Physic 231 Lecture 4
•
•
•
Main
M
i points
i t off today’s
t d ’ lecture:
l t
Example: addition of velocities
Trajectories of objects in 2
dimensions:
g = 9.8 m/s 2 downwards
vy = vy 0 − gt
Δy ≡ y − y0 =
1
(vy + vyo )t
2
1
Δy = vy0t − gt2
2
v2y − v2y 0 = −2gΔy
vx = vx0
Δx = vx0t
Example:
•
A ffootball
tb ll player
l
runs the
th pattern
tt
given
i
in
i the
th drawing
d
i by
b the
th three
th
displacement vectors A, B and C. The magnitudes of these vectors are
A=5 m, B=15 m, and C=18 m. Using the component method, find the
magnitude and direction of the resultant vector A+B+C.
A+B+C
R=
label
x-comp
y-comp
A
0
5m
B
15 m
0
C
14.7
-10.3m
R
29 7
29.7m
-5.3,m
53
C x = 18m cos(350 ) = 14.7m;
C y = 18m sin(350 ) = −10.3m;
R = 29.7 2 + 5.32 = 30.2m
tan(θ ) = −55.3
3 / 29
29.77
θ = −10.10
quiz
•
Bobb travels
B
t
l 3 km
k due
d eastt andd then
th travels
t
l 4 km
k due
d north.
th Relative
R l ti to
t
his starting position what is the magnitude of his total displacement
and the angle with respect to due east?
– a)) 7 km,
k 35 degrees
d
to
t the
th south
th off due
d east.
t
– b) 7 km, 53 degrees to the north of due east
– c) 5 km, 35 degrees to the south of due east
– d) 5 km, 53 degrees to the north of due east

R

B

A
vector
x comp.
y comp.
A
3 km
0
B
0
4 km
R
3 kkm
4 kkm
R = Rx2 + R y2 = 32 + 42 = 5km
tan θ =
Ry
Rx
= 1.333 = 53o , north of due east
relative velocity problems
•
•
Label
L
b l eachh object
bj t by
b a letter
l tt that
th t reminds
i d you off what
h t it is
i (for
(f
example p for paper, g for ground, t for truck).
Look for phases such as "the velocity of the paper relative to truck"
and
d write
it the
th velocity
l it as:


v paper _ relative _ to _ truck = v pt
•
Take the three velocities and assemble them into an equation such as;
•
Solve
S
l ffor th
the velocity
l it you want.
t N
Note
t th
thatt th
these velocities
l iti needd nott be
b
parallel. You may need to solve two equations, one in the x direction
and another in the y direction.



v pg = vtg + v pt
exercise
•
Chuckk looks
Ch
l k ahead
h d andd sees Grandpa
G d Harper.
H
He
H throws
th
him
hi a
newspaper over the top of the hood to him. The truck is moving 40
km/hr due West. Relative to the truck, the newspaper also moves West
with a velocity of 40 km/hr.
km/hr What is the velocity of the newspaper
relative to Grandpa Harper?
– a) 0 km/hr
v paper _ relative _ to _ ground
– b) 80 km/hr
k /h due
d westt
= vtruck + v paper _ relative _ to _ truck
– c) 80 km/hr due east
– d) don’t know
 v ppaper
p _ relative _ to _ gground = 40 + 40
= 80 km/hr West
vtruck
40 km/hr West
vpaper relative to truck
40 km/hr West
vpaper relative to ground
?
Example
n
north
•
y
A fferry bboatt iis traveling
t
li in
i a direction
di ti 35.1
35 1o north
th off eastt with
ith a speed
d
of 5.12 m/s relative to the water. A passenger is walking with a
velocity of 2.71 m/s due east relative to the boat. What is the velocity
(magnitude and direction) of the passenger with respect to the water?
x
east



v pw = v pb + v bw
v bw,x
(
°) = 4.19m / s
bw x = 5.12 cos(35.1
v bw,y = 5.12sin(35.1°) = 2.94m / s
vpb= 2.71 m/s
θ
35.1o
vpb= 2.71 m/s
x
y
mag
θ
vpb
2.71
0
2.71
0
vbw
4.19
2.94
5.12
35.1°
vpw
69
6.9
2 94
2.94
75
7.5
23°
v pw = 6.92 + 2.942 = 7.5m / s
(
)
94
θ = tan −1 22.94
6.9 = 23° north of east
Projectile motion in two dimensions
v y = v y 0 + a y t = v y 0 − gt
Δy ≡ y − y 0 =
1
(v y + v yo )t
2
1
1
Δy = v y 0t + a y t 2 = v y 0t − gt 2
2
2
v 2y − v 2y 0 = 2a y Δy = −2 gΔy
•
vx = vx 0 + axt = vx 0
Δx ≡ x − x0 =
1
(vx + v xo )t = vx 0t
2
1
Δx = v x 0 t + a x t 2 = v x 0 t
2
v 2x − v x20 = 2a x Δx = 0
This
hi means that
h the
h x-component off the
h velocity
l i remains
i constant. The
h
y-component reflects the gravitational acceleration.
– This is true; contrary to the presentation of Jason below:
Projectile Motion
The horizontal motion is
constant; the vertical motion is
free fall:
The horizontal and vertical
components of the motion are
independent.
Slide 3-37
example
•
A bbullet
ll t iis fired
fi d from
f
a rifle
ifl that
th t is
i held
h ld 1.6
1 6 m above
b
the
th groundd in
i a
horizontal position. The initial speed of the bullet is 1100 m/s. Find (a)
the time it takes for the bullet to strike the ground and (b) the
horizontal distance traveled by the bullet.
bullet
If upward is the
direction of positive y:
vx0 1100 m/s
vy0 0
Δy -1.6 m
Δx
Δy
clicker question
•
The diagram below shows two
successive positions of a particle; it's
a segment of a full motion diagram.
Which of the following vectors best
represents the acceleration between


vi and vf
a))
b)
c)
d)
example
•
A small
ll can is
i hanging
h i from
f
the
th ceiling.
ili A rifle
ifl is
i aimed
i d directly
di tl att the
th
can, as the figure illustrates. At the instant the gun if fired, the can is
released. Ignore air resistance and show that the bullet will always
strike the can,
can regardless of the initial speed of the bullet.
bullet Assume that
the bullet strikes the can before the can reaches the ground.
Δygravity
θ
Δycan
Reading Quiz
2. The acceleration vector of a particle in projectile motion
A. points along the path of the particle.
B. is directed horizontally.
C vanishes
C.
i h att the
th particle’s
ti l ’ highest
hi h t point.
i t
D. is directed down at all times.
E. is zero.
Slide 3-9
example
•
A motorcycle daredevil is attempting to jump across as many buses as
possible (see the drawing). The takeoff ramp makes an angle of 18.0°
above the horizontal, and the landing ramp is identical to the takeoff
ramp.
p The buses are p
parked side byy side,, and each bus is 2.74 m wide.
The cyclist leaves the ramp with a speed of 33.5 m/s. What is the
maximum number of buses over which the cyclist can jump?
33 5 m/s
33.5
_
_
2.74m
v0 33.5 m/s
a) vy = vy 0 − gt
θ
b) Δy ≡ y − y0 =
18o
1
(vy + vyo )t
2
1
c) Δy = vy0t − gt2
2
d) v2y − v2y 0 = −2gΔy
Which equation to use?
v y0 = 33.5sin(18°) = 10.35m / s
1
Δy = v y0 t − gt 2 = 0
2
1 2
2v y0
v y0 t = gt
t
2.11s
2
g
Δx = v x 0 t = 33.5cos(18°)(2.11s) = 67.25m
67.25m
Δx
N buses =
=
= 24.5 24buses
w bus
2.74m
Concept problem
•
A battleship simultaneously fires two shells at enemy ships. If
the shells follow the parabolic trajectories shown
shown, which ship
gets hit first?
–
–
–
–
a) A
b) both at the same time
c). B
d) need more information
initial vertical velocity for A exceeds
the initial vertical velocityy for B :
v y2,top = v02 − 2 gh = 0
 v02 = 2 gh; hA > hB  v0 y , A > v0 y , B
2v y 0
1 2
Δytot = 0 = v y 0ttot − gttot  ttot =
g
2
comparing the two trajectories :
v0 y , A > v0 y , B  ttot , A > ttot , B