Equilibrium & Momentum

Question 1

Define:

net force

Answer 1

• 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.

Question 2

What quantity is generally measured in newtons (N)?

Answer 2

force

1 N is equal to 1 kg*m/s².

Question 3

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

Answer 3

20 N

F = ma = (2 kg) (10 m/s²) = 20 N

Question 4

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

Answer 4

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

Question 5

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

Answer 5

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

Question 6

Define:

momentum (p)

Answer 6

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.

Question 7

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

Answer 7

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.

Question 8

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

Answer 8

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.

Question 9

Define:

impulse

Answer 9

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.

Question 10

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

Answer 10

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.

Question 11

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

Answer 11

Momentum per unit time must also be halved.

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

Question 12

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

Answer 12

Momentum also doubles.

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

Question 13

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?

Answer 13

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*t_{1 .}

Question 14

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

Answer 14

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 = F_avg*Δt, force and time are inversely proportional. Therefore, if the time increases, the force is decreased.

Question 15

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

Answer 15

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.

Question 16

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

Answer 16

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.

Question 17

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

Answer 17

The object must experience a net force.

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

Question 18

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

Answer 18

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.

Question 19

In physics, what is meant by the term translation?

Answer 19

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.

Question 20

Define:

translational equilibrium

Answer 20

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.

Question 21

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

Answer 21

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.

Question 22

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

Answer 22

• 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.

Question 23

Define:

torque

Answer 23

• 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.

Question 24

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

Answer 24

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.

Question 25

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

Answer 25

Torque will be **halved**. ## Footnote 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.

Question 26

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

Answer 26

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

Question 27

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

Answer 27

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

Question 28

# Define: rotational equilibrium

Answer 28

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

Question 29

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**?

Answer 29

No ## Footnote 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.

Question 30

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? ## Footnote Assume the seesaw itself is massless.

Answer 30

She should sit **2m** from the pivot point. ## Footnote 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.

Question 31

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

Answer 31

A **free-body diagram** can account for forces and their directions. ## Footnote Free-body diagrams represent force vectors with arrows pointing in the corresponding direction. Larger forces are represented by longer arrows.

Question 32

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

Answer 32

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) ## Footnote 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.

Question 33

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

Answer 33

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)*

Question 34

# Define: apparent weightlessness

Answer 34

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

Question 35

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

Answer 35

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

Question 36

Explain the **principle of conservation of linear momentum**.

Answer 36

**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. ## Footnote 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.

Question 37

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.

Answer 37

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

Question 38

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

Answer 38

It is possible. ## Footnote 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.

Question 39

# Define: elastic collision

Answer 39

It is one in which the mechanical (kinetic) energy and the momentum of the system are both **conserved**. ## Footnote Elastic collisions require objects to bounce off of each other perfectly, in rigid fashion, without any deformation.

Question 40

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.

Answer 40

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

Question 41

What feature or features distinguish an **inelastic collision**?

Answer 41

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

Question 42

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.

Answer 42

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

Question 43

What feature distinguishes a **totally inelastic collision**?

Answer 43

The objects will **stick together and remain connected**. ## Footnote All inelastic collisions involve a conversion of some kinetic energy to another form; momentum, however, is always conserved.

Question 44

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.

Answer 44

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

Question 45

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

Answer 45

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

Question 46

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 v_f. What equation relates their **momenta** before and after the collision?

Answer 46

m₁v₁ + m₂v₂ = (m₁+m₂)v_f ## Footnote 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.

Question 47

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 v_f. What equation relates their **kinetic energies** before and after the collision?

Answer 47

½m₁v₁² + ½m₂v₂² \> ½(m₁+m₂)v_f² ## Footnote 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.