Additional Lecture Notes (ppt file)

Report
Momentum is a vector. Suppose the mass of a block is 2kg. If the xcomponent of the momentum of a block is +4 kg·m/s and the ycomponent of the momentum of the block is -3 kg·m/s, what is the
speed of the block?
(1) 1.5 m/s
(5) 5 m/s
(2) 2.0 m/s
(6) 6 m/s
(3) 2.5 m/s
(7) 8 m/s
(4) 4 m/s
(8) 10 m/s
If m=2 kg, px = +4 kg·m/s, py = -3 kg·m/s.
What is the magnitude of the momentum of the block?
+4 kg·m/s
-3 kg·m/s
p
p p
2
x
2
y
p  4 2  (3) 2
p  5 kg  m/s
Find speed:
p = mv
5 kg·m/s = 2kg * v
v = 2.5 m/s
Could also calculate velocity components directly, then find
magnitude (speed):
px = m·vx
py = m·vy
v  vx2  v y2
4kg·m/s = 2kg *vx
-3kg·m/s = 2kg *vy
v  2 2  (1.5) 2
vx = 2 m/s
vy = -1.5 m/s
v  2.5 m/s
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


J  p
J X  p X
J Y  pY
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)



F

t

m
v

m
v

f
0

Careful with signs! VECTORS!
 Ft  m(5)  m(5)
Suppose initial velocity is -5m/s
and final velocity is + 5m/s
Ft  m(5  (5))

(presume +y is up)

 Ft  10m
A 2kg puck is traveling along at 3m/s. An impulse of 10Ns is exerted
on the puck at an angle perpendicular to its motion. Which path best
describes the motion of the puck after the impulse?
(1) q = 90°
(2) 45° < q < 90°
(3) q = 45°
(4) 0° < q < 45°
(5) q = 0°
After impulse, px = 6 kg m/s (same as before)
py = 10 kg m/s
Greater momentum in y than in x
10 kg·m/s
6 kg·m/s
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


p0  p f
• Momentum stays same – Conserved!
• System – object/group of objects considering
• Total momentum of system – add momenta
from each object.
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 with momentum conserved. pinitial to right, pfinal to right
vfinal = 0.14 m/s to the right
Two cars collide and stick together. How fast will they be traveling
after the collision?
+x
2m/s
3m/s
4000kg
2000kg
pinitial = pfinal
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
A 1kg bullet has a horizontal velocity of 200 m/s. It strikes to a
wooden block of 10 kg initially at rest and becomes embedded.
What is the velocity of the bullet and the block after the event?
A 1kg bullet has a horizontal velocity of 200 m/s. It strikes to a
wooden block of 10 kg initially at rest. If the bullet pass through the
block and its final speed is 100 m/s, what will be the velocity of the
bullet?
Clicker!
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
Clicker!
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. Was 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
m*6 = (2m) * 1m/s + m*v
v = 4 m/s
(4) 4 m/s
(8) 18 m/s
(9) 0 m/s
Collisions
•
•
•
•
All collisions conserve momentum
ELASTIC: KE conserves.
INELASTIC: KE does not conserve.
perfectly inelastic – stick together (final v same)
Elastic Collisions
KE:
KE0,1 + KE0,2 = KEf,1 + KEf,2
1 2 1 2
1 2
1 2
mv 0,1  mv 0, 2  mv f ,1  mv f , 2
2
2
2
2
p:
p0,1 + p0,2
= pf,1 + pf,2
mv0,1  mv0,2  mvf ,1  mvf ,2
Example
A bullet of 0.1 kg is fired to a wooden block of mass 5 kg hanged by a rope as
shown. After the bullet hits the block, it is embedded in the block. If the block
swings (bullet is embedded) to a height of 0.5 m due to the impact, what is the
Initial speed of the bullet?
Clicker!
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. During the collision, which of the following are
conserved?
(1) horizontal component of total momentum
of ball/block system
(2) Total KE of ball/block system
(3) Both
Problem 65 – Cutnell and Johnson
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.
(a) What is the velocity of the second block after the bullet imbeds itself?
(b) Find the ratio of the total kinetic energy after the collision to that before the
collision.
0.513m/s
KEAfter /KEBefore = 1.49e-3
Problem 65 – Cutnell and Johnson - extended
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.
(c) What is the speed of the bullet after the collision with Block 1 and before the
collision with Block 2?
(d) What is the average force of the bullet on the block if it takes 0.400ms to go
through block 1?
Three pucks collide and stick together. Initial speeds and masses are given. What is
the direction of the 3 pucks after the collision? (choose an axis only if exactly on
that axis)
Clicker!
9 if stationary
Initial momentum of system: up and to the right
Final momentum: same – up and to the right
px,i = +(2kg)(3m/s) + (1kg)(0m/s) – (1kg)(3m/s)
= + 6 kg·m/s - 3 kg·m/s
= + 3 kg·m/s
pyi = (2kg)(0m/s) + (1kg)(3m/s) + (1kg)(0m/s)
= + 3 kg·m/s
Example
Find the center of mass of the objects shown.
0.5 m
4 kg
1 kg
1m
2 kg

similar documents