Newton used the concept of momentum to explain the results of collisions Momentum = mass x velocity p = m v Units : p (kg m/s) = m (kg) x (m/s) Note since velocity is a vector quantity, (both magnitude and direction) then momentum is also a vector quantity.
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Newton used the concept of momentum to explain the results of collisions
Momentum = mass x velocity
p = m v
Units :
p (kg m/s) = m (kg) x (m/s)
Note since velocity is a vector quantity, (both magnitude and direction) then momentum is also a vector quantity
Momentum: Definition 2
Conservation of Momentum 1
Mass 75 kg
Velocity 4m/s
Mass 125 kg
Velocity ??? m/s
Mass 50 kg
Velocity 0m/s
When objects hit each other the resulting collision can be considered to be either elastic or inelastic. Momentum and total energy are always conserved in both cases.
Types of Collision 1
Elastic :
momentum conserved, kinetic energy conserved, total energy conserved
Inelastic :
momentum conserved, kinetic energy NOT conserved, total energy conserved
In an Inelastic collision some of the kinetic energy is converted to other forms of energy (often heat & Sound)
Two trolleys on an air track are fitted with repelling magnets. The masses are 0.1kg and 0.15kg respectively. When they are released the lighter trolley moves to the left at 0.24m/s. What is the velocity of the heavier trolley
A ball of 0.6kg moving at 5m/s collides with a larger stationary ball of mass 2kg. The smaller ball rebounds in the opposite direction at 2.4m/s
Calculate the velocity of the larger ball
Is the Collision elastic or inelastic. Explain your answer
Problems 1
Impulse of a Force
An object of constant mass m is acted upon by a constant force F which results in a change of velocity from u to v
From the 2nd lawF = (mv – mu )/t
Rearranging : Ft = mv – mu
Graphically.....
Force v Time Graphs
F
Area under graph “Ft” = change of momentum
force
time
t
Problems 1...
Problems 2...
Car Safety
We have seen that momentum is a vector quantity since it’s related to velocity which is a vector quantity. direction is important and therefore we need a “sign” convention to take this into account.
If we consider a ball with mass m hitting a wall and rebounding normally, (i.e. at 90°):
Rebound Impacts 1
Towards the wall we take as positive
Away from the wall we take as negative
Initial velocity = +u
Initial momentum = +mu
Rebound Impacts 2
Final velocity = -u
Final momentum = -mu
When the impact is oblique, (i.e. At an angle, not normally at 90°):
Rebound Impacts 3
Initial velocity = +u
Initial momentum = +mu
A squash ball is released from rest above a flat surface. Describe how the energy changes is i) it rebounds to the same height, ii) It rebounds to a lesser height
If the ball is released from a height of 1.20m and rebounds to a height of 0.9m show that 25% of the kinetic energy is lost upon impact
Problems 1...
Problems 2...
Repeat the last molecule question. This time the molecule strikes the surface at 60° to the normal and rebounds at 60° to the normal.
Problems 3...
Angles can be measured in both degrees & radians :
Arc
length
Radians & Degrees
The angle in radians is defined as the arc length / the radius
For a whole circle, (360°) the arc length is the circumference, (2r)
360° is 2 radians
r
Common values :
45° = /4 radians
90° = /2 radians
180° = radians
Note. In S.I. Units we use “rad”
How many degrees is 1 radian?
Angular velocity, for circular motion, has counterparts which can be compared with linear speed s=d/t.
Time (t) remains unchanged, but linear distance (d) is replaced with angular displacement measured in radians.
Angular Displacement
Angular displacement
r
Angular displacement is the number of radians moved
r
Consider an object moving along the arc of a circle from A to P at a constant speed for time t:
Angular Velocity : Definition
P
Definition : The rate of change of angular displacement with time
“The angle, (in radians) an object rotates through per second”
Arc length
r
A
r
= / t
Where is the angle turned through in radians, (rad), yields units for of rad/s
This is all very comparable with normal linear speed, (or velocity) where we talk about distance/time
The period T of the rotational motion is the time taken for one complete revolution (2 radians).
Angular Velocity : Period & Frequency
Considering the diagram below, we can see that the linear distance travelled is the arc length
Angular Velocity : linear speed
P
Arc length
r
A
r
Linear speed (v) = arc length (AP) / t
v = r /t
Substituting... ( = / t)
v = r
Angular Velocity : Worked example
Angular Velocity : Worked example
Angular Velocity : Worked example
Angular Velocity : Worked example
If an object is moving in a circle with a constant speed, it’s velocity is constantly changing....
Because the direction is constantly changing....
If the velocity is constantly changing then by definition the object is accelerating
If the object is accelerating, then an unbalanced force must exist
Velocity v
Centripetal Acceleration : Introduction
acceleration
Centripetal Acceleration : angular
Centripetal Force
Worked example 1
Worked example 1
Worked example 1
Problem 1
During the last lesson we saw that an object moving in a circle has a constantly changing velocity, it is therefore experiencing acceleration and hence a force towards the centre of rotation.
We called this the centripetal force: The force required to keep the object moving in a circle. In reality this force is provided by another force, e.g. The tension in a string, friction or the force of gravity.
Centripetal Acceleration : Recap
Consider a car with mass m and speed v moving over the top of a hill...
Over the top 1
S
mg
r
Over the top 2
Problems 1
Problems 2
Problems 1
Problems 2
Problems 3