4. CONSERVATION OF MOMENTUM (p_i = p_f)

2) Perfectly inelastic = Objects stick together after collision, and again, p is conserved but KE is not.

3) Elastic collision = both p and KE are conserved resulting in two equations simultaneously. (A game of pool is an ex of elastic collisions).

7. MORE DEMOS

1) balloon with air

2) gliders of equal mass colliding head-on and elastically exchange velocity

3) rotating stool

4) stress reliever (elastic collision)

5) Bounce and no bounce balls demonstrating elastic and inelastic collisions

6) Simulations: (Playing catch ball in space, Rocket motion, Collisions and the conservation of momentum, Inelastic car collisions, etc.)

1. MOMENTUM

A thought experiment: 1) “fast and slow” moving ball try to catch. 2) “toy car and real car” moving with same velocity and try to stop them. Consider the impact on you.

In above two thought experiments we see that both mass and velocity are factors that affect the motion of an object. We will then use these in a new concept to quantify motion. The linear momentum of an object of mass m moving with velocity v is defined:

p is a vector with the same direction as that of velocity v.

[p] = kgm/sec = Nsec

For 2-d motion:

P_x = mv_x

P_y = mv_y

2. IMPULSE - CAUSE OF MOMENTUM CHANGE

What is causing momentum change:

F = ma Þ F = m(v_f – v_i)/Dt Þ F = Dp/Dt = the time rate of change of momentum of an object equals the net force on the object

Impulse-momentum theorem:

FDt  = Impulse (is a vector)

[impulse] = Nsec = kgm/sec

The impulse of the force F, equals the change in momentum of the object. The impulse is the area under the force-time curve. The average force F provides the same impulse in Dt as the real-time varying force F.

Example 1

launching of a space shuttle.  (We will reconsider such as experiment from the conservation of momentum later)

Example 2

A 100-g ball is dropped from a height of h = 2m above the floor.  It rebounds vertically to a height of h' = 1.5m after colliding with the floor. Find the impulse on the ball by the floor by assuming no air resistance.

mgh = 1/2mv_b² Þ v_b = -Ö(2gh) = -6.26 m/sec

1/2mv_a² = mgh’ Þ v_a = Ö(2gh') = 5.42 m/sec

F t = mv_f – mv_i = 0.1[5.42 - (-6.26)] = 1.17 Kgm/sec

Applications of the impulse-momentum theorem:

1) dash boards

2) barrels filled with sand in median dividers

3) Styrofoam in packing

4) boxing gloves

5) jumping from a second story building on a concrete or on a sandy surface

In above examples: “increased collision time implies smaller force produces a particular momentum change.

3. CONSERVATION OF MOMENTUM

Proof:

F₁Dt = m₁v_1f – m₁v_1i

F₂Dt = m₂v_2f – m₂v_2i

F₁ = -F₂ Þ F₁ = – F₂ => F₁Dt = -F₂Dt => m₁v_1f – m₁v_1i = -(m₂v_2f – m₂v_2i) => m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f =>

=> p_i = p_f => conservation of momentum

Note that p_i and p_f are the total momentum of the system before and after collision.

Alternative proof of the conservation of momentum:

From the impulse-momentum theorem we have:

D(p₁ + p₂) = F_ext Dt

If the system is isolated then F_ext = 0. Therefore, D(p₁ + p₂) = 0 => (p₁ + p₂)_f = (p₁ + p₂)_i

which proves conservation of momentum.

Examples of momentum conservation:

1) air in a balloon (note the zigzag motion is due to the random motion of air atoms)

2)     Man walking on a giant skateboard on a frictionless surface

3) space shuttle

4) a game of pool

5) cars colliding

6) a squid propels itself by first filling its body with water and then expelling it with a high velocity

7) figure-skater (conservation of angular momentum)

8) astronauts move in space using a nitrogen propeller

Rotating stool analysis

By closing the hands a figure skater rearranges her mass in a way that her rotational inertia decreases and as a result her speed increases to conserve rotational momentum.

The analogous result in linear momentum is to decrease the mass of a moving object which results in an increase in the linear speed.

It is true assuming that mass somehow disappears without external forces such as e⁺ e^-

Diagram

Think of FDt = Dp Þ Dp = 0 Þ m_fv_f = m_iv_i

If F = 0(external force)

Think of a moving car where I can change its momentum when the driver throws a chair out of the window:

a)      if chair is thrown in direction of motion Þ speed is decreasing

b)      if chair is thrown against direction of motion Þ speed in increasing

c)      if chair is dropped momentum remains the same until the chair crashes on the ground and the external friction force makes the chair stop.

Example 3

A 50kg pitching machine placed on a frozen lake fires a 0.15kg baseball horizontally with a speed of 36m/sec left. What is the recoil velocity of the machine?

p_i = p_f Þ 0 = 50v + (0.15)(-36) Þ v = (0.15)(36)/50 Þ v = 0.11m/sec

4. COLLISIONS

We have seen that the total momentum is always conserved for any type of isolated collision.  However, the total KE is generally not conserved, because some of it is converted to thermal energy and internal elastic potential energy when the object deforms.

There are three types of collisions:

Inelastic collisions = one for which momentum is conserved but KE is not. 

Example: The collision of a rubber ball with a hard surface is inelastic, because some of the KE is lost when the ball is deformed during contact with the surface.

Perfectly inelastic = when the colliding objects stick together. Again, momentum is conserved but KE is not.

Elastic collision = one for which both momentum and KE are conserved.

Examples: 1) billiard ball collisions (assuming that sound energy = 0). 2) air molecules with walls of container at ordinary temperatures

Macroscopic collisions (such as billiard balls) are only approximately elastic because some permanent deformation occurs and some KE is lost.

Perfectly elastic collisions do occur on the microscopic level because the subatomic particles may collide without touching and hence no deformation occurs.  Or gravitational collisions.  (Actually even in microscopic collisions say between an e^- and an atom the collision may be inelastic because the incoming e^- may lose KE by inducing excitation in the atom without physical contact with it.)

An application of an inelastic collision is glaucoma testing = a puff of air is sent to the eye which is reflected and depending on the air’s return speed the pressure of the eye is determined.

Glaucoma = a disease in which the pressure inside the eye builds up and leads to blindness by damaging the cells. The pressure of the eye builds and makes the eye more rigid. A puff of air sent to the eye reflects upon it and its speed is measured determining the pressure in the eye.

Head-on collision = when objects move in the same or parallel (the reason I include the word parallel is understood in the Jupiter example below) line of motion before and after collision.

Examples of perfectly inelastic collisions

Example 4

A car of mass 1800kg stopped at a traffic light is struck from the rear by a 900kg car.  Cars stick together.  If the 900-kg car was moving with 20m/sec before the collision, what is the velocity of the entangled mass of the two cars just after the collision? How much KE was lost immediately after collision?

p_i + p_f Þ (900)20 + 0 = (900 + 1800)v Þ v = 1800/2700 = 2/3m/sec = 6.67m/sec

K.E._i = ½(900)20²

K.E._f = ½(900 + 1800)6.67²

Þ   K.E._i – K.E._f = lost

Example 5

A bullet is fired into a black of wood suspended from a light string.  The bullet is stopped by the wood and the entire system swings through the height h.  Find the speed of the bullet if: h = 0.05m

              m_wood = 1kg

              m_bullet = 5x10^-3kg

p_i = p_f Þ 5x10^-3v = (5x10^-3 + 1)V

½(5x10^-3 + 1)V² = (5x10^-3 + 1)gh Þ V = Ö2gh Þ v = 199m/sec

During the collision of bullet with ballistic pendulum, (note by during collision I mean while there is still a relative velocity between bullet and pendulum) the pendulum was experiencing a varying contact force and therefore started moving from zero speed to the V (final combined speed).  Thus it was rising to some height during collision but this small height it rised cannot be found unless we know details about the force it was experiencing during collision.  h in other words does not include the small height this pendulum rose during collision.

Also note that Dp_xtotal = 0 but not Dp_y ¹ 0 because along y-dir we have the external forces of the tension and weights.

Elastic collisions: Consider an elastic head-on collision. The diagram below shows one possible outcome.

1) Conservation of momentum:  m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f Þ m₁(v_1i – v_1f) = m₂(v_2f – v_2i) ------>(a)

Note that the above equation of conservation of momentum is written for a general situation and whether the velocities will be + or – will depend on an actual problem.

2) conservation of KE:  1/2m₁v_1i² + 1/2m₂v_2i² = 1/2m₁v_1f² + 1/2m₂v_2f² Þ m₁(v_1i² – v_1f²) = m₂(v_2f² – v_2i² Þ

Þ m₁(v_1i² – v_1f)(v_1i + v_1f) = m₂(v_2f – v_2i)(v_2f + v_2i) -----> (b)

Note concerning the above two equations: When the direction of motion is known, then we insert the proper sign for the velocities.  If it is not known we could either leave the unknown velocities as they appear in the equations. Or we could guess a direction.

Example 6 (Gravitational boosting flying by Jupiter)

A spacecraft is moving with a speed of 10km/sec and is approaching Jupiter which is moving in its orbit at an average speed of 13.1km/sec.  Find the speed of the spacecraft after the collision assuming that it leaves traveling in a direction “Opposite” (true of infinity) to its incoming velocity.

Note that this example assumes: same initial and final velocity of Jupiter and also head-on collision.  It is really head on because when the shuttle is at infinity it is moving parallel to the initial line of motion.

v_1i + v_1f = v_2f + v_2i Þ -13.1 – 13.1 = v_2f + 10 Þ v_2f = -36.2km/sec

More details:

y = x²  Þ dy/dx = 2x

x ® ¥ Þ dy/dx ® 0 Þ q = 90⁰

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

v_1i + v_1f = v_2i + v_2f Þ v_1f = v_2i + v_2f – v_1i

Þ 2m₁v_1i + v_2i(m₂ – m₁)/(m₁ + m₂) = v_2f Þ [2m₁v_1i – v_2i(m₁ – m₂)]/(m₁ + m₂) = v_2f

m₁ = mass of Jupiter >> m₂ = mass of shuttle

hence m₁±m₂ » m₁

Therefore

(2m₁v_1i – m₁v_2i)/m₁ » v_2f Þ 2v_1i – v_1i = v_2f Þ 2(-13.1) – 10 = v_2f Þ

Þ -36.2km/sec = v_2f

Example 7 (Elastic collision)

Two objects of equal mass m experience a head-on elastic collision. One object moves right with 30cm/sec and the other left with 20 cm/sec. Prove that they exchange velocities.

Conservation of momentum: m(30) – 20m = -mv₁ + mv₂ Þ 10 = v₂ – v₁

Conservation of energy: v_1i + v_1f = v_2f + v_2i Þ 30 – v₁ = v₂ – 20 Þ 50 = v₂ + v₁Þ 60 = 2v₂ Þ v₂ = 30cm/sec

v₁ = 20cm/sec

They exchange velocities.  Always true when objects of equal mass collide head-on elastically.

Second method:

30m – 20m = mv₁ + mv₂ Þ 10 = v₁ + v₂

v_1i + v_1f = v_2i + v_2f Þ 30 + v₁ = -20 + v₂ Þ 50 = -v₁ + v₂ Þ

Þ 60 = 2v₂ Þ v₂ = 30cm/sec

10 – v₂ = v₁ Þ 10 – 30 = v₁ Þ v₁ = -20cm/sec (moves left). It moves oppositely than what’s assumed in equations and picture.

Example 8 (Stress reliever)

Head-on elastic collision

If one ball collides at the left side, one ball will take off at the right with the same velocity.  Never one ball can cause two balls to take off because even though this does not violate conservation of momentum if we assume that each ball moves with v/2 (mv = mv/2 + mv/2), it violates conservation of KE because

1/2mv² ¹ 1/2m(v/2)² + 1/2m(v/2)² Þ

Þ v² ¹ (v/4)² + (v/4)² Þ v² ¹ v²/2

Glancing collisions

So far we considered head-on collisions. However another type of a collision is a glancing collision in which the objects rebound at some angle relative to the line of motion of the incident mass.

p_i = p_f Þ

åp_ix = åp_fx        and

åp_iy = åp_fy

Write above equations with respect to their angles in the x and y directions.

Example 9 Perfectly inelastic glancing car collision. Find the speed of the cars that stuck together.

25(1500) = (2500 + 1500)vcosq Þ 37500 = 4000vcosq

20(2500) = (2500 + 1500)vsinq Þ 50000 = 4000vsinq

Divide to get:

tanq = 1.33 Þ q = 53⁰

and