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Momentum and impulse Book page 73 - 79
© cgrahamphysics .com 2016
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
© cgrahamphysics .com 2016
• 𝑝=𝑚𝑣 and the change in momentum ∆𝑣=𝑚∆𝑣
• From Newton’s 2nd law 𝐹_𝑛𝑒𝑡=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
© cgrahamphysics .com 2016
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]
© cgrahamphysics .com 2016
V = - 7.4𝑚𝑠−1
+ 5.8 𝑚𝑠−1
Graphical interpretation
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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
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
© cgrahamphysics .com 2016
𝑣2 = 𝑢2 + 2𝑎𝑠
𝑣 = 2𝑎𝑠 = 2𝑥10 × 0.5 = 3.2𝑚𝑠−1
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
© cgrahamphysics .com 2016
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?
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
© cgrahamphysics .com 2016
• 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
© cgrahamphysics .com 2016
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
© cgrahamphysics .com 2016
Solution P = Fv = 800 x 2 = 1600W
∆𝑝 = 𝑚𝑣 = 400 × 2 = 800𝑘𝑔𝑚𝑠−1
Momentum in collisions
© cgrahamphysics .com 2016 Because they stick together, they must have equal velocity
Completely inelastic
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 • 𝑚1𝑢1
2 − 𝑚1𝑣12 = 𝑚2𝑣2
2 and 𝑚1𝑢1 − 𝑚1𝑣1 =𝑚2𝑣2 𝑚1 𝑢1
2 − 𝑣12 = 𝑚2𝑣2
2 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
© cgrahamphysics .com 2016
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
© cgrahamphysics .com 2016
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
© cgrahamphysics .com 2016
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
© cgrahamphysics .com 2016
Momentum before Momentum after
0 𝑚1𝑣1 + 𝑚2𝑣2
𝑚1𝑣1 + 𝑚2𝑣2 = 0 𝑚1𝑣1 = −𝑚2𝑣2
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
© cgrahamphysics .com 2016
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
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?
© cgrahamphysics .com 2016
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