially at rest. The collision is perfectly elastic. a) Just after the collision v
2
______ v
0
.
A) >
B) <
C) =

Example 6.9b
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
=M
1 and which is initially at rest. The collision is perfectly elastic.
Just after the collision v
1
______ 0.
A) >
B) <
C) =

Example 6.9c
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
=M
1 and which is initially at rest. The collision is perfectly elastic.
Just after the collision P
2
______ M
1 v
0
.
A) >
B) <
C) =

Example 6.9d
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
=M
1 and which is initially at rest. The collision is perfectly elastic.
At maximum compression, the energy stored in the spring is ________ (1/2)M
1 v
0
2
A) >
B) <
C) =

Example 6.9e
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
<M
1 and which is initially at rest. The collision is perfectly elastic.
Just after the collision v
2
______ v
0
.
A) >
B) <
C) =

Example 6.9f
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
<M
1 and which is initially at rest. The collision is perfectly elastic.
Just after the collision v
1
______ 0.
A) >
B) <
C) =

Example 6.9g
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
<M
1 and which is initially at rest. The collision is perfectly elastic.
Just after the collision P
2
______ M
1 v
0
.
A) >
B) <
C) =

Example 6.9h
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
<M
1 and which is initially at rest. The collision is perfectly elastic.
At maximum compression, the energy stored in the spring is ________ (1/2)M
1 v
0
2
A) >
B) <
C) =

Example 6.9i
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
>M
1 and which is initially at rest. The collision is perfectly elastic.
Just after the collision v
2
______ v
0
.
A) >
B) <
C) =

Example 6.9j
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
>M
1 and which is initially at rest. The collision is perfectly elastic.
Just after the collision v
1
______ 0.
A) >
B) <
C) =

Example 6.9k
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
>M
1 and which is initially at rest. The collision is perfectly elastic.
Just after the collision P
2
______ M
1 v
0
.
A) >
B) <
C) =

Example 6.9l
The mass M
1 enters from the left with velocity v
0 strikes the mass
M
2
>M
1 and which is initially at rest. The collision is perfectly elastic.
At maximum compression, the energy stored in the spring is ________ (1/2)M
1 v
0
2
A) >
B) <
C) =

Example
6.10
Ballistic Pendulum: used to measure speed of bullet. 0.5-kg block of wood swings up by height h = 65 cm after stopping 8.0-g bullet.
227 m/s
What was bullet’s velocity?

Example
6.11
A 5-g bullet traveling at 500 m/s embeds in a 1.495 kg block of wood resting on the edge of a 0.9-m high table. How far does the block land from the edge of the table?
71.4 cm

Example 6.12
Tarzan (M=80 kg) swings on a 12m vine by letting go from an angle of 60 degrees from the vertical.
At the bottom of his swing, he picks up Jane (m=50 kg). To what angle do Tarzan and Jane swing?
35.8 degrees
(indepedent of L or g)

Example 6.12a
Tarzan (M=80 kg) swings on a 12m vine by letting go from an angle of 60 degrees from the vertical.
At the bottom of his swing, he picks up Jane (m=50 kg). To what angle do Tarzan and Jane swing?
To calculate Tarzan’s speed just before he picks up Jane, you should apply:
A) Conservation of Energy
B) Conservation of Momentum

Example 6.12b
Tarzan (M=80 kg) swings on a 12m vine by letting go from an angle of 60 degrees from the vertical.
At the bottom of his swing, he picks up Jane (m=50 kg). To what angle do Tarzan and Jane swing?
To calculate Tarzan&Jane’s speed just after their collision (given the previous answer) you should apply:
A) Conservation of Energy
B) Conservation of Momentum

Example 6.12c
Tarzan (M=80 kg) swings on a 12m vine by letting go from an angle of 60 degrees from the vertical.
At the bottom of his swing, he picks up Jane (m=50 kg). To what angle do Tarzan and Jane swing?
To calculate Tarzan&Jane’s final height (given the previous answer) you should apply:
A) Conservation of Energy
B) Conservation of Momentum

Rotational Motion
Universal Law of Gravitation
Kepler’s Laws

Angular Displacement
• Circular motion about a fixed AXIS
• Use polar coordinates: (r,  )
• Distance r doesn’t change
• Three possible units for  :
• Revolutions
• Degrees (1 rev. = 360 deg.)
• Radians (1 rev. = 2 p rad.)


Angular Displacement, cont.
• Distance traveled by a point:
(distance r from axis) s

2 p r N
(N counts revolutions)
 r

(

is in radians)

(rad)

2 p
N (revolutions)

Example 7.1
An automobile wheel has a radius of 42 cm. If a car drives 10 km, through what angle has the wheel rotated?
a) In revolutions b) In radians c) In degrees a) N = 3789 b)  = 2.38x10
4 radians c)  = 1.36x10
6 degrees


Angular Speed
 
 
 t

 f t f
  i
 t i
(in rad/s)
• Can also be given in
• Revolutions/s
• Degrees/s

•
Linear (tangential) Speed at r v t

 s
 t
 r
 
 t v t
 r

(  in rad/s)


Example 7.2
A race car engine can turn at a maximum rate of 12,000 rpm. (revolutions per minute).
a) What is the angular velocity in radians per second.
b) If helicopter blades were attached to the crankshaft while it turns with this angular velocity, what is the maximum radius of a blade such that the speed of the blade tips stays below the speed of sound.
DATA: The speed of sound is 343 m/s a) 1256 rad/s b) 27 cm

Angular Acceleration
• Denoted by a a 
 f
  i t
•  in rad/s
• a rad/s²
• Every point on rigid object has same  and a

Rotational/Linear Correspondence:
    x

0
 f
 v
0
 v f a  a t
 t

Rotational/Linear Correspondence, cont’d
Rotational Motion
  


0
  f
 t
2
Linear Motion
 x

 v
0
 v f
 t
2
 f
 
0
 a t v f
 v
0
 at
   
0 t

1
2 a t
2
    f t

1
2 a t
2
 2 f  a  
2

 2
0
2
 x
 v
0 t

1
2 at
2
 x
 v f t

1
2 at
2 v
2 f
2
 v
0
2
2
 a
 x

Example 7.3
A pottery wheel is accelerated uniformly from rest to a rotation speed of 10 rpm in 30 seconds.
a.) What was the angular acceleration? (in rad/s 2 ) b.) How many revolutions did the wheel undergo during that time?
a) 0.0349 rad/s 2 b) 2.50 revolutions

Linear movement of a rotating point
• Distance x
 r
 
• Speed v
 r

• Acceleration a
 r a
Angles must be in radians!
Different points have different linear speeds!

Special Case - Rolling Motion
• Wheel (radius r) rolls without slipping
• Angular motion of wheel gives linear motion of car
• Distance x
 r
 
• Speed v
 r

• Acceleration a
 r a

Example 7.4
A coin of radius 1.5 cm is initially rolling with a rotational speed of 3.0 radians per second, and comes to a rest after experiencing a slowing down of a = 0.05 rad/s 2 .
a.) Over what angle (in radians) did the coin rotate?
b.) What linear distance did the coin move?
a) 90 rad b) 135 cm