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Summary Grade 12 IEB Physical Science: Momentum, Impulse, Work, Energy and Power CA$6.74   Add to cart
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Summary Grade 12 IEB Physical Science: Momentum, Impulse, Work, Energy and Power

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  • Physical Sciences
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  • 12th Grade
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  • Physical Sciences

These notes cover Section C as per the IEB Physical Science SAGS. Included are comprehensive notes and annotated examples on Linear Momentum (1D), Newton's Second Law expressed in terms of Momentum, Conservation of Momentum and Elastic and Inelastic Collisions, Impulse and Work, Energy and Power. A...

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  • Momentum, impulse, work, energy and power
  • July 11, 2023
  • July 18, 2023
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  • 2022/2023
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MOMENTUM, IMPULSE, WORK,
ENERGY AND POWER




@STUDYNOTESBYM

, C. MOMENTUM, IMPULSE, WORK, ENERGY AND POWER



1. LINEAR MOMENTUM (1D)

Linear momentum is the product of the mass and velocity of the object



p=mv
p: momentum measured in 𝑘𝑔. 𝑚. 𝑠 −1

m: mass measured in 𝑘𝑔

v: velocity measured in 𝑚. 𝑠 −1



• Momentum is a vector quantity (has a magnitude and direction) and is in the same
direction as the velocity
e.g.) if velocity is acting right, momentum will also act right


CALCULATING LINEAR MOMENTUM:
Example: A 500kg car drives along a city street at 13.4m/s. What is the car’s momentum?
Let forward be positive
1. Assign a positive
𝑝 = 𝑚𝑣 direction
= (500)(13.4) 2. Use the formula to
= 6700𝑘𝑔. 𝑚. 𝑠 −1 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 calculate the
magnitude of the
momentum
3. State a direction

, 2. NEWTON’S SECOND LAW EXPRESSED IN TERMS OF MOMENTUM
(THE CHANGE IN MOMENTUM - ∆𝑃 )

The net force acting on an object is equal to the rate of change of momentum




∆𝒑
𝑭𝒓𝒆𝒔 =
∆𝒕
When an object experiences a resultant force, it will accelerate. Because it
accelerates, the velocity changes therefore the momentum of the object will
change:

∆𝒑 = 𝒎(𝑽𝒇 − 𝑽𝒊)
Therefore:


𝒎(𝑽𝒇 − 𝑽𝒊)
𝑭𝒓𝒆𝒔 =
∆𝒕


Example:
A 500g ball, travelling at 14m/s, is kicked. The ball moves off in the opposite direction at a speed
of 18m/s. If the ball was in contact with the kicker’s foot for 0.05s, calculate the resultant force
exerted on the ball
∆𝑝
𝐹𝑟𝑒𝑠 =
∆𝑡


𝑚(𝑉𝑓 − 𝑉𝑖 )
=
∆𝑡
0,5(−18 − 14)
=
0,05
= −320
= 320𝑁 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑘𝑖𝑐𝑘𝑒𝑟 ′ 𝑠 𝑓𝑜𝑜𝑡

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