Lecture 7

Report
Vectors again

Equations for Motion Along One Dimension
v ave 
x
v  lim
a ave 
t
t  0
x
t

dx
dt
v
a  lim
t
t  0
v
t

dv
dt

Motion Equations for Constant Acceleration
• 1. v  v 0  at
• 2. x  x 0  v 0 t 
1
• 3. v 2  v 02  2 a  x
• 4. v ave 
v  v0
2
2
at
2


3 Laws of Motion
If in Equilibrium
F


If not in equilibrium
Change in Motion is Due to Force
F

0
 ma
Force causes a change in acceleration

Work
W Total   W   K
 
W  F  d  Fd cos 
W grav    U

Energy
K 
1
mv
2
2
U grav  mgy
U spring 
1
2
kx
2

Law of conservation of energy
K  U grav  U spring  K 0  U 0 grav  U 0 spring  W NC

Power
Pave 

W
t
 Fv ave
efficiency
e
Pout
Pin


If an 18 wheeler hits a car, what direction will the
wreckage move?
What is the force between the 18 wheeler and the
car?



F  ma


F  m


F 

dv
dt

dm v
dt




F 

dp
dt
Newtons 2nd law
Linear momentum


p  mv

SI
kg  m
 N s
s

Newton defined it as quantity of motion

When an object collides
with another, the forces
on the object will
momentarily spike before
returning back to zero.


dp
 F  dt


p
 F  t


 F t  p


J  p



J  p

We now define impulse, J, as the change in
momentum of a particle during a time


interval

J x   p x  Favex  t



J y   p y  Favey  t

SI unit
kg  m
s
 N s

A ball with a mass of 0.40 kg is thrown
against a brick wall. It hits the wall moving
horizontally to the left at 30 m/s and
rebounds horizontally to the right at 20 m/s.
(a) find the impulse of the net force on the
ball during the collision with the wall. (b) If
the ball is in contact with the wall for 0.010s,
find the average horizontal force that the wall
exerts on the ball during impact.

J

J

J

J

J

 p


 m v  m v0
 
 m ( v  v 0 )  0 . 4 ( 20  (  30 ))
 20 N  s

 F ave  t


J
20
F ave 

 2000 N
t
0 . 010


 d p
F 
dt
If a particle A hitsparticle B

dp B
F AonB 
dt


dp A
F BonA 
dt

If there are no external forces acting on the
system


F 0


F AonB  F BonA 

dp B
dt


dp A
dt
0

dp B
dt




dp A
dt



d ( pA  pB )
dt
Change in momentum over
time is zero
The sum of momentums is
constant


p A  p B  const
0


p A  p B  const

If there are no external
forces acting on a system,
Total Momentum of a
system conserved




p A0  p B 0  p A  p B


 p  p
0
A  Ax  A y
A x  A cos 
A y  A sin 

A marksman holds a rifle of mass 3.00 kg
loosely such that it’ll recoil freely. He fires a
bullet of mass 5.00g horizontally with
velocity relative to the ground of 300 m/s.
What is the recoil velocity of the rifle?



px  0  pB  pR


p B  m B v B  ( 0 . 05 )( 300 )  15



p R  m R v R  (3) v R   1 .5 m s

v R   0 .5 m s
m
s

Two battling robots are on a
frictionless surface. Robot A
with mass 20 kg moves at
2.0 m/s parallel to the x
axis. It collides with robot B,
which has a mass of 12 kg.
After the collision, robot A is
moving at 1.0 m/s in a
direction that makes an
angle α=30o. What is the
final velocity of robot B?


p x0  p x




p Ax 0  p Bx 0  p Ax  p Bx



m A v Ax 0  m A v Ax  m B v Bx



m A v Ax 0  m A v Ax  m B v Bx


m A v Ax 0  m A v Ax
v Bx 
mB
v Bx 
20 ( 2 )  ( 20 )( 1) cos 30
v Bx  1 . 89
12
m
s


p y0  p y




p Ay 0  p By 0  p Ay  p By


0  m A v Ay  m B v By


 m A v Ay  m B v By
v By 
v By 
 m A v Ay
mB
 ( 20 )( 1) sin 30
v By   0 . 83
12
m
s
v Bx  1 . 89
m
v By   0 . 83
s
m
s

2
2
v B  v Bx  v By

v B  2 .1 m s
  arctan
 0 . 83
1 . 89
  24 

Elastic Collisions – Collisions where the
kinetic energies are conserved. When the
particles are in contact, the energy is
momentarily converted to elastic potential
energy.

Inelastic Collisions – collisions where total kinetic
energy after the collision is less than before the
collision.

Completely Inelastic Collisions- When the two
particles stick together after a collision.

Collisions can be partly inellastic

Collisions where two objects will impact each
other, but the objects stick together and
move as one after the collision.


Momentum is still conserved
Find v in terms of v0


p0  p




p A0  p B 0  p A  p B




m Av A0  m B vB 0  m Av A  m B vB



v A  vB  v




m Av A0  m B vB 0  m Av  m B v

Assume Particle B is initially at rest




m Av A0  m B vB 0  m Av  m B v



m Av A0  m Av  m B v


m Av A0
v 
mA  mB

Kinetic Energy
1
K0 
2
2
m Av A0
 m Av A0
K  ( m A  m B ) v  ( m A  m B ) 
2
2
 mA  mB
1
K 
K
K0
1
2
m A v A 0 
1
2
2 (m A  m B )

mA
mA  mB
If B is at rest




2

At the intersection, a yellow subcompact car
with mass travelling 950 kg east collides with
a red pick up truck with mass 1900 kg
travelling north. The two vehicles stick
together and the wreckage travels 16.0 m/s
24o E of N. Calculate the speed of each of the
vehicles. Assume frictionless.




m Av A0  m B vB 0  m Av  m B v



m Av A0 x  m Avx  m B v x



m B vB 0 y  m Av y  m B v y


v x  v sin 24


v y  v cos 24

(m A  m B )v x
v A0 x 
mA
(m A  m B )v y

vB0 y 
mB
v x  v sin 24  16 sin 24  6 . 51
v y  v cos 24  16 cos 24  14 . 6
v A0 x 
vB0 y 
(m A  m B )v x
mA
(m A  m B )v y
mB
 19 . 5 m s
 21 . 9 m s

The ballistic pendulum is an
apparatus to measure the
speed of a fast moving
projectile, such as a bullet.
A bullet of mass 12g with
velocity 380 m/s is fired
into a large wooden block
of mass 6.0 kg suspended
by a chord of 70cm. (a) Find
the height the block rises
(b) the initial kinetic energy
of the bullet (c) The kinetic
energy of the bullet and
block.
m Av A0  m B vB 0  m Av  m B v
m Av A0  m Av  m B v
v

K 

m Av A0
mA  mB
Velocity after impact
1
 m A v A 0 2
2 (m A  m B )
 1 . 73 J
Kinetic energy after impact

Kinetic energy after
impact
K 

1
 m A v A 0 2
2 (m A  m B )
Converted to potential at
highest point
2
1 m A v A 0 
K 
2 (m A  m B )
U  ( m A  m B ) gy
1
 m A v A 0 2
2 (m A  m B )
 ( m A  m B ) gy
1
 m A v A 0 2
2 (m A  m B )
y 
y 
1
 ( m A  m B ) gy
 m A v A 0 2
2 g (m A  m B )
( 0 . 012 ( 380 ))
2
2 ( 9 . 8 )( 6 . 012 )
y  0 . 0293 m
2
2
K Bullet 
1
K Bullet 
1
2
2
mv B
0 . 012 ( 380 )
2
K Bullet  866 J
2


Momentum and Energy are conserved
Find v in terms of v0


p0  p




p A0  p B 0  p A  p B




m Av A0  m B vB 0  m Av A  m B vB
K0  K
K A0  K B 0  K A  K B
1
2
m Av
2
A0

1
2
mBv
2
B0

1
2
m Av 
2
A
1
2
2
m BvB


If particle B is at rest


p0  p




p A0  p B 0  p A  p B
m Av A0  m Av A  m B vB
K0  K
K A0  K B 0  K A  K B
1
2
m Av
2
A0

1
2
m Av 
2
A
1
2
mBv
2
B


If particle B is at rest
m Av A0  m Av A  m B vB
m Av A0  m Av A  m B vB
m A (v A0  v A )  m B v B
1
1
1
2
2
2
m Av A0  m Av A  m B vB
2
2
2
1
2
m Av
m A (v
2
A0
2
A0

1
2
m Av 
2
A
 v )  mBv
2
A
2
B
1
2
mBv
2
B


If particle B is at rest
m A (v A0  v A )  m B v B
m A (v A0  v A )  m B v B
2

2
2
v A0  v A  vB
m A (v A0  v A )  m B (v A0  v A )
Substitute back
m Av A0  m Av A  m B v A0  m B v A
m Av A0  m B v A0  m Av A  m B v A
(m A  m B )v A0  (m A  m B )v A
vA 
(m A  m B )v A0
(m A  m B )


If particle B is at rest
(m A  m B )v A0
vA 
(m A  m B )
v A0  v A  vB
v A0 
vB 
(m A  m B )v A0
(m A  m B )
2m Av A0
(m A  m B )
 vB
vA 
(m A  m B )v A0
vB 
(m A  m B )
2m Av A0
(m A  m B )
 If ma <<< mb
 m B v A0
vA 
 v A0
mB

vB 
2m Av A0
(m A  m B )

2m Av A0
mB
really small


If ma>>>mb
vA 
m Av A0
vB 
2m Av A0
If ma=mb
vA 
vB 
mA
 v A0
mA
 2v A0
(0)v A0
(m A  m B )
2m Av A0
2m A
0
 v A0

In a game of billiards a player wishes to sink
a target ball in the cornet pocket. If the angle
to the corner pocket is 35o, at what angle is
the cue ball deflected? (Assume frictionless)




m Av A0  m B vB 0  m Av A  m B vB
m Av A0  m B vB 0  m Av A  m B vB
2
2
Mass is the same



v A0  v A  vB

v A0  v A  vB
2
2
2
2
2



v A0  v A  vB
v A0  v A  vB


2
v A0  v A0  v A0




2
v A0  (v A  v B )  (v A  v B )
 
2
2
2
v A0  v A  2v A  vB  vB
 
0  2v A  vB
2
2
2
0  2 v A v B cos 
cos   0
  90    35
  55

Two particles with masses m and 3m are
moving towards each other along the x axis
with the same initial speeds. Particle m is
travelling towards the left and particle 3m is
travelling towards the right. They undergo an
elastic glancing collision such that particle m
is moving downward after the collision at
right angles from initial direction. (a) Find the
final speeds of the two particles. (b) What is
the angle θ at which particle 3m is scattered.


m Av A0  m B vB 0  m Av A  m B vB
m Av A0  m B vB 0  m Av A  m B vB
2
2
2
2
m Av A0  m Av A  m B vB  m B vB 0
m A (v A0  v A )  m B (v B  v B 0 )
m Av
2
A0
 m Av  m B v  m B v
2
A
2
B
m A (v A0  v A )  m B (v B  v B 0 )
2
2
v A0  v A  vB  vB 0
v A0  vB 0  vB  v A
2
2
2
B0
v A0  vB 0  vB  v A
v A 0|E  v B 0|E  v B |E  v A|E
 ( v B 0|E  v A 0|E )  v B |E  v A|E
v B |E  v A|E  v B | A
 v B 0| A 0  v B | A

In an elastic Collision, the relative velocities
of the two objects have the same magnitude

A 0.150 kg glider (puck on an air hockey
table) is moving to the right with a speed of
0.80 m/s. It has a head-on collision with a
0.300 kg glider that is moving to the left with
velocity 2.20 m/s. Find the final velocities of
the two gliders. Assume elastic collision.

A bat strikes a 0.145kg baseball. Just before
impact the ball is travelling horizontally to
the right at 50.0 m/s and it leaves the bat
travelling to the left at an angle of 30o above
the horizontal with a speed of 65.0 m/s. Find
the horizontal and vertical components of the
average force on the ball if the ball and bat
were in contact for 1.75 ms.

A 23 g bullet travelling at 230 m/s penetrates
a 2.0kg block of wood and emerges cleanly at
170 m/s. If the wood is initially stationary on
a frictionless surface, how fast does it move
after the bullet emerges?

A 90.0 kg full back running east with a speed
5.0 m/s is tackled by a 95.0kg opponent
running north at 3.00 m/s. If the collision is
completely inelastic, (a) find the velocity of
the players just after the tackle. (b) find the
mechanical energy lost during the collision.


Giancoli 7-78
A 0.25kg skeet (clay target) is fired at an
angle of 30o to the horizon with a speed of
25 m/s. When it reaches its maximum height,
it is hit from below by a 15g pellet traveling
vertically upwards at 200 m/s. The pellet is
embedded into the skeet. (a) how much
higher does the skeet go up? (b) how much
further does the skeet travel?


Objects approximated to be point particles
Objects only undergo translational motion




Real objects also undergo rotational motion
while undergoing translational motion.
But there is one point which will move as if
subjected to the same net force.
We can treat the object as if all its mass was
concentrated on a single point

Set an arbitrary origin point

rcm 




r1 m 1  r2 m 2  r3 m 3  ...
m 1  m 2  m 3  ...

 rm
m
i
i
i
Center of mass is the mass weighted average
of the particles

A simplified water molecule is
shown. The separation between
the H and O atoms is d=9.57
x10-11m. Each hydrogen atom
has a mass of 1.0 u and the
oxygen atom has a mass of
16.0 u. Find the position of the
center of mass.

For ease set origin to one of the
particles

rcm 



r1 m 1  r2 m 2  r3 m 3  ...
m 1  m 2  m 3  ...
xm
m
ym


m

x cm 
i
i

i

y cm
i
i
i


 rm
m
i
i
i
m h d cos( 52 . 5 )  m h d cos(  52 . 5 )  m o ( 0 )
mh  mh  mo
m h d sin( 52 . 5 )  m h d sin(  52 . 5 )  m o ( 0 )
mh  mh  mo
xm
m
ym


m

x cm 
i
i

i

y cm
i
i
i

2 m h d cos( 52 . 5 )
mh  mh  mo
 6 . 5 x10
 12
m
m h d sin( 52 . 5 )  m h d sin( 52 . 5 )  m o ( 0 )
mh  mh  mo
0


1) if there is an axis of symmetry, the center
of mass will lie along the axis.
2) the center of mass can be outside of the
body



The point of an object which gravity can be
thought to act.
This is conceptually different from center of
mass
For now the center of gravity of an object is
also it’s center of mass.

rcm 



r1 m 1  r2 m 2  r3 m 3  ...

 rm
m
i
i
m 1  m 2  m 3  ...
i




d rcm
d  r1 m 1  r2 m 2  r3 m 3  ... 




dt
dt  m 1  m 2  m 3  ... 




d rcm
d ( r1 m 1  r2 m 2  r3 m 3  ...)
1

dt
m 1  m 2  m 3  ...
dt





v1 m 1  v 2 m 2  v 3 m 3  ...  v i m i
v cm 

m 1  m 2  m 3  ...
 mi



v cm  m i   v i m i  p

v cm 



v1 m 1  v 2 m 2  v 3 m 3  ...
m 1  m 2  m 3  ...




d v cm
d  v1 m 1  v 2 m 2  v 3 m 3  ... 





dt
dt 
m 1  m 2  m 3  ...






a 1 m 1  a 2 m 2  a 3 m 3  ...  a i m i
a cm 

m 1  m 2  m 3  ...
 mi


(  m i ) a cm   a i m i   F
F F
 F  ( m
 F
int ernal

) a cm
i
external
ext

Center of mass
computations useful for
when mass of a system
changes with time

James and Ramon are standing 20.0 m apart
on a frozen pond. Ramon has a mass of 60.0
kg and James has mass of 90.0 kg. Midway
between the two is a mug of their favourite
beverage. They pull on the ends of a light
rope. When James has moved 6.0 m how far
has Ramon moved?


No external forces!
Center of Mass will not move!

rcm 
 rm
m
i
i

 10 ( 90 )  10 ( 60 )
i

 300
rcm 
 2m
150
90  60


Center of Mass will not move!
James moved 6m to the right

rcm 
 rm
m
i
i

i
x
 4 ( 90 )  x ( 60 )
90  60

rcm ( 90  60 )  4 ( 90 )
60
 1m
 2m

A 1200 kg station wagon is moving along a
straight highway at 12.0 m/s. Another car
with mass 1800kg and speed 20.0 m/s has
its center of mass 40.0 m away. (a) Find the
position of the center of mass of the two
cars. (b) Find magnitude of total momentum
of the system. (c) Find the speed of the center
of mass of the system. (d) Find total
momentum using center of mass.

similar documents