Equilibrium & Momentum Flashcards

From the basics of rotational equilibrium to the complexities of torque, use these cards to master the topic of equilibrium and momentum as tested on the MCAT.

1
Q

Define:

net force

A
  • It is the sum of all forces on an object, added as vectors.
  • The product of the object’s acceleration and its mass.

The SI unit of force is the newton (N). 1N = 1 kg*m/s2.

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2
Q

What quantity is generally measured in newtons (N)?

A

force

1 N is equal to 1 kg*m/s2.

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3
Q

What is the force on an object for which m = 2 kg and a = 10 m/s2?

A

20 N

F = ma = (2 kg) (10 m/s2) = 20 N

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4
Q

What is the force on an object for which m = 4kg, ∆v = 20 m/s, and ∆t = 1s?

A

80 N

Acceleration is equal to the change in velocity over time. Here:

a = (20 m/s) / (1 s) = 20 m/s2

F = ma = (4 kg) (20 m/s2) = 80 N

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5
Q

What is the force on an object for which ∆p = 10kg*m/s and ∆t = 10 s?

A

1 N

Change in momentum (∆p) is equal to the product of force and time, so force = ∆p/∆t. Here:

F = (10 kg*m/s) / (10 s) = 1 N

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6
Q

Define:

momentum (p)

A

It is a value that quantifies an object’s tendency to remain moving in a certain direction. The momentum of an object equals its mass times its current velocity.

The SI unit of momentum is kg*m/s.

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7
Q

What is the momentum of an object for which m = 2 kg and v = 15 m/s?​

A

30 kg*m/s

p = mv = (2 kg) (15 m/s) = 30 kg*m/s

Note that momentum can be positive or negative, depending on direction of velocity.

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8
Q

What is the momentum of an object for which m = 10 kg and v = 0 m/s?

A

0 kg*m/s

p = mv = (10 kg) (0 m/s) = 0 kg*m/s

Note that momentum can be positive or negative, depending on direction of velocity.

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9
Q

Define:

impulse

A

It is the change in momentum of an object due to a net force applied over some change in time. Impulse is calculated as the product of average force and time.

The units used for impulse are N*s, or kg*m/s.

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10
Q

What is the impulse on an object if it has a Δmv of 8 kg*m/s and an average force of 4 N?

A

8 kg*m/s

Impulse is equal to change in momentum, which is given here as Δmv. Note that impulse can be positive or negative, depending on direction of velocity change.

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11
Q

What change in Δp will cause the force on an object to be halved?

A

Momentum per unit time must also be halved.

Since Δp = (Favg)(Δt), momentum and force are directly proportional. Halving force requires momentum to be reduced by half as well.

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12
Q

What momentum change occurs for an object that experiences a force that doubles its velocity?

A

Momentum also doubles.

Since Δp = mΔv, a change in momentum is directly proportional to a change in velocity. Doubling one will double the other.

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13
Q

One mass experiences a certain force for a given time, while another is subjected to half of that force for twice the length of time. How do their impulses compare?

A

The impulses are equal.

Original impulse = F1*t1

Since F2 = F1/2 and t2 = 2t1, the new impulse is equal to F2*t2 = (F1/2)*(2t1) = F1*t1 .

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14
Q

Describe why airbags can limit the injuries of passengers in a car accident.

A

Airbags lengthen the amount of time that a crash victim experiences deceleration forces. This decreases the average force on the victim.

According to the formula impulse = Favg*Δt, force and time are inversely proportional. Therefore, if the time increases, the force is decreased.

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15
Q

Describe Newton’s first law of motion with regards to force.

A

For an object to change its motion, a net force must act on it.

If all forces cancel to zero, the object will not experience any change in motion. This can either cause an object to stay at rest or to continue moving with a constant velocity.

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16
Q

Describe Newton’s first law of motion with regards to momentum.

A

An object with directional momentum will continue with that momentum unless acted on by a net force.

Similarly, an object with zero momentum will continue to remain at rest until acted on by a net force.

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17
Q

With regard to force, how can an object at rest gain velocity?

A

The object must experience a net force.

Since all forces cancel to zero at rest, a net force is required to change velocity.

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18
Q

With regard to momentum, how can an object at rest gain velocity?

A

The object must have experienced an impulse, or change in momentum, from a net force.

Since all forces cancel to zero at rest, a net force is required to change velocity. Velocity will change in a way that is directly proportional to any change in momentum.

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19
Q

In physics, what is meant by the term translation?

A

It is the net movement or change in location of an object. This may be thought of as a non-zero displacement in any direction.

For example, a hockey puck experiences translation when it begins at one end of an ice rink and moves to the other end.

20
Q

Define:

translational equilibrium

A

This occurs when the sum of all of the forces acting on an object cancel to zero. The object will not accelerate, since the net force applied on it is zero.

This situation includes objects at rest as well as objects moving with constant velocity.

21
Q

A skydiver is falling from a plane and has not yet reached terminal velocity. Is the skydiver in translational equilibrium?

A

No

The skydiver is accelerating (changing velocity), which must be due to a non-zero net force. Translational equilibrium only exists when net force is zero.

Specifically, the downward force due to gravity is greater than the upward force due to air resistance.

22
Q

In a rotating system, what are the fulcrum and the lever arm?

A
  • The fulcrum is the point of an object that remains fixed during rotation.
  • The lever arm is the distance from the fulcrum to the point where a force is applied.
23
Q

Define:

torque

A
  • It is the ability of a force to create a change in the angular orientation of an object, by rotation about a fulcrum point.
  • Torque can also be thought of as the component of work perpendicular to a lever arm.

The units used to describe torque are N*m.

24
Q

What equation gives the relationship between the lever arm radius and the applied force?

A

τ = r*F sinθ

Here, r is the distance between the applied force and the fulcrum point, F is the magnitude of force applied, and θ is the angle between the lever arm and the force vector.

Only the component of force perpendicular to the lever arm will generate torque, since sin 0 = sin 180 = 0.

25
Q

How will the torque on a lever arm change if the same force is applied half the original distance from the fulcrum?

A

Torque will be halved.

Since torque = r*F sinθ, torque is directly proportional to the distance along the lever arm. Halving the distance over which the force is applied will halve the torque as well.

26
Q

Conventionally, which direction of rotation is represented as a positive torque?

A

Counterclockwise rotation

The maximum positive (counterclockwise) torque that a force can deliver is at 90 degrees to the lever arm.

27
Q

Conventionally, which direction of rotation is represented as a negative torque?

A

Clockwise rotation

The maximum negative (clockwise) torque that a force can deliver is at 90 degrees to the lever arm.

28
Q

Define:

rotational equilibrium

A

It occurs when the net torque on an object is zero. If the sum of all torques acting on an object equals zero, the object will not rotate.

This situation includes objects at rest as well as objects rotating with constant spin.

29
Q

A child on a see-saw leans back, causing her to lower and her partner to rise. Is the see-saw system in rotational equilibrium?

A

No

The see-saw is experiencing changing rotation, which must be due to a non-zero net torque. If all torques do not cancel, the system is not in rotational equilibrium.

30
Q

A child of mass 40 kg sits 1.5 m from the pivot of a seesaw. To perfectly balance, where on the opposite end should her 30kg little sister sit?

Assume the seesaw itself is massless.

A

She should sit 2m from the pivot point.

Since all forces due to gravity are perpendicular to the seesaw, torque from the 40kg child will be: (mg)r = 40(10)(1.5) = 600 Nm. To balance, the 30kg child needs to supply 600 Nm of torque as well, which requires a distance of 2m.

31
Q

What visual tool can be used to account for all forces acting on an object?

A

A free-body diagram can account for forces and their directions.

Free-body diagrams represent force vectors with arrows pointing in the corresponding direction. Larger forces are represented by longer arrows.

32
Q

Describe or draw the free-body diagram for a box with mass (m), accelerating towards the earth in free fall.

A

The free-body diagram will show:

  • A force due to gravity, pointing directly downward, with a magnitude of
    mg
  • A force due to air resistance, pointing directly upward, with a magnitude of kv2 (the negative sign is only necessary for calculations)

Note that since the mass is accelerating downwards, the arrow for gravity should be represented as larger than the arrow for air resistance.

While the MCAT will not test your drawn arrow length, of course, it can help to discern the direction that the force will accelerate.

33
Q

Describe or draw the free-body diagram for the stationary block of mass m on the slope below, providing specific values where possible.

A

The free-body diagram will show:

  • A force due to gravity, pointing directly downward, with a magnitude of mg
  • A normal force, pointing perpendicular to the slope, with a magnitude of mg cos(θ)
  • A static friction force, pointing parallel to the slope (opposing gravity), with a magnitude of mg sin(θ)
34
Q

Define:

apparent weightlessness

A

This occurs when no force opposes gravity, allowing the object to be in unencumbered free fall.

An astronaut in a low-space orbit would feel apparent weightlessness, since he is essentially falling in a large circle due to the pull from gravity.

35
Q

A skydiver leaps from a plane, and assuming there is no air resistance, are they experiencing weightlessness?

A

Yes

No force is opposing gravity, and the skydiver is not in contact with any objects that inhibit his or her inertial movement.

36
Q

Explain the principle of conservation of linear momentum.

A

Conservation of linear momentum states that in an isolated system, the initial momentum on any axis will be the same as the final momentum on that axis.

This is important when objects are moving at an angle. It means that the momenta in the x and y directions are conserved, just like the overall momentum.

37
Q

Explain why in an isolated container of gas, individual molecules can gain or lose momentum, but the system of gas doesn’t change its momentum overall.

A

Since this is an isolated system, no external forces are acting on any molecules within it. With no way to add or remove momentum from the system, its total value must be conserved.

If some molecules gain momentum, others must lose that total value.

38
Q

Explain with an example whether it is possible for objects in an isolated system to collide and stop.

A

It is possible.

Consider two cars of equal weight and equal-but-opposite velocity. The initial momentum must begin at zero, since the momenta of the cars exactly cancel out. If they collide and stick in a totally inelastic collision, then the final momentum must also equal zero and the cars will come to a stop.

39
Q

Define:

elastic collision

A

It is one in which the mechanical (kinetic) energy and the momentum of the system are both conserved.

Elastic collisions require objects to bounce off of each other perfectly, in rigid fashion, without any deformation.

40
Q

Two hockey pucks of equal mass, one traveling to the right and the other to the left, collide rigidly in space with no loss of kinetic energy. Describe the motion of the pucks after collision, assuming they began with equal speed.

A

The pucks will bounce off of each other in a perfectly elastic collision.

If no kinetic energy is lost, the collision must be elastic. The speeds and masses of the pucks will remain constant, but the signs of the velocity and momentum values will become opposite.

41
Q

What feature or features distinguish an inelastic collision?

A

It involves a conversion of some kinetic energy to another form; momentum, however, is always conserved.

These can include sound production, heat loss due to friction, or potential energy or work performed on an object by compressive forces. At least one object will be deformed, heated, broken, or otherwise changed.

42
Q

Two cars of equal mass, one traveling to the right and the other to the left, collide and deform in space. Describe the motion of the cars after collision, assuming they began with equal speed and are not stuck together afterwards.

A

The cars will bounce off of each other in an inelastic collision.

The cars will lose kinetic energy due to the collision. Their speeds will certainly decrease; masses may change if one car loses mass to the other. The signs of each velocity and momentum value will now be opposite, but total momentum remains conserved.

43
Q

What feature distinguishes a totally inelastic collision?

A

The objects will stick together and remain connected.

All inelastic collisions involve a conversion of some kinetic energy to another form; momentum, however, is always conserved.

44
Q

Two soft clay balls of equal mass, one traveling to the right and the other to the left, collide and stick in space. Describe the motion of the balls after collision, assuming they began with equal speed.

A

The balls will stick together and stop; the collision will be totally inelastic.

Total momentum began as zero, since the balls were moving in opposite directions. Because momentum is always conserved, final momentum must equal zero as well. The balls will lose kinetic energy due to the collision.

45
Q

Two balls of equal mass collide in space. Knowing nothing else, what can be said about this system after the collision?

A

Momentum will be conserved, no matter what the collision type.

If the collision is elastic, kinetic energy will also be conserved. If the collision is inelastic, kinetic energy will be lost, while a totally inelastic collision will produce one final object with double the mass of a single ball.

46
Q

Two masses, m1 and m2, are originally traveling at velocities v1 and v2. They undergo a totally inelastic collision, after which they are traveling at velocity vf. What equation relates their momenta before and after the collision?

A

m1v1 + m2v2 = (m1+m2)vf

This equation does not need to be memorized; it is simply derived from the usual conservation of momentum formula. Keep in mind that after a totally inelastic collision, the objects combine into a single mass.

47
Q

Two masses, m1 and m2, are initially traveling at velocities v1 and v2. They undergo a totally inelastic collision, after which they are traveling at velocity vf. What equation relates their kinetic energies before and after the collision?

A

½m1v12 + ½m2v22 > ½(m1+m2)vf2

This equation does not need to be memorized; it is simply discerned from the fact that any inelastic collision will cause a decrease in kinetic energy. Keep in mind that after a totally inelastic collision, the objects combine into a single mass.