#$&*
course Phy 241
"Submitting Assignment" "Phy 241" "02-68-040" "02-68-040" "Meghan"
"Nash" "Test 2 practice" "mmn276@email.vccs.edu" "December 13 around 3:30pm. I
really tried on this test and attempted them all. Maybe you can give me some feedback on
how I'm doing. Thanks so much." "Test Problems:
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Problem Number 1
A ball has a horizontal range of 5 meters when it is projected horizontally from an
altitude of 25 meters. What will be its range if it is projected at an angle of 8 degrees
above horizontal with the same initial speed? Approximately how much does its horizontal
range change per degree from horizontal?
****
Horizontal range: 5m
Altitude: 25m
8 degrees
V_x=5m @ 25m
Range: v0*sqrt(2h/g)
=5*sqrt(2(25)/9.807)
=11.29m
Then h=25m and angle=8 degrees:
R=(v0^2sin(2*theta))/g
=(5^2*sin(2(8)))/9.807
=-0.734 horizontal range/degree
range will be more than 11.29 m, probably around 12 m. Maybe an extra meter.
Dividing the extra meter by 8 deg change in angle you would get around .125 m / degree.
v0_x = v0 sin(theta)
v0_y = v0 cos(theta)
Vertical: know v0_y, a, `dy; 4th eqn -> vf = sqrt( v0_y - 2 a `dy) = sqrt( v0 sin(theta)
- 2 g `dy)
vAve = (0 + sqrt( v0 sin(theta) - 2 g `dy)) / 2
`dt = 1/2 sqrt( v0 sin(theta) - 2 g `dy)
horizontal: v0_x = v0 cos(theta) so `dx = v0_x * `dt = v0 cos(theta) * 1/2 sqrt( v0
sin(theta) - 2 g `dy)
etc.
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Problem Number 2
Given that the gravitational potential energy of a mass m at distance r from the center of
a planet of mass M is PE(r) = G M m / r, find the derivative dPE / dr.
• What is the gravitational force on mass m at distance r from the center of the
planet?
• How do your two results compare and why?
****
‘dPE_grav=weight*’dy
‘dPE=Favg*’dr
To get accurate change in potential integral, integrate F wrt distance from center:
‘dPE=Int (GMm/r) dr from r1 to r2.
That gives you G M m * ln(r1 / r2)
Integrate F `ds to get energy. Position is r, so `ds = `dr.
Work against gravity on typical r interval is F(r_hat) `dr; sum ( F(r_hat_i) `dr_i
approaches integral( F dr ) = integral (G M m / r^2 dr)
Work from infinity to R is integral ( G M m / r^2 dr, infinity, R)= G M m / R. This is
work done by gravity (force is in direction of motion from infinity to R), so PE relative
to infinity is - G M m / R.
Work against gravity from r1 to r2 is integral ( G M m / r^2 dr, r1, r2) = -G M m / r2 + G
M m / r1.
`dPE = -(work by gravity) = G M m / r1 - G M m / r2.
‘dPE=(GMm/r1)-(GMm/r2)
Grav PE=(-GM/r)
So, if I solve for dPE/dr:
dU=d(-GMm/r)
dU=GMm/r^2 (gravitational force)
dU/dr=d(-GMm/r)/dr - G M m / r^2
dU=GMm/r^2 * dr
They’re the same!
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Problem Number 3
A uniform sphere is growing in such a way that its radius increases by .7 cm every minute.
The mass density of the sphere is 5 grams per cm^3. If I(r) is the moment of inertia of
the disk when its radius is r, then what are dI / dr and dI / dt at the instant the radius
is 35 cm?
****
Moment of Inertia of a sphere: 2/5MR^2
Area of a sphere: 4*pi*r^2
M=mass density*area
=(5g/cm^3)*(4*pi*r^2)
=20*pi*r^2
I=2/5MR^2
=2/5*(20*pi*r^2)*r^2
=2/5*(20pi)*r^4
dI/dr=2/5*(20pi g/cm^2)*(4r^3)
=2/5*(20pi*4r^3)*dr/dt
=2/5*(20pi*4(35^3))*(0.7)
=2/5*(7.543*10^6)
=3017200 or 3.0172*10^6gcm
dI/dt=dI/dr*dr/dt
dI/dt=(3.0172*10^6gcm)*(0.7cm/min)
=2112040gcm^2/min
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Problem Number 4
Explain why the work required to stretch a spring or other elastic object with a linear
restoring force, of form F = - kx, from its equilibrium position to displacement x is `dW =
.5 k x^2, and why we hence say that this is the elastic potential energy of the object in
this position.
****
Elastic PE is PE stored as a result of changing or bending of an elastic object, such as a
spring. It is equal to the work done to stretch the spring, which depends on the constant
“k” and the stretched distance. Hooke’s Law says the force required to stretch the spring
will be directly proportional to the amount of stretch. Since the linear restoring force is
F=-kx, then the work done to stretch the spring at “x” distance is ‘dW=1/2kx^2. You can
also integrate (kx dx) from 0 to x to see this or you can see this by the area underneath
the FORCE curve on a graph kx vs. x.
so write down 'int( k x dx, 0, x) = 1/2 k x^2.
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Problem Number 5
If a cart coasts distances of 1.8897, 2.173155, 1.775249 and 3.366057 cm, starting from
rest each time, and requires respective times of 9.5 sec, 10 sec, 7.25 sec and 9.75 sec, is
the hypothesis that acceleration is independent of position or velocity supported or not?
****
I would need to find accelerations for each:
I know: v0, ‘dt, ‘ds
Use the equation: ‘ds=v0’dt+1/2a’dt^2
1.8897=1/2a(9.5^2)
a=0.042
2.173155=1/2a(10^2)
a=0.043
1.775249=1/2a(7.25^2)
a=0.068
3.366057=1/2a(9.75^2)
a=0.070
To an extent, all the accelerations are about the same, within reasonable uncertainty and
therefore the hypothesis that acceleration is independent of position or velocity IS
supported.
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Problem Number 6
A disk of negligible mass and radius 16 cm is constrained to rotate on a frictionless axis
about its center. The disk remains in a vertical plane with its axis horizontal. On the
disk are mounted masses of 15 grams at a distance of 12.8 cm from the center, 8 grams data
distance of 7.04 cm from the center and 38 grams at a distance of 3.2 cm from the center. A
force is applied at the rim of the disk by a mass of 4.071 grams attached to by a light
string around the rim.
• As the mass descends 111 cm, with the disk originally at rest, by how much does the
potential energy of the system change?
• What therefore will be the angular velocity attained by the disk and the velocity
attained by the descending mass?
****
Moment of Inertia of a disk: 1/2MR^2
actually could use 1/2 M R^2 + m R^2, where m is suspended mass
m R^2 = 4.07 g * (16 cm)^2 = 1012 g cm^2.
I =
Finding the moment of !!!!
Inertia of each of the 3 masses:
½(15)(12.8^2)=1228.8gcm^2
15 grams is at 12.8 cm from center. Its moment of inertia is m r^2, not 1/2 m r^2 just
because it happens to be mounted on a disk.
½(8)(7.04^2)=198.25gcm^2
½(38)(3.2^2)=194.56gcm^2
Total I is sum of all, about 6800 g cm^2.
‘dPE=mgh
=(4.071g)*(980cm/s^2)*(16cm)
=6.3833*10^4J
g cm^2/s^2 is a unit of energy but it's not a Joule, it's an erg.
J = N * m = kg m^2 / s^2.
Angular velocity: dtheta/dt
Velocity: r*angular velocity
KE = 1/2 I omega^2
`dPE = -6.3833 * 10^4 g cm^2 / s^2
Started from rest. Negligible friction -> `dW_NC = 0.
So
`dKE + `dPE = 0
`dKE = - `dPE = 6.38 * 10^4 g cm^2 / s^2
KE_0 = 0 so KE_f = 6.38 * 10^4 g cm^2 / s^2.
omega = sqrt( 2 * KE / I ) = sqrt(2 * KE / (1/2 M R^2 + m R^2) ) = sqrt( 2 * 63800 g cm^2
/s^2) / (6800 g cm^2) ) = 4 sqrt(1/s^2) = 4 s^-2 or 4 rad / sec.
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Problem Number 7
Coasting from rest down a certain hill, whose slope is variable, I reach a speed of 8.485
m/s at the bottom. If I coast from rest down the first half of the hill I reach a speed of
6 m/s. Ignoring the effects of air resistance and friction:
• How fast would I therefore be going if I coasted down the second half of the hill?
• How high would I have to climb from the bottom of the hill to reach the halfway
point?
****
I have to use energy conservation on this problem.
KNOW: (v0_A=0), (vf_B=6m/s), (v0_B=0), (vf_C=8.485m/s)
(1/2mvf^2-0)+mg(hf-h0)=0
h0-hf=1/2mvf^2/mg
(1/2mvf^2-0)+mg*’dh=0
‘dh=(-1/2mvf^2)/mg [mass cancels out]
‘dh=-1/2vf^2/g [‘dh will be negative going DOWN the hill]
=(-1/2(8.485^2))/(9.807m/s^2)
=-3.671meters
[To check ourselves: vf=sqrt(2gh)
=sqrt(2(9.807)(-3.671))
=8.485m/s]
You got the height.
Do the same for the trip down half the hill.
I think you get about 1.8 m, but it won't be exactly half the distance.
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Problem Number 8
A gun fires a bullet of mass 49 grams out of a barrel of length 40 cm. The bullet exits
the gun with a velocity of 287 m/s. The 68 kg individual firing the gun, who was standing
at rest of a frictionless skateboard before firing, rolls backward at .02058 m/s
immediately after firing.
• What is the mass of the bullet?
• Assuming that the bullet accelerated uniformly while in the barrel, how long did
the bullet spend in being accelerated down the barrel?
• From the Impulse-Momentum Theorem determine the average force exerted on the bullet
while in the barrel.
• What average force was felt by the individual firing the gun?
****
m*v=(-m*v) of person [equal and opposite]
m*287m/s=(-68kg*0.02058m/s)
mass of bullet=-0.005kg
momentum=m*v
impulse=F*’dt
bullet’s momentum: (-0.005kg)(287m/s)=(-1.435kg*m/s)
person’s momentum: (-68kg)(0.02058m/s)=(-1.400kg*m/s)
[‘dp=impulse=Favg’dt=m*’dv]
[Bullet: 287^2=2a(40), a=1029.6m/s^2, ‘dt=0.27875s]
[Bullet: *287 m/s) ^2=2a(40 cm). Deal with units. You don't get a=1029.6m/s^2,
‘dt=0.27875s]
Favg of bullet=(momentum/’dt) or (m*(‘dv/’dt))
=(-1.435kg*m/s)/(0.27875s)
=(-5.148N)
Favg of person=(momentum/’dt) or (m*(‘dv/’dt))
=(-1.400kg*m/s)/(‘dt)
Not sure if I use the ‘dt of the bullet for this or if I somehow need to find it using
another method?!
#$&*"
Here's a posted copy for you.