TestII PHY2020 Formula sheet
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 1 J = 1N*m 1 Watt=1 J/s
Comparison between metric and English units of Force: 1 lb (pound) = 4.45 N
Useful formulas:
Linear: Espring = ½ kx
Egrav. potential = mgh (h is the height)
2
EKinetic Energy = ½ mv F=ma
p(the momentum) = mv (remember, p and v are vectors, so p has both
size and – if it’s important – direction)
If there are no external forces (→ momentum is conserved) and the
collision is elastic (→ energy is conserved) if mass m1 hits a stationary
mass m2 then
v1final = v1initial (m1-m2)/( m1+m2)
and v2final = v1initial (2m1)/( m1+m2)
2
Rotational: s=rθ vlinear=r alinear=r Remember, use radians for θ,
, and  in these and all formulas below.
1 rotation = 360 o (<- don’t use!) = 2 radians
Erot. Kin. Energy = ½ I2 L(angular momentum)=I
I(moment of Inertia) is for a point mass a distance ‘r’ from the axis of
rotation=mr2 If you have more than one mass, the moment of intertia is
the sum of all the masses times their respective distances from the
common axis of rotation squared: Sum of miri2. (If you need a moment
of inertia for a special problem – as long as that’s not the question – you
will be given it.)
torque = I
also, torque = F times ‘lever arm’,
where ‘lever arm’ is the distance of closest approach to the axis of
rotation of the Force vector extended forwards and backwards.
Density is represented by the Greek letter ρ. ρ=M/V,
where M is the mass of the object and V is the volume the object takes
up.
circumference of a circle: 2r, where r is the radius
diameter = 2 * radius
Ideal gas law PV=nRT, where n=number of moles of gas and R=8.3 J/mole
Temperature in Fahrenheit/Centigrade/Kelvin. Water freezes at 32 oF, 0 oC, and 273
K, water boils at 212 oF, 100 oC, and 373 K
Test 2 Chapters 8-13
Name:_____________________ UF ID #:__________________
3 sig figs on all answers – I have given the units for each answer line
(Partial credit: minus ½ credit for wrong placement of decimal point)
1. A 5 kg object moving at +2.0 m/s runs head on into an oncoming moving object,
-
mass=10 kg, v= 1.0 m/s (see diagram). There are no external forces.
a. (2 points) If the objects stick together (the collision is inelastic), what is the final (i.
e. after the collision) total kinetic energy?
___________________________kg*(m/s)2
No external forces means the momentum is conserved. Pinitial= 5 kg*(+2 m/s) + 10
kg*(- 1 m/s) = 0 = Pfinal = 15 kg * vfinal. Since the final momentum must equal the
initial momentum, which is 0, this means that the final velocity of the two stucktogether objects (total mass=15 kg) = 0.
b. (2 points) Now consider the collision to be elastic (the objects do not stick
together); there are no external forces. What is the final (i. e. after the collision)
total kinetic energy?
__________________________kg*(m/s)2
Elastic means that the kinetic energy is conserved (=doesn’t change.)
KEinitial=KEfinal. KEinitial = ½ 5 kg *(2 m/s)2 + ½ 10 kg*(1 m/s)2 = 15 kg (m/s)2
So KEfinal = 15 kg (m/s)2
2. How many rotations per minute does a 0.5 m radius wheel of a vehicle have to
turn in order for the vehicle to move 26.8 m/s (60 miles per hour)?
_________________________rpm
v=r, with v=26.8 m/s and r=0.5 m, so =26.8 m/s/0.5 m = 53.6 radians/s. 2
radians=1 rev, so =(53.6 radian/s )/2 radians/revolution = 8.531 rev/s = 512 rpm.
3. 2 masses, each with a mass of 1 kg, are at opposite ends of a 1 meter long
massless connecting rod (see diagram). Consider that the masses are small in
comparison to the length of the rod, i. e. ignore their size/dimension. This rod and
its attached masses rotates around an axis (the dotted line in the diagram, marked
with the rotation arrow at the top) through the center of the rod. The center of the
rod is marked by an ‘X’.
a. What is the moment of inertia, I, of this object around the center of the rod?
_____________________kg*m2
I= m1r12 + m2r22 = (1 kg*(1/2 m)2 + (1 kg*(1/2 m)2 = ½ kg*m2
b. If the masses in the diagram are now put on the opposing ends of a 2 meter long
massless connecting rod, (same diagram, just change the distance between the
masses to be 2 m), what is the new moment of inertia, I, for rotation around the
center?
____________________ kg*m2
I= m1r12 + m2r22 = (1 kg*(1 m)2 + (1 kg*(1 m)2 = 2 kg*m2
4. Consider the picture below. Block #1 weighs 200 N, #2 weighs 300 N, #3 weighs
100 N; block #1 is 2 m from the fulcrum/pivot point, #2 is 1 m away (both on the left
as shown), block #3 is 1 m to the right of the fulcrum, and block #4 is 2 m to the
right of the fulcrum. If all the blocks are to be in equilibrium (no tipping or
rotating), what is the weight of block #4, in N?
#1
#2
#3
#4
^
In equilibrium means no net torques.
sum of torque = 0 = 200 N * 2 m + 300 N * 1 m – 100 N * 1 m - #4 * 2 m => #4 = 300
N
__________________ N
5. The density of gold is 19.3 g/cm3. Gold currently costs about $50/g. What
volume does $1,000,000 (1 million dollars) worth of gold occupy?
$1 x 106/($50/g) means that the gold has a mass of 20,000 g. The volume occupied by
this mass (since the density, , =M/V we calculate V=M/) is 20,000 g/(19.3
g/cm3)=1036 cm3. This is about 1 liter (=1000 cm3)
_______________________cm3
6. A piece of wood of some density is floating half submerged (see picture) in fresh
water, where the density of the water is 1 g/cm3. (The level of the water is
marked by dashed lines. The total volume of the block is 100 cm3 and, to repeat,
half of the block (i. e. a volume of 50 cm3) is submerged. What is the density of
the wood?
___________________________g/cm3
From Archimedes’ principle, the buoyancy force = the weight of the volume of
water displaced, or 50 cm3 * density of water * g = 50 cm3 * 1 g/cm3 * 9.8 m/s2 =
490 g m/s2 = 0.49 kg m/s2 = 0.49 N. So the whole block is in equilibrium, i. e. the
buoyancy force of 0.49 N balances the weight of the block. Since the volume of
the block is 100 cm3, its mass is M=V, where  is the sought-for wood density.
So =M of block/100 cm3. The 0.49 N force up matches the weight of the block
down, of Mg, so M(block)=0.49 N/g = 50 g, so  = 50 g/100 cm3 = 0.5 g/cm3
7. What temperature in oC is -400 oF?
_____________________oC
-400 oF is 400+32 oF =432 oF below freezing, which is -432/1.8 oC or -240 oC. Or,
using the formula you’re used to, temperature, T, in oC = (T in oF – 32)/1.8=
(-400 oF – 32)/1.8 = -240 oC
8. An ideal gas is warmed up from 100 oF to 190 oF, in a container where the
pressure is kept constant, and the gas either expands or contracts to change the
volume occupied by the gas at different temperatures. What is the new volume
V2 (V2 is the volume that the warmer gas occupies) in terms of V1?
______________________V1
PV1=nRT1 ; PV2=nRT2 so P=nRT1/V1 = nRT2/V2 so, canceling the
common factor of nR, we have T1/V1 = T2/V2. Now we need T in units of Kelvin:
First convert 100 oF to (100-32)/1.8 = 37.8 oC and 190 oF to (190-32)/1.8 = 87.8 oC.
Convert from Celsius to Kelvin: 37.8 oC=37.8 +273.15=310.95 K and 87.8
oC=87.8+273.15=360.95 K So V =(T /T )V = (360.95/310.95)V = 1.16 V
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