Momentum Introduction

AP PHYSICS 1 NOTES: INTRODUCTION TO MOMENTUM

DEFINITION OF MOMENTUM

• Momentum depends on two values:

  • (1) Mass

  • (2) Velocity

• Momentum describes the tendency of objects to keep going in the same direction with the same speed.

• Consider momentum a measurement of how hard it is to stop or turn a moving object.

I. MOMENTUM AS A FUNCTION OF VELOCITY

• Velocity has two components:

  • Magnitude (speed)

  • Direction

  • a. Magnitude:

  • A force is required to change the speed (magnitude) of a moving object because work must be done to change the object’s kinetic energy.

  • b. Direction:

  • A force is required to change an object’s direction of motion.

• Momentum is a property of moving mass that resists changes in a moving object’s velocity in either magnitude (speed), direction, or both.

II. MOMENTUM AS A FUNCTION OF MASS

• The more mass an object has, the more force it takes to deflect its motion.

• Inertia:

  • A property of mass that resists changes in velocity (Newton’s First Law).

  • Inertia only depends on mass, whereas momentum depends on both mass and velocity.

III. MOMENTUM AND KINETIC ENERGY

• Both momentum and kinetic energy depend on mass and velocity.

• Key differences:

  • Kinetic Energy (KE):

  • A scalar quantity.

  • Always positive or zero.

  • Momentum (p):

  • A vector quantity.

  • Can be positive or negative; direction is essential in momentum.

  • Two objects with the same mass and speed have opposite momenta if they are moving in opposite directions.

  • Example: Two spheres with the same mass and speed have the same kinetic energy but opposite momentum.

IV. MOMENTUM AND FORCES

• Force:

  • The action that changes momentum.

• Any change in momentum must create force.

V. CALCULATING MOMENTUM

• The momentum p of a particle is a vector quantity equal in magnitude to the product of its mass m and its velocity v.

  • Formula: $p = m imes v$

• Momentum uses mass and velocity to quantify an object’s motion; this motion points in the same direction as the object’s velocity.

• Units:

  • kg⋅m/s or N⋅s

  • Note: Do not confuse this with:

  • Force: kg⋅m/s²

  • Joules (energy): kg⋅m²/s²

SAMPLE PROBLEMS

• Sample Question 1:

  • A car is traveling at a velocity of 13.5 m/sec (30 mph) north on a straight road.

  • The mass of the car is 1,300 kg.

  • A motorcycle passes the car at a speed of 30 m/sec (67 mph).

  • The motorcycle (with rider) has a mass of 350 kg.

  - Calculate and compare the momentum of the car and motorcycle:

  For the car: $p<em>{car} = m</em>{car} imes v<em>{car} = 1300 ext{ kg} imes 13.5 ext{ m/s}$ For the motorcycle: $p</em>{motorcycle} = m<em>{motorcycle} imes v</em>{motorcycle} = 350 ext{ kg} imes 30 ext{ m/s}$

• Sample Problem 2:

  • How could the principles of momentum be applied to the following illustration?

• Sample Problem 3:

  • Spheres A and B have the same mass and are bombarded with the same force (1 N).

  • Which sphere, A or B, had a greater initial velocity (before the application of the force)? How do you know?

To solve Sample Problem 2, you apply the Impulse-Momentum Theorem, which connects force, time, and the resulting change in motion.

The theorem states that the change in momentum ($\Delta p$) is equal to the product of the average force ($F$) and the time interval ($\Delta t$) over which that force is applied:

$\Delta p = F \times \Delta t$

In a collision illustration, this principle can be applied in several ways:

1. Fixed Momentum Change: If an object is coming to a stop (like a car hitting a wall), its change in momentum is constant. To reduce the force ($F$) of the impact, you must increase the time ($\Delta t$) of the collision. This is the logic behind safety features like airbags and crumple zones.

2. Calculating Force: If you know the mass and the change in velocity, you can calculate the total change in momentum ($m \cdot \Delta v$). By dividing this by the duration of the impact, you can find the average force acting on the object.

3. Vector Change: Remember that because momentum is a vector, a collision that causes an object to bounce back involves a much larger change in momentum (and thus more force or time) than a collision where the object simply stops, because the velocity direction flips.

To solve Sample Problem 3, we apply the Impulse-Momentum Theorem, which states that the change in momentum ($\Delta p$) is equal to the force ($F$) multiplied by the time ($\Delta t$) it is applied. Since both spheres have the same mass and are acted upon by the same force ($1\ \text{N}$) over the same amount of time, they must experience the exact same change in momentum and, consequently, the same change in velocity ($\Delta v$).

1. Change in Velocity: Because $\Delta p = m(v<em>f - v</em>i)$, and both mass ($m$) and the impulse applied are identical for both spheres, the change in velocity ($\Delta v$) is the same for both.

2. Comparison: If Sphere A reaches a higher final velocity than Sphere B, it means it must have started at a higher initial velocity. For example, if the force adds $10\ \text{m/s}$ to each sphere, and Sphere A ends at $30\ \text{m/s}$, its initial velocity was $20\ \text{m/s}$. If Sphere B ends at $15\ \text{m/s}$, its initial velocity was only $5\ \text{m/s}$.

Conclusion: Sphere A had a greater initial velocity because it reached a higher final speed despite receiving the same increase in momentum as Sphere B.