MakeUp Test PHY2020
4/23/2008
Pick the answer closest to your answer.
g=9.8 m/s2 100 cm=1 m 39.37 inches=1 m 1 km=1000 m
12 inches=1 foot 5280 feet=1 mile 1 kg = 1000 g
1 hour = 3600 s G=6.7 10-11 Nm2/kg2
Comparison between metric and English units of Force: 1 lb (pound) = 4.45 N
Useful formulas:
f=v EKE=1/2
t=t0
Formulas: Don’t forget d = v t !!!!!
x-x0 = v0x + (1/2)axt2
y-y0 = v0y + (1/2)ayt2
2
2
2ax (x – x0) = vx – v0x
2ay (y – y0) = vy2 – v0y2
v = v0 + at F=ma Fgravity = GmM/r2 where
a = v2/r
circumference of a circle: 2r, where r is the radius
mv2
/(1-v2/c2)1/2
1. The earth revolves around the sun once every 365 days. If the distance from
the earth to the sun is 149.6 x 106 km, what is the velocity of the earth in its
orbit around the sun? (units of km/s)
a. 30 b. 50
c. 70 d. 80 e. 90
Soln.: d=vt, v=2r/(365 days * 24 hours/day * 3600 s/hr) = 29.8 km/s
2. (Cathy – this problem has to come after #1. The two problems are linked.)
What is the gravity at the earth’s orbit due to the sun’s mass? Express in units
of m/s2.
a. 6 x 10-3
b. 5.9 x 10-9 c. 5.9
d. 9.8
e. none of these
2
Soln.: a=v /r = (29.8 x 103 m/s)2/1.496 1011 m = 0.006 m/s2
3. As shown in the figure above, a force of 150 N pulls the box shown to the
right 60 o above the horizontal and a force of 100 N pulls the box to the right
50 o below the horizontal. In units of N, what is the net force in the vertical
direction due to the sum of the two forces?
a. 53
b. 139
c. –1.6
d. 50
e. 250
Soln.: Fy = 150 N * sin60 – 100 N * sin50 = 53.3 N
4. A 4 kg object moving at 3 m/s runs into a stationary object, mass=2 kg. The
collision is elastic. What is the velocity of the target (the 2 kg mass) after the
collision? Units of m/s
a. 4
b. 2 c. 3 d. 1 e. 5
Soln.: conserve p and E. pi=12 kgm/s, Ei=18 J (=1/2 mv2)
pf=pi = 4v1f + 2v2f Ei=Ef=2v1f2 + 1v2f2. Since we want v2f, eliminate v1f using the
momentum eqn.: v1f=(12 kgm/s-2v2f)/4 Put this in the energy equation:
18 J = 2((12 kgm/s-2v2f)/4)2 + v2f2 Collect terms:
v2f2(1+0.5) + v2f(-6) -18+18 = 0, so v2f=4 m/s
5. A C-note is playing in an open-at-one-end organ pipe, with f=264 Hz. Vsound=343
m/s. If the vibration in the pipe is the ‘fundamental harmonic’ (i .e. has the longest
wavelength possible for the open end organ pipe (node at the closed end, anti-node
at the open end)) how long is the organ pipe, in units of meters?
a. 0.32
`b. 1.3
c. 5.2
d. 2.6
e. 0.65
Soln.: λ=vsound/f = 343 m/s / 264 Hz = 1.300, we know from the practice problems
and class that in the length of the pipe for this standing wave L=λ/4, so L=0.32 m
6. Use Coulomb’s law (F=kq1q2/d2 , where k=9x109 Nm2/C2) to calculate the force in
Newtons between two charges, q1 = 0.1 C and q2 = 0.2 C, that are a distance ‘d’ of 10
m apart. (units of 105 N)
a. 18
b. 1.8
c. 180
d. 1800
e. 18000
Soln.: F= 9x109 Nm2/C2 * 0.1 C * 0.2 C/(10m)2 = 18 x 105 N