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Chapter 2: Two Dimensional Motion
1. Mr. Wester is driving his minivan around a curve. While rounding the curve his acceleration is 16.7 m/s2. How fast can he drive around the curve
without skidding off of the road? The radius of the curve is 150 m.
The answer is d), 50 m/s. Solution:
ac=v2
v=[acr]1/2
= [16.7 * 150]1/2 = 50 m/s
2. Curtis is spinning in his desk chair. His thumb, fixed in place by that pesky cast of his, is at the end of his extended arm. If his thumb's tangential
velocity is 2.0 m/s, and his thumb's centripetal acceleration is 3.6 m/s2 radially inward, how far is Curtis's thumb from his center of rotation?
The answer is d), 1.1 m. Solution:
ac=v2
r = v2/ac = 2.02/3.6 = 1.1 m
3. Willie is running around the track in a 200 m sprint. When he exits the curve, he has traveled the first 100 m and is now running at a speed of 8.0
m/s. The curve he has run through has a continuous radius of 40 m. What is Willie's total acceleration at the point just before he exits the curve?
Consider Willie accelerates tangentially at a constant rate all the way through the 100 m, and quits accelerating tangentially when he exits the curve.
The answer is a), 1.6 m/s2. Solution:
atotal = [atangent2 + acentrip2]1/2
Vf2-Vi2=2*atangent*disp
atangent= Vf2 - Vi2 /2 * disp
= [8.02 - 0] / 2 * 100m = 0.32 m/s2
acentrip = v2/r = 8.02/40 = 1.6 m/s2
And, oddly enough, putting both values of acceleration into the first formula gives you 1.6 m/s 2.
4. Drake is going for the state long jump record. He needs to beat 8.1 m to break the old mark. His acceleration on a 50 m after running 0.6 m/s2. He
can't jump at an angle higher than 39 degrees. Can he break the record?
The answer is no. Solution:
Vf2 - Vi2 = 2ad
since Vi=0, Vf = [2ad]1/2 = [2 * 0.6 * 50]1/2 = 7.7 m/s
xmax = Vo2* sin 2 theta = 7.72 * sin 78˚ = 5.9 m
g = 9.8 m/s/s
Not even close. That's why Drake plays soccer!
5. The AP Physics people want to put a satellite into orbit. If the intended height is 400 km, what tangential velocity must the satellite have in order
to keep it in orbit?
a) 7.7 x 103 m/s
b) 8.1 x 103 m/s
c) 7.9 x 103 m/s
d) 7.4 x 103 m/s
The answer is a). Solution:
This one is not easily solved. First, you must solve for the acceleration due to gravity at 400 km away from the surface of the earth.
So, given a radius for the earth of 6370 km, a Cavendish constant of 6.67 x 10 -11 N m2 / kg 2 (otherwise known as G), and a mass of
the earth as 5.96 x 1024 kg, you can solve for a(g) using the following equation:
F(w) = m(object)g = G * m(object) * m(earth)
[dist. b/tw centers of mass]2
The mass of the object cancels out, leaving:
g = G * m(earth) = 6.67 x 10-11 N m2 * 5.96 x 1024 kg
dist2
kg2
(6.37 x 106 + 4 x 105 m)2
g = 8.7 m/s/s
Note that factor label of the units gives N/kg, but that is equivalent to m/s/s since F=ma and the units are F (in N) = m (in kg) a (in
m/s/s). Now, using the formula for critical velocity: (Consider that g is the centripetal acceleration, and the formula becomes apparent.
While normally we don't consider critical velocities, when using vertical circles--such as twirling a yo-yo by your side--a certain
velocity must be kept, lest the acceleration due to gravity become greater than the centripetal acceler., which makes the object fall.)
v(crit) = [rg]0.5 = [6.77 x 106 * 8.7 m/s/s]0.5 = 7.7 x 103 m/s
Therefore, a tangential velocity of the satellite can be found.
6. A cockroach is crawling across Mr. Wester's kitchen table with a constant acceleration of (0.5i - 0.4j) cm/s2.
At t=0 the cockroach starts at the point (-6,3) cm with a velocity of 2.0 cm/s.
(a) What are the components of this cockroach's velocity and position vectors at any time t?
(b) What are the magnitude and direction of the velocity and position vectors at t=15.0 s?
a= (0.5i-0.4j) cm/s2
xo= (-6,3) cm
vo= 2.0 cm/s
t= 0
xi=voit+0.5ajt2+xoi
xi=0.5(0.5i)t2-6i
xi=(0.25t2-6)i cm
a) ai=0.5i
vi=vo+ait vj=voj+ajt
aj=-0.4j
vi=0.5it
vj=2.0j-0.4jt
v=0.5ti-(2.0-0.4t)j cm/s
yi=vojt+.5ajt2+yoj
yi=2.0jt+.5(-.4)t2+3j
yi=(2.0t-0.2t2+3)j cm
b) t=15.0 s
v2=[(0.25t2-6)i]2 + [(2.0t-.2t2+3)j]2
v2=[(0.25(15)2-6)i]2 + [(2.0(15)-0.2(15)2+3)j]2
v2=2525 + 144
v=51.7 cm/s
tan theta= -12/50.25
theta= -13.4˚
v=[0.5t]i-[(2-.4t)]j
v=[0.5(15)]i-[(2-.4(15)]j
v=7.5i-4j
v2=(7.5)2+(-4)2
v=8.5 cm/s
tan theta= -4/7.5
theta= -28.1˚
7. An electron is initially located at the orgin. This electron has an accelertion of 5j m/s2 and an initial velocity of 5 i m/s. Find
(a) the vector position and velocity at any time t and (b) the coordinates and speed of the electron at t=4 s.
a=5j m/s2
a)
ai vfi=voi+aot
aj=5 vfj=voj+ajt
vo=5i m/s
vfi=5i m/s
vfj=5jt m/s
t=4 s
v=(5i +5jt) m/s
xi=voit+.5ait2
xi=5it
b) v=5i+5j(4)
v2=52+202
v=20.6 m/s
yi=vojt+.5ajt2
yi=2.5jt2
x= (5t,2.5t2)
x= (5*4,2.5*(4)2)
= (20 m,40 m)
8. Mrs. Perry release a football at angle of 65 degrees above the horizontal. The football lands 30 m away from Mrs. Perry which is 20 m short of Mr.
Wester's toss. Assume Mrs. Perry releases the ball from 4 m above the ground. (a) What are the velocity components when the football hits the
ground? (b) How far away would Mrs. Perry's football land from Mr. Wester's if she throws the ball from an angle of 45 degrees above the
horizontal?
theta1=65˚
a)
xp=voxt
xp=30m
t=xp/vox
xw=50m
t=30/vo*cos65˚
y=-4m
theta2=4˚
y=voyt+0.5at2
-4=vo*sin65˚t-4.9t2
-4=30*sin65˚/cos65˚-4.9(30/vo*cos65˚)2
-4=64.3-4410/vo2
vo=8.03 m/s
vfx2=2ad+vox2
vfx2=(8.03cos65)2
vfx2=11.5 m/s
vfy2=2ad+voy2
vfy2=2(-9.8)(-4)+(8.03*sin65˚)2
vfy=60.2 m/s
b) y=vo*sin theta*t+.5at2
-4=8.03sin45t-4.9t2
t=1.65 s
xp=vo*cos theta*t
xp=8.03cos45*1.65
xp=9.37 m
x=xw-xp=50-9.37=40.6 m
9. Amit slows down his BMW as it rounds an extremely sharp horizontal curve. He slows down from 85 km/hr to 60 km/hr in the 20 s it takes to
round the bend. The radius of this curve is 100 m. Find the acceleration at the moment Amit's BMW reaches 60 km/hr.
vo=85 km/hr=23.6 m/s
at=del v/ del t=(16.7-23.6)/20 = -0.345 m/s2
vf=60 km/hr=16.7 m/s
t=20 s
r=100 m
ar=v2/r=16.72/100=2.79 m/s
a=[ar2+at2]0.5
a=[(2.79)2+(-.345)2]0.5
a=2.81 m/s2
10. Curtis is the pilot of an airplane and wants to fly due west in wind that's blowing 80 km/hr toward the north. If the speed of Curtis' plane in the
abscence of wind is 250 km/hr, (a) what direction should Curtis go and (b) what should be the plane's speed relative to the ground?
vw= 80 km/hr
a) sin theta= 80/250
vp= 250 km/hr
theta= 17.7˚
b) tan 17.7= 80/R
R=251 km/hr
11) Find the direction and magnitude of the average velocity vector of the tip of a 10.0 cm long minute hand as the time changes from 4:30 to 4:45.
R2 = 102 + 102
R2 = 200
(R2)1/2 = (200)1/2
R = 14.1cm
delta t = (15 min)(60 s / 1 min) = 900 s
ave V = d /(delta t)
ave V = (14.1/900)
ave V = 1.57 * 10-2
tan x = 10/10
x = 45˚
x = 90 + 45 = 135˚
12) The speed of a nanobot moving in a circle 4 m in radius increases at the constant rate of 6 m/s^2. At some instant, the magnitude of the total
acceleration is 10 m/s^2. At this instant, find (a) the centripetal acceleration of the nanobot and (b) its speed.
A)
a = ( (ar)2 + (at)2 )1/2
10 = ( (ar)2 + 62 )1/2
10 = ( (ar)2 + (36) )1/2
100 = (ar)2 + 36
(ar)2 = 64
ar = 8 m/s2
B)
ar = (v2) / r
8 = (v2) / 4
v2 = 32
v = 5.7 m/s
13) A Stealth submarine, having lost its way after surfacing, crosses a river with a width w = 320 m in which the current flows with a uniform speed
of 3 m/s. The steersman maintains a bearing (i.e., the direction in which his boat points) perpendicular to the river and a throttle setting to give
constant speed of 4 m/s with respect to the water. (a) What is the velocity of the boat relative to a stationary shore observer? (b) How far downstream
from the initial position is the boat when it reaches the opposite shore?
w = 320 m
vr = 3 m/s
vb = 4 m/s
(a)
32 + 42 = R2
9 + 16 = R2
R = 5 m/s
tan x = 4/3
tan x = 53.1˚
tan 53.1 = 320/L
L = 320 / tan 53.1
L = 240 m
14) A Corey in danger of drowning in a river is being carried downstream by a current that flows uniformly with a speed of 5 km/h. Corey is 0.12 km
from shore and 0.16 km upstream of a hovercraft landing when a rescue hovercraft sets out. (a) If the rescue hovercraft proceeds at its maximum
speed of 40 km/h with respect to the water, what heading relative to the shore should the hovercraft captain take? (b) What angle does the boat
velocity v make with the shore? (c) How long will it take the hovercraft to reach Corey? (This problem will require a vector diagram)
vr = 5 km/h
dy = 0.12 km
dx = 0.16 km
vb = 40 km/h
(a)
tan x = 0.12/0.16
x = 37˚
(b)
tan x = 40 sin 37 / (40 cos 37-5)
x = 42˚
(c)
delta t = (d/v)
t = 0.12/24 =0.005*60/1 = 0.3 min
15) A William Sandusky Mauldin is riding on the flatcar of a train travelling along a straight horizontal track at a constant speed of 20 m/s. William
Sandusky Mauldin throws a ball into the air along a path that he judges to make an initial angle of 30 degrees with the horizontal and to be in line
with the track. William Sandusky Mauldin's teacher, Mr. Vector oops, I mean Mr. Wester who is standing on the ground nearby, observes the ball
rise vertically. How high does the ball rise?
vox = 20 m/s
vo cos 30 = vox
vo = (10/(cos 30))
vo = 12 m/s
voy = vo sin x
voy = 40 sin 60
voy = 35 m/s
vfy2 - viy2 = 2ad
(vfy2 - viy2)/2a = d
-(352)/2(-9.8) = d
d = 63 m
Ch.3: The Laws of Motion
1. Three forces are acting on a 6 kg mass. The forces are 30 newtons at 25 degrees, 45 newtons at 112 degrees, and 65 newtons at 270 degrees. What
is the net force acting on the mass?
a. 65N at 90˚
b. 10.6 N at 314˚
c. 27.2 N at 22˚
d. 14.8 N at 314˚
1.
Vertical components:
F2cos 22+F1sin25-65=10.6N down
Horizontal components:
F2sin22-F1cos25=10.3N left
(10.32+10.62)1/2=14.8N
tan x = 10.3/10.6
x=46˚
2. A man with a mass of 90 kg is trying to accelerate a cart with a mass of 200 kg at 3 m/s 2. What force must the man use to do this?
a. 600N
b. 270N
c. 870N
d. 200N
90Kg+200Kg=290Kg
F=ma
F=290kg*3m/s2
F=870N
3. A mass of 5kg is sitting on top of a 25 degree incline that is frictionless. If the mass is released from rest, what is its final velocity after it has
moved 5 meters down the incline?
a. 6.6 N
b. 4.14 N
c. 20.7 N
d. 49 N
g sin 0= a (down the incline)
2ax=vf2-vi2
2(5m)(4.14 m/s2)=vf2
vf=6.6m/s
4. A 5 kg mass is sitting on top of a 15 kg mass and the coefficient of friction between the two masses is 0.3. There is no friction between the 15 kg
mass and the surface on which it is sitting. What is the acceleration of the masses if a force of 20 N acts on the 15 kg mass?
a. 14.7 N
b. 2.94 N
c. 44.1 N
d. 1 N
F=ma
20N=20kg*a
a=1m/s2
5. Anderson is lying on a Teflon-coated surface (no friction). John David kicks Anderson's feet applying a force of 45 N. If Anderson's mass is 85 kg,
what is his acceleration?
a. 0.49 m/s2
b. 0.31 m/s2
2
c. 0.53 m/s
d. 0.74 m/s2
Fnet=ma
a=Fnet/m = 45 N/85 kg = 0.53 m/s2
6. A block rests on a frictionless incline. Derive an equation for acceleration in terms of the gravitational acceleration and the angle of inclination.
a. a=g sin theta
b. a=g*cos theta
c. a=g tan theta
d. a=g/cos theta
Fnet x=m * g * sin theta= m * ax
Mass is same on both sides of the equation, and therefore not a factor one whit.
ax=g * sin theta
7. Two masses are connected by a string of negligible mass that is run over a frictionless pulley. Mass A rests on a flat surface. Mass B hangs by a
string over the edge of the surface in mid-air. If mass A is 20 kg, and mass B is 17 kg, and the coefficient of friction is 0.5, what is the acceleration of
the system?
a. 10.8 m/s2
b. 19.6 m/s2
2
c. 19.3 m/s
d. 26.1 m/s2
For Mass A:
Tension - Ff A = 20 ax
FN A=FW A
For Mass B:
No forces act along x-axis
FW B - Tension = 17 ay
Tension in system is same for both masses, so set both equations equal
to T and then to each other.
Ff A + 20 ax = FW B - 17 ay
ax from A is equal to ay from B.
Ff A - FW B = -3 a
0.55 * 9.8 * 20 - 9.8 * 17 = -3a
a = 19.6 m/s2
8. What is the coefficient of friction necessary to keep a system in equilibrium if a 40 kg mass is subjected to a 78.4 N force?
a. 0.20
b. 0.40
c. 0.10
d. 0.25
Fnet x = m ax
Fapplied = Ffriction
Fnet y = m ay
Fnormal = Fweight
Fapplied = mu * Fnormal
Fapplied = mu * mg
mu = Fapplied / mg = 78.4 N / 392 N = 0.20
A baseball in a pitching machine accelerates from rest to 100 mph (27.8 m/s) before it leaves the barrel, which is 1 meter long. The
ball has a mass of 100 grams. What is the force applied to the ball?
Vf² - Vi² = 2ad
27.8² = 2 (1m) × a
772.84 m²/s² / 2m = a = 386.42 m/s²
Force = m × a = 0.1 kg 386 m/s² = 38.6 N
A 4 kg sack of rice is too heavy for a toddler. She drops the sack on her mother's foot and crawls away to play with the flour. What
force is applied to the mom's foot? Is it greater than the force applied to the baseball in the previous problem?
F=ma
Weight = mg
W = (4 kg) (9.8 m/s²)
W = 39.2 N
Yes, it is slightly greater than the force applied to the baseball.
Two masses of 5 kg and 4 kg, mass1 and mass2 respectively, are connected by a cord which passes over a pulley. This setup is called
an Atwood machine. What is the acceleration of the system and in what direction is it?
m1g - m2g = (m1 + m2) a
49 N - 39.2 N = 9kg × a
a=1.09m/s²
The system is accelerating 1.09 m/s² towards mass 1.
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A 70 kg man is standing on a bathroom scale in an elevator. The elevator is moving upwards at 5 m/s². Find the man's apparent weight
Fnormal - Fweight = ma
Fnormal - 70*9.8= 1036 N
The same man is in the same elevator. This time the elevator is accelerating upwards at 2m/s² when the cord snaps and the man
experiences freefall. What is the change in his apparent weight?
When he is accelerating:
Fnormal - Fweight = ma
Fnormal – 70*9.8= 826 N
When he is in freefall:
The Fnormal does not exist (Fnormal = 0) and so he appears to be weightless.
Therefore, the change in his apparent weight is -826 N.