Conservation of momentum in one dimension

Conservation of momentum in one dimension

Lessons

In this lesson, we will learn:

  • Meaning of conservation of momentum
  • Deriving conservation of momentum from Newton's third law
  • Meaning of closed/open systems and identifying conservation of momentum problems
  • Problem solving with conservation of momentum in 1 dimension

Notes:

  • Momentum is a conserved quantity: the total momentum of a set of objects before an event (like a collision or explosion) is equal to total momentum after.
  • Conservation of momentum (pi=pf\sum\vec{p}_i = \sum\vec{p}_f)is a result of Newton's 3rd
  • Law (FA=FB\vec{F}_A = -\vec{F}_B).
  • When talking about conservation of momentum, system is a name given to an object or set of objects that are being examined.
    • A closed system is an object or set of objects that are not affected by external forces (i.e. their motion is not changing due to forces from outside the system). Momentum is conserved in a closed system.
      • A system can have external forces acting on it while still being closed, as long as those forces are balanced. A ball rolling across a frictionless table would be considered a closed system even though gravity and normal force act on the ball. Gravity and normal force are balanced, so they are not causing the ball to accelerate, and momentum is conserved.
    • An open system is an object or set of objects that are affected by external forces (i.e. their motion does change due to forces from outside the system). Momentum is not conserved in an open system.
      • If the ball from the earlier example rolls off the edge of the frictionless table and is accelerated by gravity towards the earth, then it is no longer considered a closed system. The system is open and gaining momentum from the acceleration due to gravity.

Momentum

p=mv:\vec{p} = m \vec{v}: momentum, in kilogram meters per second (kg∙m/s)

m:m: mass, in kilograms (kg)

v:\vec{v}: velocity, in meters per second (m/s)


Conservation of Momentum

pi=pf\sum\vec{p}_i = \sum\vec{p}_f

pi:\vec{p}_i: initial momentum, in kilogram meters per second (kg·m/s)

pf:\vec{p}_f: final momentum, in kilogram meters per second (kg·m/s)


Impulse

J=FΔt=Δp=m(vfvi\vec{J} = \vec{F} \Delta t = \Delta \vec{p} = m( \vec{v}_f - \vec{v}_i)

J:\vec{J}: impulse, in newton seconds (N∙s)

F:\vec{F}: net force acting on an object, in newtons (N)

Δt:\Delta t: the length of time for which the force acts, in seconds (s)

Δp:\Delta \vec{p}: change in momentum of an object, in kilgram meters per second (kg∙m/s)

  • Introduction
    Introduction to conservation of momentum in one dimension:
    a)
    Meaning of conservation of momentum

    b)
    Deriving conservation of momentum from Newton's third law

    c)
    Meaning of closed/open systems and identifying conservation of momentum problems


  • 1.
    pi=pf:\bold{\sum\vec{p}_i = \sum\vec{p}_f}: Objects that bounce apart after collision
    a)
    A 0.200 kg billiard ball travelling 7.00 m/s to the right collides with a stationary 0.0450 kg golf ball. The velocity of the billiard ball after collision is 4.43 m/s [right]. What is the velocity of the golf ball after collision?

    b)
    A 2.20 kg ball rolling 1.50 m/s [E] collides head-on with a ball rolling 2.40 m/s [W]. The 2.20 kg ball rolls away at 1.00 m/s [W] after the collision, and the second ball rolls 1.80 m/s [E]. What is the mass of the second ball?

    c)
    4500 kg train car A travels at 6 m/s [E]. It collides with 3750 kg train car B which is travelling 3.2 m/s [E] on the same track. After the collision, train car A travels at 3.5 m/s [E]. Find the final velocity of train car B.

    d)
    A rifle points [E] at a 3.40 kg wood block. It fires a 22.0 g bullet travelling at 460.0 m/s, which passes through the block and continues at 299 m/s. Find the velocity of the wood block when the bullet leaves.


  • 2.
    pi=pf:\bold{\sum\vec{p}_i = \sum\vec{p}_f}: Objects that stick together after collision
    a)
    A rifle points [E] at a 3.40 kg wood block. It fires a 22.0 g bullet travelling at 460.0 m/s, which embeds in the block. Find the velocity of the bullet and the wood block after embedding.

    b)
    An 80.0 kg hockey goalie is gliding 0.230 m/s [W] when he catches an incoming 162 g puck travelling [W]. If the goalie then travels at 0.300 m/s [W], what is the velocity of the puck before he catches it?

    c)
    A 2575 kg truck travelling at 15.0 m/s right collides with a 1895 kg car reversing at 1.00 m/s to the left. The vehicles stick together after the collision.
    1. What is the velocity of the vehicles after the collision?
    2. If the crashed cars skid for 4.30 s after crashing, what is the average force of friction from the road acting on them?


  • 3.
    pi=pf:\bold{\sum\vec{p}_i = \sum\vec{p}_f}: Objects that explode apart
    a)
    A 68 000 kg rocket initially at rest in space fires a thruster which expels 43 kg of gas at 730 m/s [left]. Find the final velocity of the rocket.

    b)
    Two paddlers dock their canoe. The canoe is stationary when 72 kg paddler A steps out at 2.4 m/s forwards. This causes the canoe and 59 kg paddler B to move at a speed of 1.8 m/s. Find the direction that paddler B is moving in after paddler A steps out, and the mass of the canoe.

    c)
    A 3400 kg rail car carries a 200 kg cannon and a 15 kg cannonball along a level rail at 9.5 m/s [E]. The cannon fires the cannonball at 478 m/s [E]. Find the velocity of the rail car and cannon after firing.