PHYS1115 Lecture Notes Flashcards

Momentum

• Momentum (p) is defined as the product of mass and velocity: p = mv
• The unit for momentum is kg·m/s.
• The more momentum an object has, the harder it is to stop.
• A moving object can have a large momentum if it has a large mass, a high speed, or both.

Example 1: Momentum Calculation

• What is the momentum of a 3000 kg truck traveling at 25 m/s? (Note: the slide uses 15 m/s in the calculation but states 25 m/s in the problem.)
• Given: m = 3000 \text{ kg}, v = 15 \text{ m/s}
• p = mv = (3000 \text{ kg})(15 \text{ m/s}) = 45000 \text{ kg m/s}

Example 2: Velocity Calculation

• A 1500 kg ferryboat has a momentum of 30,000 kg·m/s. What is the velocity of the ferryboat?
• Given: m = 1500 \text{ kg}, p = 30000 \text{ kg m/s}
• v = \frac{p}{m} = \frac{30000 \text{ kg m/s}}{1500 \text{ kg}} = 20 \text{ m/s}

System Momentum

• If a system contains more than one object, the total momentum is the vector sum of the momenta of those objects.
• p{\text{system}} = p1 + p2 + p3 + …
• p{\text{system}} = m1v1 + m2v2 + m3v_3 + …

Example: System Momentum

• Determine the momentum of a system of two objects: m1 with a mass of 6.0 kg and a velocity of 13 m/s towards the east, and m2 with a mass of 14 kg and a velocity of 7.0 m/s towards the west.
• Given: m1 = 6.0 \text{ kg}, v1 = +13 \text{ m/s}, m2 = 14 \text{ kg}, v2 = -7.0 \text{ m/s}
• Choose "east" as positive.
• \Sigma p = (6 \text{ kg})(13 \text{ m/s}) + (14 \text{ kg})(-7 \text{ m/s}) = 78 \text{ kg m/s} - 98 \text{ kg m/s} = -20 \text{ kg m/s}

Impulse

• Impulse: A force sustained for a long time produces more change in momentum than does the same force applied briefly.
• Both force and time are important in changing an object's momentum.
• The quantity force x time interval is called impulse.
• \text{impulse} = F \times t
• The greater the impulse exerted on something, the greater will be the change in momentum.
• \text{impulse} = \text{change in momentum}
• Ft = \Delta(mv) = mvf - mvi
• There is no specially named unit for impulse; we use the product of the units of force and time: N·s or kg·m/s. This is the same unit used for momentum.

Example 1: Impulse and Change in Velocity

• A constant force of 12 N acts for 5.0 s on a 5.0 kg object. What is the magnitude of the impulse delivered to the system? What is the change in the object's velocity?
• Given: F = 12 \text{ N}, t = 5.0 \text{ s}, m = 5.0 \text{ kg}
• I = F \Delta t = (12 \text{ N})(5 \text{ s}) = 60 \text{ N s}
• \Delta p = m \Delta v \implies \Delta v = \frac{I}{m} = \frac{60 \text{ N s}}{5 \text{ kg}} = 12 \text{ m/s}

Example 2: Average Force on a Golf Ball

• A 0.030 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.0030 s. What is the average force on the ball by the golf club?
• Given: m = 0.030 \text{ kg}, vf = 34 \text{ m/s}, vi = 0 \text{ m/s}, t = 0.0030 \text{ s}
• I = F \Delta t = \Delta p = mvf - mvi
• I = (0.03 \text{ kg})(34 \text{ m/s}) - (0.03 \text{ kg})(0 \text{ m/s}) = 1.02 \text{ N s}
• F = \frac{I}{\Delta t} = \frac{1.02 \text{ N s}}{0.003 \text{ s}} = 340 \text{ N}

Conservation of Momentum

• The law of conservation of momentum states that, in the absence of an external force, the momentum of a system remains unchanged.
• Momentum is conserved for all collisions as long as external forces don't interfere.
• Whenever objects collide in the absence of external forces, the net momentum of the objects before the collision equals the net momentum of the objects after the collision.
• \text{net momentum}{\text{before collision}} = \text{net momentum}{\text{after collision}}
• Objects in an isolated system can interact with each other in two basic ways:
  • They can collide.
  • If they are stuck together, they can explode (push apart).
• In an isolated system, both momentum and total energy are conserved. But the energy can change from one form to another.
• Conservation of momentum and change in kinetic energy predict what will happen in these events.

Types of Collisions

• Inelastic collisions: Two objects collide, converting some kinetic energy into other forms of energy such as potential energy, heat, or sound.
  • Perfect inelastic collisions: Two objects collide, stick together, and move as one mass after the collision, transferring some of the kinetic energy into other forms of energy.
    • m1v1 + m2v2 = (m1 + m2)v'
  • General inelastic collisions: Two objects collide and bounce off each other, transferring some of the kinetic energy into other forms of energy.
    • m1v1 + m2v2 = m1v1' + m2v2'
• Elastic collisions: Two objects collide and bounce off each other while conserving kinetic energy—energy is not transformed into any other type.
  • (v1 - v2) = -(v1' - v2')
• Explosions: An object or objects break apart because potential energy stored in one or more of the objects is transformed into kinetic energy.
  • (m1 + m2)v = m1v1' + m2v2'

Collision Types Summary

Example 1: Railroad Freight Car Collision

• A 15,500 kg railroad freight car travels on a level track at a speed of 5.4 m/s. It collides and couples with a 30,000 kg second car, initially at rest and with brakes released. What is the speed of the two cars after collision?
• Given: m1 = 15500 \text{ kg}, v1 = 5.5 \text{ m/s}, m2 = 30000 \text{ kg}, v2 = 0 \text{ m/s}
• Since the cars couple, this is a perfectly inelastic collision: m1v1 + m2v2 = (m1 + m2)v'
• v' = \frac{m1v1}{m1 + m2} = \frac{(15500 \text{ kg})(5.5 \text{ m/s})}{15500 \text{ kg} + 30000 \text{ kg}} = 1.9 \text{ m/s}

Example 2: Skater Throwing a Ball

• A 55 kg skater at rest on a frictionless rink throws a 3 kg ball, giving the ball a velocity of 8 m/s. What is the velocity of the skater immediately after?
• Given: m1 = 55 \text{ kg}, v1 = 0 \text{ m/s}, m2 = 3 \text{ kg}, v2 = 0 \text{ m/s}, v_2' = 8 \text{ m/s}
• (m1 + m2)v = m1v1' + m2v2'
• 0 + 0 = m1v1' + m2v2'
• v1' = -\frac{m2v2'}{m1} = -\frac{(3 \text{ kg})(8 \text{ m/s})}{55 \text{ kg}} = -0.4 \text{ m/s}