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Tue Oct 13
• Conservation of Momentum
• Angular Velocity
• Assign 8 – Friday
Questions about grading: write note at
top and hand back in at end of class
Conservation of Momentum
If no net external force,
pinitial = pfinal
Do not apply a grade to any one
exam. 70% with 95% on homework,
lab, etc... gives about a B-
A tennis ball cannon is mounted on sled in the middle of a frozen
pond. The total mass of the unit (unloaded) is 50kg. The speed of
each ball after being fired is 30m/s and the mass of each ball is 0.10
kg. How fast is the unit moving after it shoots 10 tennis balls to the
right. Presume it starts at rest.
(1) 0.06 m/s left
(4) 0.6 m/s right
+x to right
(2) 0.06 m/s right
(5) 6 m/s left
(3) 0.6 m/s left
(6) 6 m/s right
pinitial = pfinal
0 = 10*pball,f + pcannon,f
0 = 10*mballvball + mcannonv
0 = 10*(0.10 kg)*(30m/s)+(50kg)v
v = -30/50 = -0.6m/s
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 + pblock1,f + pblock2,f
mbullet vi,bullet = mbulletvbullet + mBlock1 vBlock1 + mBlock2 (0)
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
Rotational Kinematics – Describing Rotation
• Angular Displacement Δθ
– measure in radians, degrees, or
revolutions
• Angular velocity
ω = Δθ / Δt
– rad/s, rpm, deg/sec
! Sect 6.1
Radians
• Δθ(rad) = arc length/radius = Δs/r
– ratio: dimensionless
– rad simple placeholder
• Full circle:
Δθ(rad) = circumference/radius = 2πr/r = 2π
• 360˚ = 2π rad
180˚ = π rad
• 135˚ = ?
(3/4)π rad or 2.36 rad
Angular Velocity:
⎛ 2π rad ⎞⎛ 1 min ⎞
1rev/min (rpm)⎜
⎟⎜
⎟ = 0.1047 rad/s
⎝ 1 rev ⎠⎝ 60 sec ⎠
A wheel undergoes an angular displacement of π/3 radians. What is
this in degrees?
(1) 15
(6) 90
(2) 30
(7) 105
(3) 45
(8) 120
(4) 60
(9) 135
(5) 75
(0) 150
Two objects are sitting on a rotating turntable. One is much further out
from the axis of rotation. Which one has the larger angular velocity?
1) the one nearer the disk center
2) the one nearer the disk edge
3) they both have the same angular velocity
All points on rigid object have same angular displacement (Δθ),
same angular velocity (ω), and same angular acceleration (α)
This is why angular quantities are so useful!
Linear quantities
• Arc length or linear distance travelled
Δs = r Δθ
• Tangential velocity or speed
(Δs/Δt) = r (Δθ/Δt)
vt
=rω
An object is rotating at an angular velocity 0.5 rad/s. What
is the total angular displacement after rotating for 10
seconds?
(1) 0.05 rad
(4) 10 rad
(2) 0.5 rad
(5) 20 rad
Δθ = ω * Δt
Δθ = (0.5 rad/s) * 10 s = 5 rad
(3) 5 rad
(6) 50 rad
Uniform Circular Motion
• If constant speed, but change direction –
changing velocity – so non-zero acceleration.
• If acceleration, net force.
• If motion in circle at const speed, acceleration
towards center.
• Can calculate this acceleration in terms of v and r
Δv vΔt
=
v
r
Δv v2
a=
=
Δt
r
Centripetal (center seeking) acceleration
ac = v2 / r