Momentum and impulse Book page 73 - 79

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Definition • The rate of change of linear momentum is directly

proportional to the resultant force acting upon it and takes place in the direction of the resultant force

• 𝑝 = 𝑚𝑣 and the change in momentum ∆𝑣 = 𝑚∆𝑣

• From Newton’s 2nd law 𝐹𝑛𝑒𝑡 = 𝑚𝑎 and

𝑎 =𝑣−𝑢

∆𝑡 we get the expression

𝐹 =𝑚(𝑣 − 𝑢)

∆𝑡=

𝑚∆𝑣

∆𝑡=

∆𝑝

∆𝑡

• The longer the time of collision, the smaller the force exerted

• 𝐹∆𝑡 = 𝑚∆𝑣 = ∆𝑝

• This quantity is called Impulse J

• Momentum is a vector quantity

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Example • A ball of mass 0.25kg is moving to the right at a speed of 7.4𝑚𝑠−1.

It strikes a wall at 900 and rebounds from the wall with a speed of 5.8𝑚𝑠−1. Calculate the change in momentum. Solution

• ∆𝑝 = 𝑚 𝑣2 − 𝑣1

• ∆𝑝 = 0.25 ( 5.8 – (- 7.4)) = 3.3𝑘𝑔𝑚𝑠−1 [left]

V = - 7.4𝑚𝑠−1

+ 5.8 𝑚𝑠−1

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Graphical interpretation

Real graph Idealized graph

• The area under the graph = 𝐹∆𝑡= Impulse If we know the impulse we can find the change in speed

• Impulse J = 𝑚∆𝑣

∆𝑣 =𝐽

𝑚

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Example • Water is poured from a height of 0.50 m on to a top pan balance at

the rate of 30 litres per minute. Estimate the reading on the scale of the balance.

Solution

• h = 0.50m rate = 30l /min = 30kg / min = 0.6 kg / sec

• Assumption: water bounces off the top of the balance horizontally

• 𝑚𝑔ℎ =1

2𝑚𝑣2 OR

𝑣 = 2𝑔ℎ = 2 × 10 × 0.5 = 3.2𝑚𝑠−1

• ∆𝑝 = 𝑚∆𝑣 = 0.5 × 3.2 = 1.6𝑁

• Reading on scale = m = 𝑊

𝑔=

1.6

10= 0.160𝑘𝑔

𝑣2 = 𝑢2 + 2𝑎𝑠

𝑣 = 2𝑎𝑠 = 2𝑥10 × 0.5 = 3.2𝑚𝑠−1

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Example

• 𝐾𝐸 =1

2𝑚𝑣2 =

𝑚×𝑚×𝑣2

2𝑚

KE = 𝑝2

2𝑚

• The graph right shows how the momentum of an object of mass 40kg varies with time

• A) calculate the force acting on the object

• B) The change in KE over the 10s of motion

Solution

Gradient = ∆𝑝

∆𝑡= 𝐹 =

200−0

10−0= 20𝑁

∆𝐾𝐸 =𝑝2

2𝑚=

2002

2 × 40= 500𝐽

Can you think of examples where theory of impulse has been used to good effect?

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Law of conservation of momentum • In a closed system, linear momentum is always conserved • Closed system: no external forces are acting • If the net force of a system is zero, then there is no change of

momentum of the system • Momentum before = momentum after • The momentum gained by one object is equal to the momentum lost

by another object

• 3rd law: 𝐹𝐴𝐵 = −𝐹𝐵𝐴 • 2nd law: 𝑚𝐴𝑎𝐴 = −𝑚𝐵𝑎𝐵

•𝑚(𝑣𝐴−𝑢𝐴)

∆𝑡= −

𝑚(𝑣𝐵−𝑢𝐵)

∆𝑡

• Since time is the same for both • 𝑚𝐴𝑣𝐴 − 𝑚𝐴𝑢𝐴 = 𝑚𝐵𝑢𝐵 − 𝑚𝐵𝑣𝐵 • 𝑚𝐵𝑢𝐵 + 𝑚𝐴𝑢𝐴 = 𝑚𝐴𝑣𝐴 + 𝑚𝐵𝑣𝐵 • Momentum before = momentum after • This approach cannot always be used

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Example

• Sand is poured vertically at a constant rate of 400 kg s-1 on to a horizontal conveyor belt that is moving with constant speed of 2.0 m s-1.

• Find the minimum power required to keep the conveyor belt moving with constant speed.

• Rate = 400 kg s-1

• 𝑣𝑏𝑒𝑙𝑡 = 2.0 m s-1

• Force on belt = 800N = force of friction between sand and belt

• This force accelerates the sand to the speed of the surveyor belt

Solution P = Fv = 800 x 2 = 1600W

∆𝑝 = 𝑚𝑣 = 400 × 2 = 800𝑘𝑔𝑚𝑠−1

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Momentum in collisions

Because they stick together, they must have equal velocity

Completely inelastic

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Case 5: Head on collisions of two identical masses where one is at rest perfect elastic collision

• 𝑚1𝑢1 + 0 = 𝑚1𝑣1 + 𝑚2𝑣2

• KE is conserved: 1

2 𝑚1𝑢1

2 + 0 = 1

2 𝑚1𝑣1

2 + 1

2 𝑚2𝑣2

2

• 𝑚1𝑢12 = 𝑚1𝑣1

2 + 𝑚2𝑣22

𝑚1𝑢1 = 𝑚1𝑣1 + 𝑚2𝑣2

• Rearrange

• 𝑚1𝑢12 − 𝑚1𝑣1

2 = 𝑚2𝑣22 and 𝑚1𝑢1 − 𝑚1𝑣1 = 𝑚2𝑣2

𝑚1 𝑢12 − 𝑣1

2 = 𝑚2𝑣22 and 𝑚1(𝑢1−𝑣1) = 𝑚2𝑣2

• Take ratios

•𝑚1 𝑢1

2−𝑣12

𝑚1(𝑢1−𝑣1)=

(𝑢1−𝑣1)(𝑢1+𝑣1)

(𝑢1−𝑣1)=

𝑚2𝑣22

𝑚2𝑣2

• 𝑣2 = 𝑢1 + 𝑣1 and 𝑣1 = 𝑣2 − 𝑢1

• Substitute back into 𝑚1𝑢1 − 𝑚1𝑣1 = 𝑚2𝑣2

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continued • 𝑚1𝑢1 − 𝑚1𝑣1 = 𝑚2𝑣2

• 𝑣2 = 𝑢1 + 𝑣1 and 𝑣1 = 𝑣2 − 𝑢1

• 𝑚1𝑢1 − 𝑚1𝑣1 = 𝑚2(𝑢1 + 𝑣1) and 𝑚1𝑢1 − 𝑚1(𝑣2 − 𝑢1)= 𝑚2𝑣2

• 𝑚1𝑢1 − 𝑚2𝑢1 = 𝑚1𝑣1 + 𝑚2𝑣1 and 2 𝑚1𝑢1 = 𝑚1𝑣2 + 𝑚2𝑣2

• 𝑢1(𝑚1 − 𝑚2) = 𝑣1(𝑚1 + 𝑚2) and 2 𝑚1𝑢1 = 𝑣2(𝑚1 + 𝑚2)

• 𝑣1 =𝑚1−𝑚2

𝑚1+ 𝑚2× 𝑢1

• 𝑣2 =2 𝑚1𝑢1

𝑚1+ 𝑚2

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Inelastic collisions – energy lost • Energy is lost

• 1 mass is stationary

• 𝑚1𝑢1 = (𝑚1+𝑚2)𝑣1

• 𝑣1 =𝑚1𝑢1

𝑚1+𝑚2

• Energy loss:

𝐸𝑖 =1

2 𝑚1𝑢1

2

• 𝐸𝑓 =1

2 (𝑚1+𝑚2)𝑣1

2 =1

2(𝑚1+𝑚2)

𝑚12𝑢1

𝑚1+𝑚22

2

=1

2

𝑚12

𝑚1+𝑚2𝑢1

2

• Take ratio of energies

•𝐾𝐸 𝑖𝑛𝑖𝑡𝑖𝑎𝑙

𝐾𝐸 𝑓𝑖𝑛𝑎𝑙=

1

2 𝑚1𝑢1

2

1

2

𝑚12

𝑚1+𝑚2𝑢1

=(𝑚1+𝑚2)

𝑚1

• If we know the mass, we can find the ratio of energies

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Two initially stationary masses gain energy

• If the masses are equal, then the speeds are the same with one velocity the negative of the other

• If the masses are not equal, then

𝑚1

𝑚2= −

𝑣2

𝑣1

Momentum before Momentum after

0 𝑚1𝑣1 + 𝑚2𝑣2

𝑚1𝑣1 + 𝑚2𝑣2 = 0 𝑚1𝑣1 = −𝑚2𝑣2

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Example • A railway truck B of mass 2000Kg is at rest on a horizontal track. Another

track A of the same mass is moving with a speed of 5.0𝑚𝑠−1 collides with the stationary truck and they link up and move together. Find the speed with which the two trucks move off and the loss of KE on collision

• 10000 = 4000v v = 2.5𝑚𝑠−1

• 𝐾𝐸𝑏𝑒𝑓𝑜𝑟𝑒 =1

2𝑚𝑣2 =

1

2× 2000 × 52 = 25000𝐽

• 𝐾𝐸𝑎𝑓𝑡𝑒𝑟 =1

2(𝑚1 + 𝑚2)𝑣2 =

1

2× 4000 × 2.52 = 12500𝐽

• ∆𝐾𝐸 = 25000 − 12500 = 12500𝐽

Momentum before Momentum after

2000 x 0 (𝑚1 + 𝑚2) v

2000 x 5 (2000+2000)v

Energy lost - Dissipated to

surroundings - Sound

- Most heat up coupling b/w trucks

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Example • A billiard ball of mass 100g strikes the cushion of a billiard

table with 10𝑚𝑠−1 at an angle of 450 to the cushion. It rebounds at the same speed and angle to the cushion. What is the change of momentum of the billiard ball?

45 45

mu mv

∆𝑝 = 𝑚𝑣 − 𝑚𝑢 ∆𝑝 is a vector need vector addition

∆𝑝 = 𝑚𝑣 + (−𝑚𝑢) Reverse mu

-mu

mv

R

𝑅 = 𝑚𝑣 2 + 𝑚𝑢 2 = 𝑚2(𝑣2 + 𝑢2)

= 0.1 102 + 102 = 0.2 200 = 1.4𝑘𝑔𝑚𝑠−1 At 900 away from the cushion

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