Recap: For situations involving an impact or a collision, where large forces exists for a very small...

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Recap: •For situations involving an impact or a collision, where large forces exists for a very small time we define: Impulse = F x Δt (units: N.s) - a vector where F is the force and Δt is the time of action. •By Newton’s 2 nd law (F = m.a) we determined: Impulse = F. Δt = m. Δv or Impulse = Change in momentum (ΔP) Result: Large impulses cause large changes in motion! nservation of Momentum & Collision (Chapter 7) impulse (F. Δt) m P = m. Δv) momentum change

Transcript of Recap: For situations involving an impact or a collision, where large forces exists for a very small...

Page 1: Recap: For situations involving an impact or a collision, where large forces exists for a very small time we define:  Impulse = F x Δt (units: N.s) -

Recap:• For situations involving an impact or a collision, where

large forces exists for a very small time we define: Impulse = F x Δt (units: N.s) - a vector

where F is the force and Δt is the time of action.• By Newton’s 2nd law (F = m.a) we determined: Impulse = F. Δt = m. Δv

or Impulse = Change in momentum (ΔP)

Result: Large impulses cause large changes in motion!

Conservation of Momentum & Collisions(Chapter 7)

impulse(F. Δt)

m(ΔP = m. Δv)

momentum change

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Momentum (P) is the product of the mass of an object and its velocity:

P = m.v (units: kg.m/s)• Momentum is a vector acting in same direction as velocity

vector. Example 1: A 100kg boulder rolling towards a castle gate

at 3m/s.Momentum boulder: P = m.v = 100 x 3 = 300 kg.m/sExample 2: A 1 kg missile flying towards the castle gate at 300 m/s (speed of sound).Momentum missile: P = m.v = 1 x 300 = 300 kg.m/s

• Result: Different objects can have the same momentum (quantity of motion)

• But it’s the change in momentum of an object that is important as this equals the impulse:

ΔP = m. Δv = impulse

Momentum

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An impulse acting on an object causes a change in its momentum.

• The change in momentum is equal in magnitude and direction to the applied impulse.

• Impulse = m. Δv• A new way of looking at Newton’s 2nd law!Example: A 50kg rock is hurled by a giant catapult with a

force of 400 N applied for 0.5 sec.Impulse = F. Δt = 4000 x 0.5 = 2000 N.s

Thus: Change in momentum: = 2000 Ns = m. Δv or Δv = = 40 m/s ( ≈ 140 km/hr)Note: When the initial velocity is zero:

• change in momentum = objects momentum• change in velocity = objects velocity

impulseF. Δt

mΔv

200050

Impulse - Momentum Principle

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Ex: Ball’s Change in Momentum1. Assume no energy is lost,

therefore KE of the ball is the same before and after impact. (KE = ½m.v2)

2. On impact the momentum of ball is decreased to zero.

P2=m.v

ΔP

before during after

impulse

P1=m.v

3. The total change in momentum ΔP is:

ΔP = P2 – P1 = m.v – (- m.v)

ΔP = 2 m.v• The impulse required to change the direction of ball is therefore equal to twice momentum of the impacting ball.

(ie. Twice as large as what is needed to simply stop the ball.)(As P1 opposite to P2)

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Summary:• Rewriting Newton’s 2nd law shows that:

Impulse = Change in momentum (ΔP)

Impulse = F. Δt = m. Δv

• Large impulses produce large changes in momentum -resulting in large velocity changes.

Where:

Impulse = F x Δt (units: N.s) - a vector

Momentum: P = m.v (units: kg.m/s) - a vector.

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Conservation of Momentum• A new principle for studying collisions which results from

Newton’s 3rd law applied to the impulse/ momentum equation.

• Conservation of momentum enables us to understand collisions and to predict many results without a detailed knowledge of time varying forces.

Consider: Two sticky objects moving towards each other…they meet in mid-air and after colliding stick together and move as one body.

• During the moment of impact there is a strong force acting for a time Δt.

• By Newton’s 3rd law, an equal and opposite force ‘F’ acts back (remember forces occur in pairs).

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• As Δt is the same for both forces, the impulses they produce (= F.Δt) are the same magnitude but in opposite directions.

• By Newton’s 2nd law:

Impulse = Change in momentum (ΔP = m. Δv)- Thus the change in momentum experienced by both objects must be the same…but in opposite directions.

- The total change in momentum of the system (i.e. both objects combined) is therefore ZERO!

- In other words , the total momentum of the system is conserved (i.e. changes of momentum within system cancel each other out).

v2v1

- impulse impulse

Conservation of Momentum

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Conservation of Momentum The total momentum of the system is

conserved (if no other external forces acting).• However, different parts of the system can exchange

momentum (but the total remains the same).

Note: If a net external force acts on system, then it will accelerate and its momentum will change.

• Conservation of momentum allows us to examine interesting impact situations…

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Example: Custard pie fights!

vc = -10 m/s vt = 0.5 m/s

Pc = mc.vc Pt = mt.vt

Pc = -1 x 10 = -10 kg.m/s Pt = 100 x 0.5 = 50 kg.m/s (negative sign as opposite direction)

Total momentum = Ptarget + Pcustard pie = 50 - 10 = 40 kg.m/s

Question: What is the velocity of target and pie after impact?

Total mass = mt + mc = 101 kg Total momentum = 40 kg.m/s (P = m.v) v = P / m = 40 / 101 ≈ 0.4 m/s

Result: The unsuspecting target has a larger initial momentum, so his direction of motion prevails but the pie reduces his forward velocity (briefly)!

mc = 1 kg mt = 100 kg

positive direction

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• A special case of conservation of momentum when the initial velocity of the interacting bodies is often zero.E.g.: - two ice skaters “pushing off” - firing a gun - rocket propulsion…

• We have already looked at what happens to the ice skaters motion (using Newton’s 3rd law), but now we can use conservation of momentum to determine their velocities…

Example: Initial momentum = 0

Thus: total momentum after “push off” = 0

The momentum of each person must therefore be equal but opposite in direction and P2 = -P1.

But, as P1 = m1.v1 and P2 = m2.v2, the velocities will be in opposite directions and will depend on their

masses.

Eg. If m1 = 3m2 then v2 will be 3v1 in opposite direction!

Recoil

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Momentum of bullet = Momentum of gun m1v1 = - m2v2

• Mass of bullet is small but its velocity is high…creating a large recoil.

• To reduce velocity of recoil (v2), hold gun with locked arms so the mass m2 becomes mass of gun + your body.

• Similarly a very massive cannon will “jump back” much less than a light one…for the same shot.

Rocket propulsion: • Exhaust gasses have large momentum (light molecules but

very high velocity).• Momentum gained by rocket in forward direction equals

momentum of exhaust gasses in opposite direction.• This is why rockets (i.e. recoil) work in outer space… as

gasses and rocket push against each other as gasses expelled.

Firing a Gun (initial momentum = zero)

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• Two main types: Elastic and Inelastic…

• Different kinds of collisions produce different results…e.g. sometimes objects stick together and other times they bounce apart!

• Key to studying collisions is conservation of momentum and energy considerations…

Questions:

- What happens to energy during a collision?

- Is energy conserved as well as momentum?

Collisions

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• E.g. Two objects collide head on and stick together, moving as one after collision (only one final momentum / velocity to compute).

• Ignoring external forces (which are often low compared with large impact forces), we use conservation of momentum.

E.g. Coupling train trucks (low rolling friction)

Before: System momentum: mv = 2x104x10 = 2x105 kg.m/sAfter: Final system mass = (20+10+5+15)x103 kg = 50x103 kg

As final momentum = initial momentum

vfinal =

• Thus total momentum of system has remained constant but the colliding truck’s velocity has reduced (i.e. momentum shared).

Perfectly Inelastic Collisions (Sticky ones!)

2x104 kg 104 kg 5x103 kg 15x103 kg

v = 10 m/s 3 stationary trucks

Pfinal

total mass=

2x105

5x104= 4 m/s

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Question: What happens to the energy of this system?

Total energy = Kinetic Energy = ½.m.v2 (i.e. no PE change)

Before impact: KEtot = KEtruck + KE3trucks

= ½(2x104)(10)2 + 0 = 106 Joules (1MJ)

After impact: KEtot = ½.mtot.v2tot = ½(5x104)(4)2

= 2x105 JEnergy differences = (10 - 2)x105 J = 8x105 J (i.e. 80% loss)

Results: Energy is lost in an inelastic collision (heat, sound…) and the greatest portion of energy is lost in a perfectly inelastic collision when objects stick together!

Extreme example:Pie (or bullet) hitting a wall… All KE is lost on impact!

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• If objects bounce off one another rather than sticking together, less energy is lost in the collision.

• Bouncing objects are called either “elastic” or “partially inelastic”. The distinction is based on energy.

Elastic Collisions:• No energy is lost in an elastic collision. E.g. A ball bouncing off a wall / floor with no change in its

speed (only direction). Partially Inelastic Collisions:• In general most collisions are “partially inelastic” and

involve some loss of energy… as they bounce apart.• Playing pool:• Very little energy is lost when balls hit each other and the

collision is essentially elastic. In such cases: Momentum and energy are conserved.

Bouncing Collisions

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• In an elastic collision we need to find the final velocity of both colliding objects.• Use conservation of momentum and conservation of energy considerations…

Example:

• Answer: The cue ball stops dead on impact and red ball moves forward with the same velocity (magnitude and direction) as that of the cue ball prior to impact!

• Why?...Because both KE(= ½.m.v2) and momentum (m.v) are conserved on impact.• As the masses of both balls are the same the only solution to conserve both KE and

momentum is for all the energy and momentum to be transferred to the other (red) ball.• It’s a fact…try it for yourself!!!

Question: What happens to red ball and cue ball?

P1

P2

cue ball(no spin)

Page 17: Recap: For situations involving an impact or a collision, where large forces exists for a very small time we define:  Impulse = F x Δt (units: N.s) -
Page 18: Recap: For situations involving an impact or a collision, where large forces exists for a very small time we define:  Impulse = F x Δt (units: N.s) -

Impulse is the average force acting on an object multiplied by its time interval of action.

Impulse = F . Δt (units = N.s)

Note: Since instantaneous force may vary during impact we must use average force.

• Impulse is a vector acting in the direction of average force• The larger the force (F) and the longer it acts (Δt) the

larger the impulse. ( Impulse is therefore a measure of the overall effect of the force.)However: Impulse = F. Δt = m. Δv

• So an impulse causes a change in velocity (Δv) in magnitude and direction.

• In Newton’s words the product “m.Δv” is the change in the “quantity of motion”.

• We now term this the change in momentum.

Impulse

Page 19: Recap: For situations involving an impact or a collision, where large forces exists for a very small time we define:  Impulse = F x Δt (units: N.s) -

Other forms of transport: - Aircraft move by pushing against air.

- Boats push against sea. - Walking / driving pushing against land. - Rockets: self contained and pushes against its own exhaust gasses.

Note: - Rockets gain momentum gradually (rather than

through a single brief impulse)… but can be thought of as a continuous series of small impulses.

- As their mass decreases during flight, the resultant velocity is more difficult to calculate.