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Physics GOIUYS > Advanced Mechanics > Flashcards

Advanced Mechanics Flashcards

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1
Q

In Projectile Motion, horizontal motion (u cosθ) is always (unless stated otherwise);

A) Constant.
B) Changing.

A

A) Constant.

because there is no reason to change motion

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2
Q

In Projectile Motion, vertical motion (u sinθ) is always (unless stated otherwise);

A) Constant.
B) Changing.

A

B) Changing.

because gravity is always acting on it

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3
Q

In Projectile Motion, at two same points of height (one going up and one going down) velocity is (unless stated otherwise);

A) The same.
B) Different.

A

A) The same.

because it needs to reach a certain velocity to beat gravity, but after beating it, it falls down with whatever gravity it had to beat

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4
Q

In Projectile Motion, at the maximum height of the projectile, the vertical component of velocity is;

A) 0ms^-1.
B) constant.

A

A) 0ms^-1.

the upwards velocity will finally equal to the downwards working gravity, causing a net velocity of zero

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5
Q

In Circular Motion, the velocity tangential to the circle’s circumference at a given moment is NOT;

A) angular velocity.
B) orbital velocity.
C) linear velocity.
D) rotational velocity.

A

A) angular velocity and D) rotational velocity

angular/rotational velocity has units rads^-1 and is denoted with ω

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6
Q

In Circular Motion, the period if defined as;

A) the number of revolutions it can make in a second.
B) the amount of time it takes to make one revolutions (in seconds).

A

B) the amount of time it takes to make one revolutions (in seconds)

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7
Q

In Circular Motion, for an object spinning on a string vertically (like in the diagram) the force of tension is at the TOP and BOTTOM respectively (say tension force = Ft, centripetal force = Fc, gravity force = mg) is;

Diagram:
https://www.scienceabc.com/wp-content/uploads/ext-www.scienceabc.com/wp-content/uploads/2018/10/stone-thread.jpg-.jpg

A) Ft = Fc - mg and Ft = Fc - mg.
B) Ft = Fc - mg and Ft = Fc + mg.
C) Ft = Fc + mg and Ft = Fc + mg.
D) Ft = Fc + mg and Ft = Fc - mg.

A

B) Ft = Fc - mg and Ft = Fc + mg.

because while the object is spinning, centripetal force will always (unless stated otherwise) be constant, as will gravity.
To get it to move up and fight gravity, tension needs to be stronger. To get it to move down, both forces are acting side to side, so tension can ease up a bit

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8
Q

In Circular Motion, for a car moving around in a circular road that is tilted from the surface, the formula for the centripetal force is (diagram below);

Diagram:
https://physicsteacher.in/wp-content/uploads/2020/08/image-4.png

A) Fc = mg.
B) Fc = mg cosθ.
C) Fc = mg sinθ.
D) Fc = mg tanθ.

A

D) Fc = mg tanθ.

looking at the diagram, we can conclude Fn sinθ = Fc. We can also see that Fn cosθ = mg, which can be rearranged to Fn = mg/cosθ.
Putting mg/cosθ into Fn sinθ, we can see that Fc = mg tanθ

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9
Q

In Circular Motion, for a car moving around in a circular road that is tilted from the surface, the car when moving TOO FAST and TOO SLOW respectively will;

A) leave the circle path, sink into in the path.
B) sink into the circle path, leave the path.
C) no difference.

A

A) leave the circle path, sink into in the path.

because, when moving at max velocity, the force will be too weak to hold in the car and it will go out
Alternatively, when moving too slow, it will sink into the centre of the circle

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10
Q

In Circular Motion, for a car moving around in a circular road that is tilted from the surface, the formulas for when the car move TOO FAST and TOO SLOW (say normal force = Fn, centripetal force = Fc, friction force = Ff) is;

A) Fc = Fn sinθ + Ff cosθ and Fc = Fn sinθ - Ff cos.
B) Fc = Fn sinθ - Ff cosθ and Fc = Fn sinθ + Ff cos.
C) no difference.

A

A) Fc = Fn sinθ + Ff cosθ and Fc = Fn sinθ - Ff cos.

remember, friction is a reactive force which opposes the motion of the object. So as stated before, if the car is too fast it will leave, too slow it will sink.
So the friction for a fast car sinks and the friction of a slow car leaves.
Looking at the formula, friction will oppose the motion by moving opposite to the motion of the car, hence it will add to the Fc if moving too fast and be subtracted from Fc if moving too slow

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11
Q

In Circular Motion, when an object is moving in a circle, placing two points in near the centre and further away from the centre, the linear velocity for each is;

A) the same.
B) outer is slower, inner is faster.
C) outer is faster, inner is slower.

A

C) outer is faster, inner is slower.

because they move at the same T, the one outer has to travel a greater distance than the inner one but at the same time. Seen when v = 2πr/T, period T is constant but r is not (inner is smaller, outer is greater). So v for the lesser r is smaller than the v for the bigger r

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12
Q

In Circular Motion, torque is positive if;

A) it moves clockwise.
B) it moves anti-clockwise.

A

B) it moves anti-clockwise.

a general rule of thumb for torque can be seen with your actual thumb. Give a thumbs up. Whatever direction your fingers move, the torque is represented by where your thumb points. Or righty tighty, lefty loosey

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13
Q

In Circular Motion, net torque is real;

A) true.
B) false.

A

A) true.

remember that torque can act on either side of the pivot point and to add up the resultant pivot.
In this diagram of equal force move around the pivot on opposite sides but in the same directions.
The resultant torque is 2Fr

Diagram:
https://i.stack.imgur.com/hiPiN.jpg

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14
Q

In Gravitational Motion, to see the acceleration of gravity on a single object, the formula of F = GMm/r^2; can be rearranged into (saying acceleration of gravity is a);

A) a = Gm/r^2.
B) a = GM/r.
C) a = GM/r^2.
D) .a = GMm/r.

A

C) a = GM/r^2.

looking at F = GMm/r^2, you can divide both sides by m and since F = ma, you can find a of the acceleration at which an object is pulled into the centre of mass M from a distance of r with a = GM/r^2

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15
Q

In Gravitational Motion, orbital velocity (linear velocity in space!!!) is seen in the formula;

A) v = √(2GM/r).
B) v = (GM/r).
C) v = (GM/r)^2.
D) v = √(GM/r).

A

D) v = √(GM/r).

looking at F = GMm/r^2 and F = mv^2/r, we can see that GMm/r^2 = mv^2/r. Cancel the r and m and square root both sides to get v = √(GM/r)

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16
Q

In Gravitational Motion, a geostationary satellite is one that;

A) has a period of 24 hours and scans one specific latitude.
B) has a period of 24 hours and scans one specific spot.
C) has a period of 24 hours and scans the entire earth..

A

B) has a period of 24 hours and scans one specific spot.

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17
Q

In Gravitational Motion, Low-earth orbit satellites (LEOs);

A) have a longer period than GSO and are slower.
B) have a longer period than GSO and are faster.
C) have a shorter period than GSO and are slower.
D) have a shorter period than GSO and are faster.

A

D) have a shorter period than GSO and are faster.

to stop them from crashing into the centre, they must move very fast to beat gravity. This makes there periods shorter than GSO. They also have a shorter altitude

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18
Q

In Gravitational Motion, for r^3/T^2 = GM/4π^2, the M is representative of;

A) the smaller masses (which encircle a bigger mass).
B) the bigger mass (which is encircled by smaller masses).

A

B) the bigger mass (which is encircled by smaller masses).

this is to show that for all the masses that surround one big mass, it will be the same or a constant, in terms of their distance from this big mass and the period which takes to orbit it

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19
Q

In Gravitational Motion, Kepler’s 1st Law states;

A) orbits are never perfect circles.
B) equal time means equal areas.
C) r^3/T^2 = GM/4π^2.
D) orbits are always perfect circles.

A

A) orbits are never perfect circles.

that’s just the way God intended things to be… It has a distinct major and minor axis, the biggest and smallest distances form each end blah blah blah

20
Q

In Gravitational Motion, Kepler’s 2nd Law states;

A) orbits are always perfect circles.
B) time of orbit is inversely proportional to area.
C) equal time means equal areas.
D) r^3/T^2 = GM/4π^2.

A

C) equal time means equal areas.

for the time it takes a planet to around a mass, the area made from the radius from the start and end of the path, for an equal area, the time will be equal. This is because the gravity will force objects to either move faster or slower, stronger gravity therefore faster speed when closer and weaker when further therefore slower speed when further away

21
Q

In Gravitational Motion, Kepler’s 3rd Law states;

A) orbits are always perfect circles.
B) equal time means equal areas.
C) r^3/T^2 = GMm.
D) r^3/T^2 = GM/4π^2.

A

D) r^3/T^2 = GM/4π^2.

or more specifically, Kepler witnessed this phenomena where for objects surrounding a certain mass, the radius of these objects from the mass cubed over the time it took for one revolution squared were constant for all objects surrounding a particular mass. It is also seen in Newton’s equations wow!

22
Q

In Gravitational Motion, the formula for gravitational potential energy, U = -GMm/r is negative because;

A) it will naturally move to M due to gravity against the direction of +r.
B) because it is opposite M.
C) conservation of energy means that K + U = 0 and K = -U.
D) as you m further away from M (r → ∞, or U gets closer to zero).

A

D) as you m further away from M (r → ∞, or U gets closer to zero).

if r → ∞, then U gets to 0. If 0 is the max potential energy with max distance, than clearly as the number gets closer to 0 and away from negative values, the potential energy is increasing. So a small amount of potential energy would be -10000, but a bigger amount would be -10 because the latter is closer to 0 than the former

23
Q

In Gravitational Motion, kinetic energy formula (of the thing moving around a mass) is (because);

A) GMm/r, because K + U = 0.
B) -GMm/r, because K = U.
C) GMm/2r, because orbital velocity is put into kinetic energy formula.
D) GMm/2r, because kinetic energy opposes gravity pulling objects in.

A

C) GMm/2r, because orbital velocity is put into kinetic energy formula.

kinetic energy is 1/2mv^2 and orbital velocity v = √(GM/r). Putting that in, we get GMm/2r which is the kinetic energy of the thing orbiting the mass. So this formula will be applicable to find the energy of a thing like a satellite orbiting the earth

24
Q

In Gravitational Motion, work to get an object in orbit is seen in the formula (say W = total work = E, K = kinetic energy);

A) W = K + U.
B) W = K + ΔU.
C) W = ΔK + ΔU.
D) W = ΔK + U.

A

B) W = K + ΔU.

to get a satellite from point A (lower) to point B (higher up), we need to use energy to get it up there or put work into it or ΔU. But that’s not all, we also need to give it additional energy to actually move and shit or K

25
Q

In Gravitational Motion, work to get a satellite up and into orbit is;

A) the energy gained AND the total energy.
B) the energy gained NOT the total energy.
C) NOT the energy gained, but the total energy.
D) neither of these options.

A

B) the energy gained NOT the total energy.

the equation W = K + ΔU doesn’t say how much energy is in the satellite, it only tells us the necessary energy to get the satellite into the position and movement of orbit and stuff

26
Q

In Gravitational Motion, which diagram reflects the way potential gravity increases?

Diagram:
https://www.reddit.com/user/la_li_luu_le_lo/comments/m1rmkh/graphs_for_physics_stuff/

A)
B)
C)
D)

A

D)

as potential energy increases, that is when the object is moving higher and higher or away from the mass, seeing formula U = -GMm/r, the closer it gets to 0, the higher its altitude will be

27
Q

In Gravitational Motion, for escape velocity, Sir Isaac Newton imagined a cannon firing off a mountain. What does this have to do with escape velocity?;

A) if the cannon is fired fast enough, it will leave the influence of gravity.
B) if the cannon is fired high enough, it will leave the influence of gravity.
C) event if the cannon is fired fast enough, it will always coming crashing eventually.
D) if the cannon is fired fast enough, it will always form an ellipse that will encircle space.

A

A) if the cannon is fired fast enough, it will leave the influence of gravity

to beat the acceleration due to gravity, an object in very fast motion can overcome gravity. In most cases however, that is when it’s too slow, it will crash into the earth because it cannot beat the gravity

28
Q

In Gravitational Motion, for escape velocity, the formula is (check answer to see how it’s derived);

A) v = √(2GMm/r).
B) v = √(GMm/r).
C) v = √(2GM/r).
D) v = √(GM/2r).

A

C) v = √(2GM/r)

the formula itself is derived from both GMm/r and 1/2mv^2. At max potential, r = ∞ and v = 0. Likewise at least potential, r = radius of planet and v = escape velocity.
So, GMm/r = 1/2mv^2, more specifically ΔU = ΔK, and even more specifically GMm/∞ - GMm/r = 1/2m0^2 - 1/2mv^2 → v = √(2GM/r)

29
Q

Formulas for Projectile Motion?

A

Horizontal V = Ucosθ
Vertical V = Usinθ - gt

Horizontal S = Ucosθt
Vertical S = Usinθt - gt^2/2

30
Q

Formula for Centripetal Force?

A

Fc = mv^2/r

31
Q

Formula for Centripetal Acceleration?

A

Ac = v^2/r

32
Q

Formula for Linear Velocity?

A 
v = 2πr/T
v = ωr

[T = time for 1 revolution (seconds)]

33
Q

Formula for Angular Velocity?

A 
ω = 2π/T
ω = θ/t

[θ = radian (θ)]

34
Q

Formula for Tension Force of spinning thing (at top)?

A 
Ft = Fc - mg
Ft = mv^2/r - mg
35
Q

Formula for Tension Force of spinning thing (at bottom)?

A 
Ft = Fc + mg
Ft = mv^2/r + mg
36
Q

Formula for Centripetal Force on banked surface (no friction)?

A

Fc = mgtanθ

37
Q

Formula for Centripetal Force on banked surface (friction, going too fast)?

A 
Fc = mgtanθ + Ffcosθ
mv(max)^2/r = mgtanθ + Ffcosθ

[Ff = friction force]

38
Q

Formula for Centripetal Force on banked surface (friction, going too slow)?

A 
Fc = mgtanθ - Ffcosθ
mv(min)^2/r = mgtanθ - Ffcosθ

[Ff = friction force (N)]

39
Q

Formula for Torque?

A

T = Frsinθ

[r = distance from pivot (m)]
[θ = angle between F and device]
40
Q

Formula for Gravitational Force?

A

F = GMm/r^2

[r = distance between two bodies (m)]
[G = gravitational constant (6.67 x 10^-11)]
41
Q

Formula for Gravity of M?

A

g = GM/r^2

[G = gravitational constant (6.67 x 10^-11)]

42
Q

Formula for Orbital Velocity?

A

v = √(GM/r)

[G = gravitational constant (6.67 x 10^-11)]

43
Q

Formula for Kepler’s Constant?

A

r^3/T^2 = GM/4π^2

[G = gravitational constant (6.67 x 10^-11)]

note this is a constant only shared by the masses (m) surrounding a greater mass (M)

44
Q

Formula for Potential Gravitational Energy?

A

U = -GMm/r

[G = gravitational constant (6.67 x 10^-11)]

45
Q

Formula for Work (to get body into orbit)?

A 
W = ΔU + K
W = GMm/re - GMm/2ro
W = GMm(re + 2h)/2re(re+h)
[G = gravitational constant (6.67 x 10^-11)]
[h = altitude (m)]

note W is the energy needed to get it into orbit and moving NOT the total energy

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