1oCi.f)
ic’Is
;;:
7r
21tr)
)(
+
‘
=
_.O
lO
/
9
J
3
rZ
-
V
5
1.
2c4oo
(Q_
)f,O&t
gO’
sot.
cfr.b9
a-i
-
QJ3. 8(1
.
‘b4.1
‘A
/
23
Stone:
k3.OOkg ball drops from rest at a height of 40.Om. If the impact between
the ball and the floor lasts O.800s, what is the net force exertea by the ball
on the ground?
Answer: 105N
Woik:
Momentum
P = rn v) Use kinematics to find change in time
O.5atlL2
=
xfxi tvt+
40= 4.9t”2
t 2.86s
v= a(t) = 9.8m!s
2(2 66$:; = 2SrWs
6
Monentum = rn(j 3’28m!s) = 84kgmls
se = = F(t,
1
lmpu
F = J!t = (84kgmisZ3.Ss = 105N
Abhinav Raghunathan
Period 8 Physics HNS
Momentum
A frictionless spring is compressed 0.2 m while a box is pushed against it. The spring constant is
300 N/rn. The box has a mass of 5.0g. What is the momentum of the box after it has been
released?
k= 300 N/rn
m = 5.Og = .0050 kg
x = 0.2 meters
Conservation of Energy:
PEspr,ng = KEbOX
2
i
ft
X
/2
P E spring i,’
KEbOX / m
2
2 = (0,5)(0.0050 kg)(v)
(0.5)(300 N/m)(0.2m)
v = 49 m/s
*
—
*
—
*
*
Definition of Momentum:
p = my
p (0.0050 kg) (49 m/s)
p = 0,25 kgm/s
Alex L, Eric Z, Francis S, Madhav P
1. Charge problem
Positively charged sphere B is placed between two neutral spheres A and C. We cut connection
of A and C with ground. If we putA closer to the first electroscope and touch C to the sphere of
second electroscope, find the type of charge electroscopes have.
A and C are negatively charged by induction. Thus, leaves of both electroscopes are negatively
charged.
2. Coulombs law
Suppose that two point charges, each with a charge of +1.00 Coulomb are separated by a
distance of 1 .00 meter. Determine the magnitude of the electrical force of repulsion between
them.
Felect= k’Ql •Q2/d2
Felect = (9.0 x 109 N•m2/C2)
Felect
•
(1.00 C)
(1.00 C) / (1.00 m)”2
9.0 x 10A9 N
=
3. Electric field
At what distance from a negative charge of 5.536 nC would the electric field strength be 1.90 x
105 N/C?
e
=
kq/rA2
2
q
qOKlOLrn(
ox1O
553
d=
-
r=1.6 cm
c)
4k
Michelle Giang, Neha Arsha, Sanjana Arsha, Tejashri Nandan
Kinematics
1. You are designing an airport for small private aircraft. A Cessna 182 has a stall
speed of about 27.8 mIs and can accelerate at a rate of 2 mIs
2 How long must a
runway be in order to accommodate this plane?
.
Solution:
Given: v
= 0 mIs
0
a = +2m 152
= 27.8m Is at liftoff
v + 2a(x-x.
2
0
)
3
mls
1
)
(2*2m1s
) =193.2 m.
(v v
)I(2a)= x-x.
2
0
3 = (27.8 2
It needs to be 193.2 m long to accommodate this plane.
=
-
2. A baseball is popped straight up into the air and has a hang-time of 6.25 s.
Determine the height to which the ball rises before it reaches its peak. (Hint: the
time to rise to the peak is one-half the total hang-time.)
a=-9.8m1s
2
= 0 mIs
t=3.13s
v + at
1
)*(3.13 s)
2
o mIs = v1 + (-9.8 mIs
o m/s = v1 30.7 mIs
= 30.7 m/s (30.674 mIs)
= v + 2*a*d
)*(d)
2
(0 mIs)
2 = (30.7 m/s)
2 + 2*(9.8 mIs
/s = (940 2
m
02
1s + (-19.6 m/s.)*d
m
)
)*d
2
-940 2
1s = (-19.6 m/s
m
=
-
(-940 2
1
m
)
I(-19.6
s m/s
)
2
d = 48.0 m
=
d
3. A motorcyclist traveling 33.3 mis (75 mph) passes a constable in a 55 mph
zone. The constable immediately begins pursuit at a constant acceleration of 2.8
. How much time will it take for the constable to overtake the motorcyclist
2
mis
(who does not know that he is being pursued). How fast will the constable be
traveling?
Solution: 67.2 mis
3=v
x-x,
t + (%)at
0
2 and v,= v
0 + at
Motorcyclist: x-= (33.3 mis) (t)+(%) (O)(t
)
2
Constable: x-= (O)(t) + (%) 2
(2.8mi
)
)
(t
s
Set the motorcyclist and constable equations together and solve for t.
Plug in t=23.8 s for the velocity equation —‘ 0 + (2.8 mis
2 )(24 s)= 67.2 mis
Celine, Kimberly, Sahana, falia
Physics Final Project: Newton’s Laws Answers
Answer to Problem 1:
F(tanSion)s(theffi)
a
—
a
IF(tension)cos(thsffi)
I
F(gre*)
4.fter drawrng the free-body diagram, find the measure of the anglc between F(tension) and
F (tcnson)eos(theta) Notice that f(tension)eos(theta) and F(tension)sin(theta) are in different
places then where they usually are Using 60 degrees a theta use 50114 11- fOA o rcalv e
v F v thsa two have switched places
Vnee tfe elephant ( F ci is at the center of this free-body diagram) s a equilibrium, t c
xterna forces a ting or the elephant must have a iet fore of zero Mark d nvn the force pair
which mst qual on a othcr in rder o mak the net force or the e ephant zero rhesc are
I (i Id
I tcr sion) in(theta), and F tensionkos(thetaj F (gras ty) Bec usc we have Looked at
both F te is on)sm(theta) and F(tension)os(theta) wc have taken into account botf tht
eompon°r ts of F (tension and thus do not have to match F (tension) with another force as we d
w tF t e other forces
F id (gray ty)
F( ravity) mg
(;ravityi = (5000 kfl9 R m/cThi
F(gravity) 49000 N
ual o F tens on)eo theta
g rv y)
gravt )is Findi(t nsor)
Fheta here is 60 degrees We have just found what
F(gravity) F(tension)cos(theta)
49000 = F(tension)eos(60 degrees)
F(tension) = 98000 N
F(tension)sin(theta) is equal to F(field). which is what we are trying to find. Theta, here, is 60
degrees, and we just found what F(tension) is. Find F(fieid).
F(tension)sin(theta) = F(field)
(98000 N)(sin 60 degrees) F(field)
F(field 84870 N
The answer is approximately 84870 N.
Ansiier to Problem 2:
F(normal)
52N
F
I
24N
F(griPj)
After drawing the free body diagram. think about the problem I he agon is moing forward
not hack. and not up and down. This means that F(gravity) = F(normal). and that the net force on
the wagon is acting to push it forward. Remember that there is still a resistance force on the
wagon. though. that must be taken into account!
Find the net force acting on the wagon by finding the net horizontal force (since the vertical
forces amount to zero in total, since the wagon is not moving up or down).
52 N: forward —24 N back = 28 N forward
2 and that w = mg (we
To find the total mass of the wagon. remember that gravity is 9.8 mis
already know that the wagon has a weight of 304 N, because we were told so in the problem).
mg
)
2
304 N = (m)(9.8 mis
m-3l.0kg
Finding the total acceleration of the wagon is simple once we know the net force acting upon it
and its total mass. We just have to remember that in physics, when we use equations to find the
total of part of a problem. all the variables in the equation have to be of the total of the problem.
For instance, if we want to find the acceleration of the wagon using the formula F = ma, we can’t
(31 kga), because 52 N is not the net force acting on the wagon. We must
write out 52 N
separate the problem into separate systems total, just the resistance forcc, just the normal force,
—
in order to get proper, correct numbers,
F = ma
28 N (31 0 kgka)
2 forward
a 0,90ms
etc
—
[he ar swer is 28 N forward, 31.0 kg and 0 90 m
52
forward.
Answer to Problem 3:
[he Nexuon’s law pairs are Sahana and the elephant (the force exerted by Sahana on the
elephant and thc elephant on Sahana), the elephant and the ball (the force exerted by the elephant
on the ball and the ball on the elephant), and the ball and the sand (the force exerted by the ball
on the sand and the sand on the ball) (or, the ball and the earth).