Thur Oct 8
• Assign 7 – Friday
• Exams back Tuesday
Impulse and Momentum
Conservation of Momentum
• Work (W): force over distance related to change in KE
• Impulse (J): force over time
related to change in momentum
• Momentum of an object = p = mv
• Impulse and momentum VECTOR quantities
Power
• “Rate at which work is done”
• “Rate at which energy is expended”
– J/s – watt (W)
P=
Work
Time
P=
Change in Energy
Time
Work, Energy, and Power in Humans
Efficiency (eff) = Useful Energy or Work Output / total Energy Input
What is the minimum energy needed for you to climb a 20m-tall cliff?
Assume a mass of 70kg. Also assume that you are at rest before and
after the climb.
(1) 14.3 J
(3) 196 J
(5) 686 J
(2) 140 J
(4) 463 J
(6) 13700 J
KEi + PEi + WNC = KEf + PEf
Pick reference at base
0J + 0J + WNC = 0 J + mgh
WNC = (70kg) (9.8m/s2)(20m) = 13700J
What is the minimum power output if this climb is accomplished over
a time of 3 minutes?
(1) 76.2 W
(2) 2380 W
(3) 4570 W (4) 8290 W
(5) 12850 W (6) 41200 W
Power = Energy or Work / time
= 13700 J / 180 s
= 76.2 W
If your body is only 0.20 (20%) efficient at creating useful energy
from your energy input (food), what would be the energy input
needed for the climb?
(1) 2740 J
(3) 13700 J
(5) 68500 J
(2) 10960 J
(4) 17100 J
Eff = Useful Energy / Energy Input
Energy Input = Useful Energy / eff
= 13700 J / 0.20
= 68500 J
How much ‘non-useful’ energy would be produced?
(typically in the form of thermal energy)
Energy input = Energy useful + Energy not useful
ENOT USEFUL = 68500 J – 13700 J = 54800 J
A car is traveling fast. A large truck is moving slowly. Which has the
greatest momentum?
1.
2.
3.
4.
The car
The truck
Both have the same momentum
Not enough information
Momentum depends on both mass and velocity. In this case we’d
need the numbers to determine the greatest momentum.
F versus t graph
• Impulse equal to area under F vs t graph
• Vector quantity: need to account for direction
• Impulse in x direction will only affect
momentum in x
– will not change momentum in y
I apply the same impulse to 3 objects which are initially at rest. Which
of the three has the greatest momentum after the impulse has been
applied?
(1)
(2)
(3)
(4)
A small toy car
An empty shopping cart
A full shopping cart
All three the same
Same impulse, same increase in momentum!
I apply the same impulse to 3 objects which are initially at rest. Which
of the three is traveling the fastest after the impulse has been applied?
(1)
(2)
(3)
(4)
A small toy car
An empty shopping cart
A full shopping cart
All three the same
Same impulse, smallest mass will have greatest increase in speed.
If mass constant
Two balls of equal mass are dropped on the table from the same
height. Ball A bounces back up to roughly the same height from
which it was dropped. Ball B does not bounce back up at all. Which
ball receives the greater impulse from the table?
1. Ball A
2. Ball B
3. Both feel the same impulse
Greater change in momentum (not only stopped, but reversed
direction)
Careful with signs!
VECTORS!
Suppose initial
velocity is -5m/s and
final velocity is + 5m/s
(presume +y is up)
Collisions
• During collision: F12 = - F21 during whole time
• |J12| = |J21|
• (Sum ext forces + sum int forces)Δt = pf – p0
• Sum int forces always zero
• If sum of average external forces = 0
• Momentum stays same – Conserved!
• System – object/group of objects considering
• Total momentum of system – add momenta
from each object.
"Explosions”/Recoil
• Total initial momentum = 0
• Total momentum: add up momentum of all objects
in system – remember vector nature
• If net external force is zero, then total final
momentum = 0
• Doesn't mean momentum of each individual object
is zero
For two objects in one dimension with pinitial= 0:
pinitial = pfinal
0 = p1,f + p2,f
0 = m1v1,f + m2v2,f
m1v1,f = -m2v2,f
Examples: firecrackers, rockets, recoil, …
Somehow you got a hold of an old cannon
(mass=600kg). You place it on a frozen lake
(ignore friction) and fire a 5kg cannon ball at
a speed of 300m/s to the right. What is the
velocity of the cannon after it has been fired?
The cannon starts at rest.
(1) 0.5 m/s to left
(2) 0.5 m/s to right
(3) 2.5 m/s to left
(4) 2.5 m/s to right
(5) 5.0 m/s to left
(6) 5.0 m/s to right
(7) 10 m/s to left
(8) 10 m/s to right
pinitial = pfinal
0 = pc,f + pb,f
0 = mcvc,f + mbvb,f
mcvc,f = -mbvb,f
600kg vc,f = - 5kg * 300m/s
vc,f = -2.5m/s
Two small railroad cars are traveling at each other with the masses
and speeds shown below. If you take the two cars as the system, what
is the initial total momentum of the system?
(1) 11000 kgm/s to the left
(2) 1000 kgm/s to the left
(3) 0 kgm/s
(4) 1000 kgm/s to the right
(5) 11000 kgm/s to the right
6000 kg m/s to right; 5000 kg m/s to left
After the two cars collide, they stick together. Which direction will
they be traveling? (How fast will they be traveling?)
(1) to the left
(2) they will be stationary
(3) to the right
Collision: momentum conserved.
pinitial to right, pfinal to right
pinitial = pfinal
p1,i + p2,i = p12,f
m1v1,i + m2v2,i = (m1+ m2)vf
2000kg·3m/s + 5000kg·(-1m/s) = (7000kg)vf
1000kg·m/s =(7000kg) vf
Vf =+0.14 m/s
Two cars collide and stick together. How fast will they be traveling
after the collision?
+x
2m/s
3m/s
4000kg
2000kg
pinitial = pfinal
(1)
(2)
(3)
(4)
(5)
(6)
(7)
0 m/s
0.33 m/s to the right
0.33 m/s to the left
2.3 m/s to the right
2.3 m/s to the left
2.5 m/s to the right
2.5 m/s to the left
p1,i + p2,i = p12,f
m1v1,i + m2v2,i = (m1+ m2)vf
4000kg·2m/s + 2000kg·(-3m/s) = (6000kg)vf
2000kg·m/s =(6000kg) vf
Vf =+0.33 m/s
Three railroad cars of equal mass are on a track. The one on the left is
approaching at 6m/s. Ignoring friction, when they have all linked up
as one, what is their speed?
(1) 1 m/s
(5) 5 m/s
(2) 2 m/s
(6) 6 m/s
pinitial = pfinal
m*6 = (3m) * v
v = 2 m/s
(3) 3 m/s
(7) 12 m/s
(4) 4 m/s
(8) 18 m/s
(9) 0 m/s
Three railroad cars of equal mass are on a track. The one on the left is
approaching at 6m/s. Ignore friction. At the end the two left cars are
linked and traveling at 1m/s, but the rightmost car didn’t connect
properly and is free. What is the speed of the separate unlinked car?
(1) 1 m/s
(5) 5 m/s
(2) 2 m/s
(6) 6 m/s
(3) 3 m/s
(7) 12 m/s
pinitial = pfinal
p1,i + p2,i + p3,i = p12.f + p3,f
m*6 + 0 + 0 = (2m) * 1m/s + m*v
v = 4 m/s
(4) 4 m/s
(8) 18 m/s
(9) 0 m/s
Example
Two low-friction carts at
rest on a track.
Total momentum before
the explosion = 0.
One of the carts is
equipped with a springloaded plunger that can be
released by tapping on a
small pin.
The spring is compressed
and the carts are placed
next to each other.
Classifying Collisions
•
•
•
•
All collisions conserve momentum
If also conserve KE, ELASTIC
Otherwise, INELASTIC
perfectly inelastic – stick together (final v same)
Conserved?
Momenta
m,v
Not Conserved?
Impulse
Initial
F, Δt
Momenta
m,v
Final
Elastic Collisions – 2 Objects
KE:
KE0,1 + KE0,2 = KEf,1 + KEf,2
1 2 1 2
1 2
1 2
mv0,1 + mv0, 2 = mv f ,1 + mv f , 2
2
2
2
2
p:
p0,1 + p0,2 = pf,1 + pf,2
mv0,1 + mv0, 2 = mv f ,1 + mv f , 2
For elastic collisions with object 2 at rest: (Example)
• If m1 > m2 then both head off in same direction, 2 faster than 1
• If m2 > m1 then 1 bounces back and 2 is slower than 1
• If m2 = m1 then velocities swap! 1 stops and 2 moves forward
1
2
A 4.00-g bullet is moving horizontally with a velocity of +355 m/s,
where the + sign indicates that it is moving to the right. The mass
of the first block is 1150 g, and its velocity is +0.550 m/s after the
bullet passes through it. The mass of the second block is 1530 g.
If the system is the bullet and both blocks, what is the total initial
momentum?
(1) 1.4 kg·m/s to the right
(3) 410 kg·m/s to the right
(5) 950 kg·m/s to the right
(2) 1.4 kg·m/s to the left
(4) 410 kg·m/s to the left
(6) 950 kg·m/s to the left
Choose +x to the right
pinitial = sum of all initial momentum
= pi,Bullet + pi, Block 1 + pi,Block 2
= 0.004kg * 355m/s + 1.15kg * 0 m/s + 1.53kg * 0 m/s
= + 1.4 kg·m/s
If the system is the bullet and both blocks, what is the total final
momentum after the bullet has lodged in the second block?
(1) 1.4 kg·m/s to the right
(3) 410 kg·m/s to the right
(5) 950 kg·m/s to the right
(2) 1.4 kg·m/s to the left
(4) 410 kg·m/s to the left
(6) 950 kg·m/s to the left
Collision - no external impulses – momentum conserved
pfinal = pinitial
(a) What is the velocity of the second block after the bullet imbeds
itself?
pinitial = pfinal
pi,bullet = pbullet,f + pblock2,f+ pblock1,f
mbullet vi,bullet = (mbullet + mBlock2 )vf + mBlock1 vBlock1
1.42 kg m/s = 1.534 kg vf + 1.150kg * 0.550 m/s
vf = 0.513 m/s
(b) Find the ratio of the total kinetic energy after the collision to that
before the collision.
KEBefore = ½ mBulletv2Bullet= 252J
KEAfter = ½ m1v2Block1+ ½ (m2+mBullet) v2f = 0.376J
KEAfter /KEBefore = 1.49e-3
(c) What is the speed of the bullet after the collision with Block 1 and
before the collision with Block 2?
pinitial = pfinal
pi,bullet = pbullet,f + pblock2,f+ pblock1,f
mbullet vi,bullet = mbulletvbullet + mBlock2 (0) + mBlock1 vBlock1
1.42 kg m/s = 0.004 kg vf + 1.150kg * 0.550 m/s
vf = 197 m/s
(d) What is the average force of the bullet on the block if it takes
0.400ms to go through block 1?
Fon block1 Δt = Δpblock1
Fon block1 (4x10-4 s) = pBlock1,f – pBlock1,I
F (4x10-4 s) = mBlock1vBlock1 – 0
F = 1580N to the right
Ballistic Pendulum (future lab)
• Measure velocity of projectile
• Perfectly inelastic collision
• Use conservation of energy to determine KE just after collision but
before change in height.
• KE gives velocity after collision
• Use conservation of momentum to find velocity before collision
A ball is attached to a wire, held horizontally, and dropped. It strikes
a block that is sitting on a horizontal, frictionless surface. Air
resistance is negligible and the collision is elastic. The block is
more massive than the ball. Which of the following are conserved
as the ball swings down?
(1) Ball’s Kinetic Energy
(2) Ball’s Momentum
(3) Ball’s Total Mechanical Energy
(4) 1 and 2
(5) 1 and 3
(6) 2 and 3
(7) 1, 2, and 3
A ball is attached to a wire, held horizontally, and dropped. It strikes
a block that is sitting on a horizontal, frictionless surface. Air
resistance is negligible and the collision is elastic. The block is
more massive than the ball. During the collision, which of the
following are conserved? (ignore any changes of height during the
collision)
(1) horizontal component of total momentum
of ball/block system
(2) Total KE of ball/block system
(3) Both
A ball is attached to a wire, held horizontally, and dropped. It strikes
a block that is sitting on a horizontal, frictionless surface. Air
resistance is negligible and the collision is elastic. The block is
more massive than the ball. Which direction do you expect the ball
to be traveling after the collision?
(1) To the left
(2) To the right
(3) Ball will be stationary
Bullet and Block
• Projectile motion
• Collision
• Could you find the velocity
of the block just after the
collision as it is flying off
the table?
Multiple Concept Problems