2. RE-CAP from last week…
• We discussed vector and scalar quantities
• A scalar quantity is one that has only the magnitude
• A vector quantity is on that consists of the magnitude AND direction
• Examples of Scalars: length, area, volume, speed…etc
• Examples of vectors: displacement, velocity, acceleration…etc
4. By the end of this lesson, you should be able
to:
• Define Momentum and Define Impulse
• Calculate impulse and momentum using the correct formulas
• Apply impulse to safety consideration in everyday life
5. MOMENTUM
• Momentum is the product of the mass of a particle and its velocity
• Momentum: p= mv, where m is the mass of the object in kg and v is
the speed at which the object is moving in m/s−1
• Unit: kgms-1
• Momentum calculations: When doing calculations concerning
momentum, we must keep the vector nature of momentum in mind.
A 5000kg bus moves due west at 10 m•s-1.
• Find its momentum.
• p = mv = 5000 x 10 = 50 000 kg •m•s-1
6. Change in momentum
• If a non-zero net force acts on an object, then its velocity changes and
so does its momentum.
• The change in momentum of a body is the difference between its
final and its initial momentum.
• We calculate the change in momentum of a body with the equation:
∆p = mΔv = m(vf –vi)
7. Worked example
• A batsman hits a 160g cricket ball approaching him at 30 m•s-1 so that it
moves in the opposite direction at 40 m•s-1.
• Calculate the change in momentum of the ball.
• Take approach as positive
• m = 0,16 kg
• vi = 30 m•s-1 and vf = -40 m•s-1
• ∆p = m(vf –vi)
• = 0,16 ( -40 -30)
• = - 11,2 kg•m•s-1
• = 11,2 kg•m•s-1 away from the batsman
8. Your Turn!!!!
• 1. Calculate the momentum of a body if its mass is 5 kg and its
velocity is 25 m•s-1 due east.
• 2. A 1100 kg car accelerates along a straight road and its velocity
increases from 10 m•s-1 to 20 m•s-1. Calculate its change in
momentum.
10. IMPULSE
• Isaac Newton realized that the larger the change of momentum and
the shorter the time that it takes place in, the larger the force that is
necessary to bring it about.
• In terms of Newton’s Second Law of Motion:
• The rate of change of momentum of a body is directly proportional to
the net force acting on it and the change of momentum is in the
direction of the force.
11. IMPULSE CONTINUED…
• ∆Fnet = ∆p ∆t Rearranging, Fnet∆t = ∆p The product Fnet∆t is called the
impulse.
• Impulse has the unit N•s
• Worked example The driver of a car of mass 1200 kg travelling at 20 m•s-1
along a straight road, slows down uniformly for 4s. The brakes apply a net
force of 4800 N.
• a) Calculate the impulse exerted by the brakes. Impulse = Fnet∆t = -4800 x
4 = -19200 N•s
• b) What is the change in momentum of the car? Change in momentum =
impulse = -19200 kg•ms-1
• c) Calculate the velocity, vf, of the car after 4s. Fnet∆t = m(vf – vi) -19200 =
1200(vf -20) vf = 4 m•s-1
12. IMPULSE & MOTOR VEHICLE SAFETY
• During a collision in which a car is brought to rest, the people in the
car must also be brought to rest.
• From the impulse equation (Fnet∆t = ∆p) we can see that if it
increases, the net force acting on the people to bring about the
required decrease in momentum is smaller thus injury is reduced.
• During a front-end collision the front of the car is designed to
crumple. While the front crumples the strong central cab, with the
people in it, is able to move for a little longer while slowing down.
This increases it thus decreasing the force.
• When an airbag is deployed, it begins to exert a retarding force on the
person. As the air in the airbag compresses it is able to exert this force
over a period of time thus decreasing the required force
14. ASSESSMENT TASK
In groups of 3, do practical 1 on our practical manual, attached below, to
be submitted on the 30th of September.
https://nect.org.za/materials/physical-sciences/term-2/practical-
booklets/gr-12-term-2-2018-ps-practial-booklet-b.pdf
