Mcat Phys 1 Topical

Front 1

The force of air resistance due to friction is proportional to the product of velocity and the object's surface area. If the force of friction were only dependent on surface area, how will this affect the amount of time needed for the object to reach its terminal velocity?

Back 1

Since Ff no longer depends on velocity, the object will never reach terminal velocity.

Front 2

If an object is shot out of a cannon at angle 1 and angle 2, and angle 2 > angle 1, how will this affect the position of the object's peak in its trajectory?

Back 2

Greater theta means Vy of angle 2 will be greater than Vy of angle 1 (greater theta --> greater sin value).

But for Vx, greater theta means lower cos value.

Thus the peak will be closer and higher.

Front 3

If a clown is shot out of a cannon at a velocity of V and angle of theta, and there were a hoop at the peak of his/her trajectory, how long would the rope that holds the hoop in place have to be to keep the clown from hitting the ceiling?

Back 3

At peak of trajectory:
Vy = 0
Vx = vcostheta
KE at time of impact = 1/2m(vcostheta)^2, which is converted into PE:
1/2m(vcostheta)^2 = mgh
Solve for h

Front 4

If a boat travels at 20 m/s and must traverse a river with stream flow of 10 m/s, at what angle must the boat travel to reach the opposite side of the shore?

Back 4

Vector problem where 10 is southward, and 20 is northeastward (hypotenuse).
costheta=10/20=1/2
arccos(1/2)=theta=60

Front 5

If two blocks of m1 and m2, where m1>m2, are released from rest on an inclined plane where m1 is released at theta1, and m2 is released at theta 2 (theta 1>theta 2), and mew sub k is the same for both , which will reach the bottom of the plane first and why?

Back 5

M1 because it has a greater theta.

Front 6

For a roller coaster traveling over a looped track how would the force of friction affect the normal force? Provide a proof.

Back 6

Fn = Fc
N - mgcostheta = mv^2/r
N = mv^2/r + mgcostheta=
m(v^2/r + gcostheta)

Taking Ff into account, the velocity would decrease, and thus Fn would decrease as well.

Front 7

For a roller coaster traveling over a looped track, what would be the kinetic energy of the cart at the peak of the loop?

Back 7

Fn = mg = Fc = mv^2/r
v = (rg)^1/2
KE = 1/2mv^2 = mrg/2

Front 8

If no approximation were made with regard to distance, how would the graph of acceleration of a falling object vs height appear?

Back 8

F = mg = GMe*m/(Re + h)^2
g is proportional to 1/(Re+h)^2
g vs h appears as inverse square (straight horizontal line at first, then exponential increase)

Front 9

According to Kepler's third law, the square of the period of any satellite is proportional to the cube of the semimajor axis of orbit (R). If the radius of orbit is increased by a factor of 4, how will the period of orbit be affected?

Back 9

T^2 is proportional to R^3
T^2/R^3 = x/(4R)^3 = x64R
thus:
(64T^2)^1/2=8T
Period will increase by a factor of 8

Front 10

If the amount of energy required to move a mass m from the surface of the Earth to a distance R from the Earth's center = GmMe(1/Re-1/R), what is the minimal velocity required to escape the Earth's gravitational field?

Back 10

Mass must be infinite distance from the Earth's center:
E = GmMe(1/Re-1/infiniti)
(1/infiniti = 0)
= GmMe/Re } KE req'd to escape Fg

Thus KE = 1/2mv^2 = GmMe/Re
v=(2GmMe/Re)^1/2

Front 11

For a massless rod of length 4m with m1 hanging from the left end and m2 hanging from the right, where m1 = 2m2, where must the fulcrum be placed for the system to remain in equilibrium?

Back 11

Torque=Frsintheta
Torque1 must = Torque2
sintheta=1

thus:
2mgx=mg(4-x)
2x=4-x
x=4/3=1.33m from left

Front 12

What is the work done on the Earth in one complete orbit by the gravitational force of the sun? Assume orbit to be circular. Provide 3 valid arguments.

Back 12

1) PE dependent only on position, planet as same position at end as it was at beginning, so change in U = 0. Since total mechanical energy is conserved, deltaK=0.

2) Gravitational force is conservative, thus work done is path-independent, so starting and ending at same place means W = 0.

3) For a circular orbig:
W=Fdcostheta, where theta = 90 and cos90 =0

Front 13

If U = -GmM/r and K=-U/2, what is the total energy for a circular orbit?

Back 13

E = K + U = -U/2 + U
=-(-GmM/2r) + (-GmM/r)
=-GmM/2r

If K=-U/2, then U = -2K
thus:
E = K + U = K-2K = -K = -1/2mv^2

Front 14

What is the ratio of the acceleration of a black hole to its companion star?

Back 14

F = ma
a = F/m
F= GmM/r^2

Newton's Third states that the magnitude of the Fstar must equal the magnitute of the Fblackhole

Thus: Fbh=GmM/r^2
a(bh)=GmM/r^2M = GM/r^2

a(star)=GmM/r^2m = GM/r^2

aBH/astar=(Gm/r^2)/(GM/r^2)=m/M

Front 15

If a block is dropped from a height h on a looped frictionless track from rest and hits a ball of equal mass (also at rest) at bottom of track, in terms of h, determine the max height reached by the block and ball after the collision. The collision is inelastic.

Back 15

Since collision is inelastic, kinetic energy is NOT conserved, but momentum is.

Inital E converted to KE from PE:
mgh=1/2 mv^2
v=(2gh)^1/2

pi = mv = m(2gh)^1/2
pi=pf (momentum conserved)
m(2gh)^1/2=2mvf
vf=(2gh)^1/2/2

Immediately after collision, PE = 0, thus:
KE = 1/2mv^2=1/2(2m)((2gh)^1/2)/2)^2 = 2mgh/4 = mgh/2} will convert completely to PE at peak of travel, so:
mgh/2 = 2mgh'
h' = h/4

Front 16

Two balls undergo head-on collision. M1 has twice the mass of M2. Ball 2 is at rest. Post collision, ball 1 travels with a speed of v/3. What type of collision has occurred?

Back 16

Momentum is always conserved, so:
pi=pf
2mv=2mv/3 + mv'
mv'=2mv-2mv/3
mv'=4mv/3

Now check to see if KE is conserved:
KEi=1/2(2m)v^2=mv^2
KEf=1/2(2m)(v/3)^2 + 1/2(m)(4v/3)^2
= mv^2/9 + 16mv^2/18
=18mv^2/18=mv^2=KEi

KE is conserved, thus collision is elastic

Front 17

If an electron and a positron lose 260MeV of energy per revolution, and the frequency of revolution is 10,000 Hz, how much power is required to keep the total energy constant at 180 GeV?

Back 17

Just need to make sure the total energy is constant, thus Work lost=Work req'd

If 260 meV lost per revolution per particle and f is 10,000 Hz, in one second, 10,000 revolns are made and total energy lost in one second by ONE particle =
2.6 x 10^2 x 10^4 = 2.6 x 10^6 MeV

But since there are two particles, multiply by 2:
2 x 2.6 x 10^6 = 5.2 x 10^6 MeV/s = power req'd