Transcript with momentum - Cloudfront.net

Momentum & Impulse
Day #1: Introduction
HW #7
Momentum & Collisions:
Define Momentum:
Momentum, p, is defined as the product of mass and
velocity.


p  mv
m kg m
Units: momentum = (mass) (velocity)  kg 
s
s
Momentum is a vector, and the direction of the
momentum is the same as the velocity.
Define Impulse and Relation to Newton’s 2nd Law:
Newton’s original statement of his 2nd law of motion:
The time rate of change of the momentum of an
object equals the net force impressed upon the
object.


p
Fnet 
t
Approximately true for real
differences, considered true
when used in calculus.
Is this the same as F = ma?
Fnet
p

t
 mv 

t
Fnet
mv vm


t
t
vm
 ma 
t
For all the situations so far, mass has been constant.
In that limit, F = ma. When mass changes, the
problems are considerably harder.
Impulse is defined as the change of momentum:
Fnet
p

t
p  Fnet t
Units: impulse = (force) (time)
 Ns
The two units for momentum are equivalent, and
both are used.
kg m
1Ns  1
s
Example #1: (a) What is the momentum of a 10.0 kg mass that travels
at 20.0 m/s?
p  mv
 10.0 kg 20.0 ms 
 200 kgsm
(b) What would be the speed of a 40.0 kg mass with the same
momentum?
p  mv
p
v
m
200 kgsm

40.0 kg
 5.00 ms
(c) Which object has the larger kinetic energy?
KEa  mv
2
KEb  mv
2
1
2
1
2
1
m 2
 10.0 kg 20.0 s 
2
 2000 J
1
m 2
 40.0 kg 5.00 s 
2
 500 J
2
In general:
1  p
KE  mv  m 
2 m
p2
KE 
2m
For a given momentum, a smaller
mass will have greater KE
1
2
2
Example #2: A 255 gram baseball is thrown at 40.0 m/s towards the
right. The ball is struck by a bat and subsequently the ball travels at
50.0 m/s towards the left. The ball is in contact with the bat for 2.00 ms
(millisecond). (a) What is the impulse delivered to the ball?
Take towards the right as the positive direction.
vo  40.0 ms
m  255 g  0.255 kg
v  50.0 ms
Impulse = p  Fnet t
 mv f  mvo
p  0.255 kg  50.0 ms    40.0 ms   22.95 Ns
(b) What is the net force on the ball?
p  Fnet t
Fnet
p

t
 22.95 Ns

2.00  10 3 s
 11,500 N
Application of Momentum: Collisions.
Initially, each mass has
velocity and momentum.
FB on A = Applied Force
During the collision, masses exert
equal but opposite forces on one
another.
Finally, each mass has
velocity and momentum.
FA on B = Applied Force
Each mass will have an initial momentum:
p1i  m1v1i = initial momentum of mass m1.
p2i  m2v2i = initial momentum of mass m2.
Each mass will have an final momentum:
p1 f  m1v1 f = final momentum of mass m1.
p2 f  m2 v2 f = final momentum of mass m2.
During the collision, each mass exerts a force on the other.
By Newton’s 3rd law, these are equal and opposite.
F1on 2   F2 on1
Impact takes a time t. The impulse on each mass is:
p2  F1on 2 t   F2 on1 t   p1
m v
2 2f
 m2 v2i   p2  p1  m1v1 f  m1v1i 
With a little rearranging, get:
m1v1i  m2 v2i  m1v1 f  m2 v2 f
This is known as Conservation of Total Momentum
Example #3: A 534 kg cannon fires an 6.39 kg projectile at 1342 m/s.
(a) What is the total initial momentum?
Before the projectile is fired,
everything is at rest.
v1i  0
v2i  0
Thus the total momentum is
zero.
m1v1i  m2v2i  0
(b) Determine the recoil velocity of the cannon after the projectile is fired.
m1v1 f  m2 v2 f  m1v1i  m2 v2i  0
v1 f  
m2 v2 f
m1
6.39 kg1342 ms 

534 kg
v1 f  16.1 ms
The ( – ) sign means the cannon recoils in the direction
opposite to the projectile.
Momentum & Impulse
Day #2: Collisions
Conservation of Momentum
HW #7
Example #4: Four railroad cars, each of mass 2.50 × 104 kg, are
coupled together and coasting along horizontal tracks at speed vo
toward the south. A very strong but foolish movie actor riding on the
second car uncouples the front car and gives it a big push, increasing
its speed to 4.00 m/s south. The remaining three cars continue moving
south, now at 2.00 m/s. (a) Find the initial speed of the cars.
vo  ?
before
v1 f  2.00 ms
after
v2 f  4.00 ms
Momentum Conservation:
4 Mvo  3Mv1 f  Mv2 f
vo 
3v1 f   v2 f 
4
32.00 ms   4.00 ms 

4
vo  2.50 ms
(b) How much work did the actor do?
Work done = change of KE:
W

1
2
3M v1 f 
2

1
2
W  KE  KE  KEo
M v2 f 
2

1
2
4M vo 
2
KE f
W  2.50 10 kg
4
 32.00 
1
2
m 2
s
KEo

1
2
4.00 

m 2
s
1
2
W  3.75 104 J  37.5 kJ
42.50  
m 2
s
Example #5: A 10.0 kg mass travels towards the right at a speed of
6.00 m/s and collides with and sticks to a 20.0 kg initially at rest. (a)
What is the speed of the combined object after the collision?
v1i  6.00 ms
m1  10.0 kg
m2  20.0 kg
v1 f  v2 f  v f  ?
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1v1i  m2 v2i  m1  m2 v f
m1v1i  m2 v2i  m1  m2 v f
0
m1v1i

10.0 kg6.00 ms 
vf 

m1  m2
10.0 kg  20.0 kg
 2.00 ms
(b) What is the amount of kinetic energy after the collision? Write it as a
percentage of the initial kinetic energy.
KE f
KEi

1
2
m1  m2 v f 
2
1
2 m1 v1i 
2
30.0 kg2.00 

10.0 kg6.00 
m 2
s
m 2
s
KE f
KEi
 0.333  33.3%
Example #6: A 10.0 kg mass travels towards the right at a speed of
6.00 m/s and collides with and sticks to a 20.0 kg moving initially
towards the left at 4.50 m/s. (a) What is the speed of the combined
object after the collision?
v1i  6.00 ms
v2i  4.50 ms
m1  10.0 kg
m2  20.0 kg
v1 f  v2 f  v f  ?
m1v1i  m2 v2i  m1  m2 v f
m1v1i  m2 v2i 10.0 kg 6.00 ms   20.0 kg  4.50 ms 
vf 

m1  m2
10.0 kg  20.0 kg
vf 
10.0 kg6.00 ms   20.0 kg 4.50 ms 
10.0 kg  20.0 kg
v f  1.00 ms
(b) What is the amount of kinetic energy after the collision? Write it as a
percentage of the initial kinetic energy.
KE f
KEi
KE f
KEi
m1  m2 v f 

2
2
1
1
2 m1 v1i   2 m2 v2 i 
1
2
2
30.0 kg 1.00 

m 2
m 2
1
1
2 10.0 kg 6.00 s   2 20.0 kg  4.50 s 
1
2
KE f
KEi
m 2
s
 0.0392  3.92%
Example #7: A mass of 5.00 kg slides towards the left at a speed of
10.0 m/s. At the same time, a mass of 15.0 kg slides towards the right
at 5.00 m/s. The two masses collide with one another and stick
together after the collision. (a) Determine the velocity (speed and
direction) of the combined mass after the impact.
m1v1i  m2 v2i  m1  m2 v f
m1v1i  m2 v2i
vf 
m1  m2 
vf
5.00 kg 10.0 ms   15.0 kg 5.00 ms 

5.00 kg  15.0 kg
v f  1.25 ms
(b) Determine the amount of kinetic energy that is lost in this collision.
{Hint: Subtract the initial KE from the final KE.} Write the kinetic energy
lost as a fraction of the initial kinetic energy.
KE f  KEi
%KE lost =

KEi
1
2
1
2

KE f
KEi
1
m  m v 
2
1
m1 v1i 
2
2
f
 12 m2 v2i 
2
1
20.0 kg1.25 

1
2
2
5.00 kg10.0 ms   15.0 kg5.00 ms 
m 2
s
% KElost  0.964  96.4%
The ( – ) sign means
KE is lost.
Example #8: A 2000 kg car was traveling towards the east when it
collided with and stuck to a 4000 kg car initially traveling towards the
north. The wreckage slid at an angle of 55.2 degrees NE for a distance
of 18.15 meters before stopping. Officer Bogart determines the
coefficient of friction between the types of tires on the vehicle and the
asphalt to be 0.600 for the conditions present that day. Determine the
speeds of the cars before the collision. Which, if any, were driving over
the 40 mph speed limit?
Find vf from energy:
E f  Ei  Wnc
0  KEi  Fk x cos180
0  mv f  k mgx
1
2
2
v f  2 k gx
 14.61 ms
Momentum conserved in each component direction.
2000 kgv1i   6000 kg14.61 ms cos55.2
v1i  25.0 ms  56 mph
4000 kgv2i   6000 kg14.61 ms sin 55.2
v2i  18.0 ms  40 mph
Example #9: A ballistic pendulum is shown at right. A heavy mass
(M = 2.00 kg) hangs vertically downwards at rest. A small projectile
(m = 0.0400 kg) travels with initial speed vo and collides with and sticks
to the heavy mass. After the collision, the heavy mass swings out and
rises up vertically 12.0 cm.
(a) What was the speed of the combined block and
projectile immediately after the collision?
Order of Events:
Bullet embeds in block. Solve for final
speed with momentum conservation.
Final speed of collision becomes initial
speed of block rising. Use energy
conservation.
MIT Ballistic Pendulum
Start with energy conservation.
Ebottom  Etop
KEbottom  PEtop
1
2
mtot vbottom  mtot gh
2
vbottom  2 gh  29.80 m s 2 0.120 m 
vbottom  1.53 ms
(b) What was the initial velocity of the projectile?
Next use momentum conservation.
m1v1i  m2 v2i  m1v1 f  m2 v2 f
0
m1v1i  m2v2i  m1  m2 vbottom masses stick together
v1i

m1  m2 vbottom

m1
2.04 kg 1.53 ms 

0.0400 kg
v1i  78.2 ms
Elastic Collisions
Elastic Collisions, 1-dim
HW #6: Handout
#18 – 21, 24 – 26.
{on the schedule…}
Elastic Collisions in One Dimension:
Definition of Elastic Collisions:
All collisions conserve momentum!
Elastic collisions also conserve total
kinetic energy.
Momentum conservation is always true. KE is only
conserved for elastic collisions.
{common source of error!}
Inelastic collisions only conserve momentum.
Usually KE is lost.
inelastic = not elastic
Equation for momentum conservation:
m1v1i  m2 v2i  m1v1 f  m2 v2 f
Equation for kinetic energy conservation:
1
2
m1v1i  m2v2i  m1v1 f  m2v2 f
2
1
2
2
1
2
2
1
2
Two equations means two unknowns!
2
Derivation of reduction of kinetic energy equation:
Show
v1i  v2i  v2 f  v1 f
Note: You would be asked to derive this only on a bonus
question…
Start with momentum:
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1v1i  m1v1 f  m2 v2 f  m2 v2i
m1 v1i  v1 f   m2 v2 f
 v2 i 
Eq. #1
Next KE:
1
2
m1v1i  m2v2i  m1v1 f  m2v2 f
2
2
1
2
2
1
2
1
2
m1v1i  m1v1 f  m2v2 f  m2v2i
2
m1 v1i
2
2
 v1 f
2
2
  m v
2
2f
m1 v1i  v1 f v1i  v1 f   m2 v2 f
2
 v2i
2
2

 v2i v2 f  v2i 
Note: The underlined parts are equal from equation #1.
2
v1i  v1 f  v2 f  v2i
v1i  v2i  v2 f  v1 f
Note: This is the simplified form of Conservation of Total
Kinetic Energy. This is only used with Elastic collisions.
Example #1: A mass m1 = m travels towards the right at a speed v1i = vo
and collides with an identical mass m2 = m initially at rest. If the two
masses collide elastically, what is the final velocity of each mass?
Start with momentum:
m1v1i  m2 v2i  m1v1 f  m2 v2 f
mvo  0  mv1 f  mv2 f
vo  v1 f  v2 f
Next use KE:
v1i  v2i  v2 f  v1 f
vo  v2 f  v1 f
Solve the simultaneous equations:
vo  v1 f  v2 f

vo  v2 f  v1 f
vo  v1 f  v2 f

vo  v2 f  v1 f
2 vo  2 v 2 f
2vo  2v1 f
vo  v2 f
v1 f  0
The particles completely exchange velocities.
Example #2: A 10.0 kg mass (m1) travels towards the right at 15.0 m/s
and collides elastically with a 20.0 kg mass (m2) initially at rest. (a)
Determine the final velocity of each mass.
Start with momentum:
m1v1i  m2 v2i  m1v1 f  m2 v2 f
10.0 kg 15.0 ms   0  10.0 kg v1 f  20.0 kg v2 f
v1 f  2v2 f  15.0 ms
Next use KE:
v1i  v2i  v2 f  v1 f
v2 f  v1 f  15.0 ms
v1 f  2v2 f  15.0 ms
Solve:

v2 f  v1 f  15.0 ms
3v2 f  30.0 ms
v2 f  10.0 ms
Pick one equation to solve for v1f:
v1 f  2v2 f  15.0 ms
v1 f  15.0 ms  210.0 ms 
v1 f  15.0 ms  2v2 f
v1 f  5.00 ms
(b) What percentage of the initial kinetic energy has been transferred to
the second mass?
KE2 f
KE1i  KE2i

1
2
1
2
m2 v2 f
2
m1v1i  0
2
20.0 kg10.0 

10.0 kg15.0 
m 2
s
m 2
s
 0.889  88.9 %
Example #3: A 20.0 kg mass (m1) travels towards the right at 15.0 m/s
and collides elastically with a 10.0 kg mass (m2) initially at rest.
Determine the final velocity of each mass.
Start with momentum:
m1v1i  m2 v2i  m1v1 f  m2 v2 f
20.0 kg 15.0 ms   0  20.0 kg v1 f  10.0 kg v2 f
2v1 f  v2 f  30.0 ms
Next use KE:
v1i  v2i  v2 f  v1 f
v2 f  v1 f  15.0 ms
2v1 f  v2 f  30.0 ms
Solve:

v2 f  v1 f  15.0 ms
3v1 f  15.0 ms
v1 f  5.00 ms
Pick one equation to solve for v2f:
2v1 f  v2 f  30.0 ms
v2 f  30.0 ms  25.00 ms 
v2 f  30.0 ms  2v1 f
v2 f  20.0 ms
Example #4: A cart with a mass of 340 grams moving on a frictionless
surface at initial speed of 1.2 m/s collides elastically with a second mass
at rest. The mass of the second object is unknown. After the elastic
collision, the first cart travels in its original direction at 0.66 m/s.
Determine the mass of the unknown cart and the speed of the unknown
cart after the collision.
Start with the KE equation, only one unknown!
v1i  v2i  v2 f  v1 f
1.2 ms  0  v2 f  0.66 ms
v2 f  1.86 ms
Next use momentum to solve for the mass:
m1v1i  m2 v2i  m1v1 f  m2 v2 f
0.340 kg1.2 ms   0  0.340 kg0.66 ms   m2 1.86 ms 
0.340 kg1.2 ms   0.340 kg0.66 ms 
m2 
1.86 ms 
m2  0.0987 kg  98.7 gram
Example #5: A steel ball of mass 0.500 kg is fastened to a cord 70.0 cm
long and fixed at the far end, and is released when the cord is horizontal,
as shown in the figure. At the bottom of its path, the ball strikes a 2.50
kg steel block initially at rest on a frictionless surface. The collision is
elastic. (a) Find the speed of the block after the collision.
Start with energy conservation to find the
speed of the steel ball at the bottom before
the impact.
Next solve the elastic collision to find the velocities of the
two objects after the collision.
Finally use energy conservation again to solve for the height
to which the steel ball will rise.