mechanics - PHSI

LECTURE 1 - KINEMATICS

• Question from 2005 PHSI 110 final exam:

  • Car A travels at a constant velocity of $15 \text{ m s}^{-1}$.

  • Car B is stationary at $t = 0$.

  • Car B accelerates at a constant rate of $0.1 \text{ m s}^{-2}$.

  • Question: At what speed will Car B be traveling as it passes Car A (in $\text{m s}^{-1}$)?

    • (A) 10 (B) 15 (C) 25 (D) 30 (E) Car B will never pass Car A as the acceleration is too small.

• In 2005, less than 50% of students got the question right, indicating guessing or lack of understanding.

Kinematics – Learning Goals

• What is kinematics and why is it important?

• Fundamental Kinematic Quantities

  • Time, distance, speed, acceleration.

• Vector quantities

  • Displacement, velocity, acceleration.

• Average and instantaneous quantities

• Relative displacement, velocity, acceleration.

Kinematics

• Description of motion

  • What quantities are needed to adequately describe motion?

• Motion = change in position with time

  • Need to measure time and position

• Rates of change

  • position

  • rate of change of the rate of change?

Time, Distance, Speed

• Example: A car moves along a straight road at $5 \text{ m s}^{-1}$.

  • How far does it go in 3 seconds?

  • The car moves 5 m in each second so in 3 seconds it moves 15 m.

  • Time Position

    • 0 s 0 m

    • 1 s 5 m

    • 2 s 10 m

    • 3 s 15 m

• The speed of an object is the rate of change of the position of that object in a given time.

• To find the distance an object travels in a given time if it’s traveling at a constant speed,

  • $v = \frac{\Delta x}{\Delta t}$

  • $\Delta x = v \Delta t$

• For the car traveling at $5 \text{ m s}^{-1}$ for 3 seconds

  • $\Delta x = 5 \text{ m s}^{-1} \times 3 \text{ s} = 15 \text{ m}$

  • Δ means “change in”.

Average Speed

• This equation gives the average speed,

  • $v_{av} = \frac{\Delta x}{\Delta t} = \frac{\text{Total distance travelled}}{\text{Total time it takes}}$

• This is not the speed of an object at a particular time.

• At constant acceleration, the average speed is also given by,

  • $v<em>{av} = \frac{v</em>f + v_i}{2}$

Instantaneous Speed

• The speed of an object at a particular time is called the instantaneous speed.

• The average speed gets closer to the instantaneous speed as we reduce Δx and Δt.

• Example

  • If you drive 300 km to Christchurch in 5 hours, what is your average speed?

    • You may have traveled at a constant $60 \text{ km h}^{-1}$ the whole way or you might have travelled at $40 \text{ km h}^{-1}$ up the Kilmog, had a half hour lunch in Oamaru, then driven at $90 \text{ km h}^{-1}$ across the plains….

    • This will not change the average speed for the trip to Christchurch.

  • Your instantaneous speed is the speed at which you are traveling at a particular instant, e.g., at exactly 12:30 on the day of your trip.

  • $v_{av} = \frac{\text{Total distance travelled}}{\text{Total time it takes}} = \frac{300 \text{ km}}{5 \text{ h}} = 60 \text{ km h}^{-1}$

Acceleration

• A car starts from rest ($v_i = 0$) and increases its speed along a straight road by $2 \text{ m s}^{-1}$ in every second.

  • How fast will it be going after 4 seconds?

  • This is an acceleration of 2 metres per sec per sec. and is written as $a = 2 \text{ m s}^{-2}$ (or $2 \text{ m/s}^2$)

  • This acceleration is called uniform (or linear) because the change in velocity per second is always the same: $2 \text{ m s}^{-1}$.

  • Time Speed

    • 0 s 0 m s-1

    • 1 s 2 m s-1

    • 2 s 4 m s-1

    • 3 s 6 m s-1

    • 4 s 8 m s-1

• Acceleration is the rate of change of the speed

• For a car, accelerating from rest for 4 s, with $a = 2 \text{ m s}^{-2}$,

• To find the change in speed for a given uniform acceleration in a given time,

  • $a = \frac{\Delta v}{\Delta t}$

  • $\Delta v = a \Delta t$

  • $\Delta v = 2 \text{ m s}^{-2} \times 4 \text{ s} = 8 \text{ m s}^{-1}$

Velocity-Time graph Example

• A train increases speed uniformly (constant acceleration) from $3 \text{ m s}^{-1}$ to $11 \text{ m s}^{-1}$ over 100 s.

  • What is the average speed of the train?

    • $v<em>{av} = \frac{v</em>f + v_i}{2} = \frac{11 + 3}{2} = 7 \text{ m s}^{-1}$

  • How far does the train travel in 100 s?

    • $\Delta x = v_{av} \Delta t = 7 \text{ m s}^{-1} \times 100 \text{ s} = 700 \text{ m}$

Uniform Acceleration

• If an object has a uniform acceleration,

• Or If the object starts from rest,

  • $\Delta x = v<em>{av} \Delta t = \frac{1}{2} (v</em>i + v<em>f) \Delta t = \frac{1}{2} (v</em>i + v_i + a \Delta t) \Delta t$

  • $\Delta x = v_i \Delta t + \frac{1}{2} a \Delta t^2$

  • $\Delta x = \frac{1}{2} a \Delta t^2$

Vector Quantities

• Displacement is the name for the vector which describes both the distance and the direction to the object.

  • e.g., 4 m east, 3 m north

• Velocity is the rate of change of the displacement

  • e.g., $12 \text{ km h}^{-1}$ south, $10 \text{ m s}^{-1}$ NE, $5 \text{ m s}^{-1}$ NE

• Acceleration is the rate of change of the velocity

• Deceleration is an acceleration in the opposite direction to the velocity vector

Adding Velocities

• Suppose an aeroplane can fly $100 \text{ km h}^{-1}$ in still air and the wind travels at $30 \text{ km h}^{-1}$.

  • How fast does the aeroplane travel?

    • Wind and Plane in the same direction

      • Resultant $130 \text{ km h}^{-1}$

    • Wind and Plane in the opposite direction

      • Resultant $70 \text{ km h}^{-1}$

• A bird, which flies at $3 \text{ m s}^{-1}$ in still air, is pointing due north while the wind blows at $4 \text{ m s}^{-1}$ to the east.

  • What is the resultant velocity of the bird?

    • $d^2 = 3^2 + 4^2 = 25$

    • $\therefore d = 5 \text{ m}$

    • $\therefore \theta = 37^o \text{ N. of E.}$

    • $tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}$

    • The resultant velocity is $5 \text{ m s}^{-1}$ at $\theta = 37^o \text{ N. of E.}$

Acceleration due to Gravity

• Every object in free fall has its downward speed increased by $9.8 \text{ m s}^{-1}$ in every second:

  • $a = g = 9.8 \text{ m s}^{-2} \approx 10 \text{ m s}^{-2}$

  • This quantity, g, is called the acceleration due to gravity.

• $\Delta v = a \Delta t$

• $\Delta v = g \Delta t$

• Example: after 5 s of free fall from rest

  • $\Delta v = g \Delta t = 10 \text{ m s}^{-2} \times 5 \text{ s} = 50 \text{ m s}^{-1}$

• Galileo found (and countless experiments since have shown) that all objects falling freely towards the Earth have the same acceleration.

  • Time Velocity

    • 0 s 0 m s-1

    • 1 s 10 m s-1

    • 2 s 20 m s-1

    • 3 s 30 m s-1

    • 4 s 40 m s-1

    • 5 s 50 m s-1

• Velocity after 5 s = $50 \text{ m s}^{-1}$

• Start velocity = $0 \text{ m s}^{-1}$

• ∴ Distance fallen in 5 s is

  • $v_{av} = \frac{0 + 50}{2} = 25 \text{ m s}^{-1}$

  • $25 \text{ m s}^{-1} \times 5 \text{ s} = 125 \text{ m}$

• If you drop a cricket ball from a plane, how far will it fall in 5 seconds?

  • $\Delta x = \frac{1}{2} g \Delta t^2 = \frac{1}{2} \times 10 \times 5^2 = 125 \text{ m}$

• If you throw a cricket ball straight up at $30 \text{ m s}^{-1}$, how high will it go?

  • Gravity will reduce its upward velocity by $10 \text{ m s}^{-1}$ every second.

  • When its upward velocity is zero, it's reached its highest point.

  • ($\Delta v = g \Delta t \Rightarrow \Delta t = \frac{\Delta v}{g}$)

  • $\Delta t = \frac{30 \text{ m s}^{-1}}{10 \text{ m s}^{-2}} = 3 \text{ s}$

  • (to the highest point)

  • $v<em>f = v</em>i + a \Delta t$

  • $\therefore 0 = 30 - 10 \Delta t$

  • $\therefore 10 \Delta t = 30 \Rightarrow \Delta t = 3 \text{ s}$

  • ($v<em>{av} = \frac{v</em>f + v_i}{2} = \frac{(30 + 0)}{2} = 15 \text{ m s}^{-1}$)

  • Hence $\Delta x = v_{av} t = 15 \times 3 = 45 \text{ m}$

• Hence same time ( 3 s) to come down as to go up.

• Downward velocity at ground:

• Free fall from rest (at highest point) for 3 s at $10 \text{ m s}^{-2} \Rightarrow$ ($\Delta v = g \Delta t$)

  • $v = 10 \text{ m s}^{-2} \times 3 \text{ s} = 30 \text{ m s}^{-1}$

• Same speed as it was thrown up (at same height, i.e., ground).

• How long will the ball take to reach the ground again? (i.e., falling from rest at a height of 45 m)

  • vi = 0, $\Delta x = 45 \text{ m}$, g = $10 \text{ m s}^{-2}$ freefall:

• $\Delta x = \frac{1}{2} g \Delta t^2$

  • $\therefore 45 = \frac{1}{2} \times 10 \times \Delta t^2 = 5 \Delta t^2$

  • $\therefore \Delta t^2 = 9 \Rightarrow \Delta t = 3 \text{ s}$

Vertical and Horizontal Motion

• Both balls fall together, independent of sideways motion.

• Vertical and Horizontal motions are independent.

• Horizontal velocity = $10 \text{ m s}^{-1}$

• Vertical and horizontal motions are independent.

Example – Relative Motion

• A polar bear is 200 m behind a man who is running straight towards his car.

  • If the man is running at $7 \text{ m s}^{-1}$ and the bear is running at $11 \text{ m s}^{-1}$, how long will it take the bear to catch the man (in s)?

    • (A) 10 (B) 20 (C) 30 (D) 50 (E) 70

LECTURE 6 - WAVES AND OSCILLATIONS

Waves and Oscillations– Learning Goals

• Hooke’s Law

• Simple Harmonic Motion

  • Energy, period, frequency

  • Springs and the simple pendulum

• Wave Motion

  • Relation to SHM

  • Transverse and longitudinal waves

  • Superposition

  • Interference

Hooke's Law

• For springs and many materials…

• The length change is proportional to the (restoring) force:

  • $F = -kx$

    • k is called the spring constant

• Example

  • If $k = 300 \text{ N m}^{-1}$, what is the restoring force for x = +0.1 m, +0.2 m, − 0.2 m?

  • If a bone bends 4 mm when a 2 kg mass is suspended from it, what is its spring constant, k?

Energy in Hooke's Law Deformations

• As the spring is stretched (or compressed) by a force, F:

  • $W = F \times d = \frac{1}{2} kx \times x$

• The work done on the spring by the force is, $\frac{1}{2} kx^2$

• This work is stored as potential energy in the spring,

  • $PE = \frac{1}{2} kx^2$

Example

• The spring in a toy gun has $k = 50 \text{ N m}^{-1}$, and is compressed 0.15 m to fire a 2 g plastic bullet.

  • With what speed will the bullet leave the gun?

Simple Harmonic Motion (SHM)

• A mass attached to a spring will always experience a restoring force back towards its equilibrium position, if displaced.

  • If $F = -kx$ (very common), it will oscillate (sinusoidally) with simple harmonic motion.

• The time for each full cycle of oscillation is called the period, T (eg 0.5 seconds).

• The frequency, f, is the number of cycles per second: eg if T = 0.5 s, then f = 2 cycles/s = 2 Hz (hertz).

  • $f = \frac{1}{T}$

  • $T= \frac{1}{f}$

Energy Conservation

• Total Energy in the (general) position shown = PE + KE

  • i.e $E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2$

• At ends, i.e., x = ± A, no KE ⇒ E = $\frac{1}{2} kA^2$

• At centre, i.e., x = 0, no PE ⇒ E = $\frac{1}{2} mv_{max}^2$

• So, for total energy:

  • $E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} kA^2 = \frac{1}{2} mv_{max}^2$

Period & Frequency of SHM

• Period

  • $T = 2 \pi \sqrt{\frac{m}{k}}$

• Frequency

  • $f = \frac{1}{T} = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$

The (simple) Pendulum

• The restoring force:

  • $T_H \approx -kx$ where $k = \frac{mg}{L}$

• and, since $T = 2 \pi \sqrt{\frac{m}{k}}$ for SHM,

• $T = 2 \pi \sqrt{\frac{L}{g}}$ for a simple pendulum

• Example

  • How long is a simple pendulum with a period of exactly 1 s?

SHM and Wave Motion

• Sign convention used here: positive downwards ↓, negative upwards ↑.

Waves: frequency, wavelength, velocity

• In one period, T, a crest will have moved distance, λ , (from P to Q).

• $\therefore v_{wave} = \frac{\lambda}{T} = f \lambda$ (since 1/T = f )

• Generally, $v = f \lambda$ for any wave.

Transverse Waves

• The oscillation is transverse (perpendicular) to propagation direction.

Longitudinal Waves

• Oscillation is in direction of propagation

  • eg sound in air

Superposition of Waves

• (Pure) Constructive Interference

  • Waves 1 & 2 in-phase

• (Pure) Destructive Interference

  • Waves 1 & 2 out-of-phase

  • Waves 1 & 2 are 180° out-of phase

Superposition & Interference of Waves

• Two sources, S1 and S2, in phase.

• At P, if $D<em>2 - D</em>1 = \Delta D = m \lambda$ (m = 0,1,2….) then we have constructive interference at P

• Two sources, S1 and S2, in phase.

• At P, if $\Delta D = D<em>2 - D</em>1 = (m+0.5) \lambda$ (m = 0,1,2….) then we have destructive interference at P

Reflection and Standing Waves

• Suppose the string is fixed to the wall at its far end…

• What happens when the wave gets to the wall and the string can't move?

  • The wave reflects with a 180° phase change….

Standing Waves

• Nodes (places where string is not moving)

• Antinodes (places where string is moving)

• $L = n \frac{\lambda}{2}$

  • eg n = 1

Interference: Contructive interference

• Beats

Interference: Destructive interference

• NOT EXAMINABLE

Energy in Waves

• Size of wave: Generally energy & power increase as the square of amplitude

  • recall $PE = \frac{1}{2} kA^2$ (A is the oscillation amplitude of a spring)

The special role of sine waves in nature

• A wave of frequency, f, often has sine waves of frequency 2f, 3f, 4f, 5f … etc associated with it (as well as a sine wave of frequency, f ).

• The sine wave of frequency, f , is known as the fundamental.

• The sine waves of freq., 2f, 3f , 4f etc are the 2nd, 3rd …harmonics.

• NOT EXAMINABLE

Periodic Waves

• NOT EXAMINABLE

LECTURE 2 - DYNAMICS

Dynamics - Learning Goals

• Newton’s Laws of Motion and the definition of the concept of Force

• Understand the relation to kinematic concepts (Newton’s 2nd Law)

• Weight and Mass

• Vector character of forces

• Force as interaction (Newton’s 3rd Law)

• Forces acting on surfaces (Normal force)

• Motion in a circle

  • Velocity

  • Centripetal acceleration and force

Newton’s Laws of Motion

• Newton's 1st Law: Any object continues at rest, or at constant velocity, unless an external force acts on it.

  • i.e., at constant speed in a straight line

• Newton's 2nd Law: An external force gives the object an acceleration which is proportional to the force.

  • F = ma

• Newton's 3rd Law: Forces come in symmetric pairs – equal in magnitude, opposite in direction.

Newton’s Laws of Motion

• Newton's 1st Law: Any object continues at rest, or at constant velocity, unless an external force acts on it.

  • i.e., at constant speed in a straight line (Galileo)

• Newton's 2nd Law:

  • F = ma

    • a is the acceleration ($\text{m s}^{-2}$)

    • m is the mass (kg)

    • F is the force (N) (N = newton)

  • The unit of force is the Newton (N).

    • One newton (1 N) is the force needed to accelerate a 1 kg mass at $1 \text{ m s}^{-2}$ (i.e., to cause its velocity to increase by $1 \text{ m s}^{-1}$ in every second.)

    • m (kg) a ($\text{m s}^{-2}$) F (N)

      • 1 1 1

      • 1 2 2

      • 2 1 2

      • 2 2 4

      • 4 5 20

Weight and Mass

• Since any object of mass, m, falls with acceleration, g, it must be acted on by a force:

  • $F = ma \Rightarrow F = mg$

    • This downward force is due to gravity and is generally called the object's Weight (in N).

    • $W = mg$

    • A 1 kg mass has a weight of ~10 N.

    • A 20 kg mass has a weight of ~200 N.

Forces are Vectors

• Components of a Force, F:

  • Any force, F, is equivalent to the vector sum of two other suitable forces.

  • Fx and Fy are called components of F because, acting together, they have the same effect as F.

  • $F_x = F \cos{\theta}$

  • $F_y = F \sin{\theta}$

• They have both magnitude and direction and can be added by the parallelogram rule:

Example

• If a force, F, has a magnitude of 10.0 N and is directed at an angle of $30^o$ from the vertical, what are the vertical and horizontal components of the force?

Newton's 3rd Law:

• For every action there is an equal and opposite reaction.

• Forces come in symmetric pairs – equal in magnitude, opposite in direction.

  • Action = force applied by 1 on 2 - accelerates 2.

  • Reaction = force applied by 2 on 1 - accelerates 1

  • Same “types” of force.

  • Always act on different objects.

Example

• Case 1: A student pushes a loaded sled so that it moves with constant velocity.

  • Suppose that there is a frictional force of f = 20 N between the sled and the floor.

  • Calculate the size of the forces indicated on the diagram.

• Case 2: A student pushes a loaded sled so that it accelerates at $a = 0.5 \text{ m s}^{-2}$.

  • Suppose that there is a frictional force of f = 20 N between the sled and the floor.

  • Suppose also that the sled has a mass of 20 kg, and the student has a mass of 70 kg.

  • Calculate the size of the forces indicated on the diagram.

The Normal Force between surfaces

• Flower pot at rest on table:

  • Newton’s First law says that there must be no net vertical force:

    • $N = W = mg$

  • If m = 30 kg,

    • N = W = 300 N

  • The force is due to the deformation of the table surface.

• Use Newton’s Third Law to identify action-reaction pairs in the diagram.

Motion in a Circle

• Assuming constant speed, the magnitude of the velocity is constant, but the velocity is changing because its direction is changing….

• Time to go fully around is the period, T.

• Distance fully around is $2\pi r$.

• Hence speed, v

  • $v = \frac{2 \pi r}{T}$

  • E.g., if r = 2 m, T = 3 s, then:

    • $v = \frac{2 \times 3.14 \times 2}{3} = 4.2 \text{ m s}^{-1}$

Centripetal Acceleration

• Even when the speed is constant, if an object is moving around in a circle its velocity is changing because its direction is changing.

• Hence there is always acceleration towards the centre of the circle

• It can be shown that

  • $a = \frac{v^2}{r}$

    • This is the magnitude of the centripetal acceleration of an object moving at speed v, round a circle of radius, r.

  • We can write this in terms of the period of the motion around a circle,

    • $a = \frac{v^2}{r} = \frac{(\frac{2 \pi r}{T})^2}{r}$

    • $a = \frac{4 \pi^2 r}{T^2}$

Centripetal Force

• Because F = ma, this centripetal acceleration means there must be a centripetal force:

  • $F = m \frac{v^2}{r}$

    • whenever an object moves in a circle at constant speed.

Example

• A steel ball, with a mass of 1 kg, is attached to the end of a 0.5 m long cord (which has negligible mass) and is swung so that it travels in a circle with a period of 2 s.

  • What is the velocity and acceleration of the ball at a given instant?

  • What is the magnitude and direction of the force exerted on the ball by the cord?

Sources of Centripetal Force

• An object moving round a circle because a centripetal force acts on it.

• Where does this centripetal force come from?

  • Friction (rubber on road), or

  • Normal force (banked road), or

  • Tension in string (toy car), or

  • Gravitation (moon/planet in orbit), or

Example – Banked Road

• A car has a mass of 1000 kg and drives around a banked corner without slipping.

  • If the angle of banking is $\theta = 30^o$ and the car travels at $v = 25 \text{ m s}^{-1}$ ($90 \text{ km h}^{-1}$), what is the radius of curvature of the bend?

LECTURE 5 - MOMENTUM

Momentum – Learning Goals

• Concept of Momentum

• Conservation of Momentum

• Collisions

  • Inelastic

  • Elastic

• Approach and Recoil velocity

• Collisions in 2D

Linear Momentum

• Momentum is a vector quantity defined by Newton as mass times velocity

  • $p = mv$

  • Symbol for momentum p

• Newton’s 2nd law:

  • $F = ma = m \frac{\Delta v}{\Delta t} = \frac{\Delta (mv)}{\Delta t} = \frac{\Delta p}{\Delta t}$

• In words: “The (time) rate of change of momentum is proportional to the net external force”

• Net sum of external forces

• cf Newton's 3rd Law of Motion………

  • From above: F∆t = ∆p "impulse"

FR and FB

• The blue ball coming from the left collides with the red ball:

• During a short time interval ∆t (while the balls are colliding):

  • The blue ball exerts a force on the red ball ($F<em>R$) and the red ball exerts a force on the blue ball ($F</em>B$).

  • Newton's 3rd Law ⇒ $F<em>R = -F</em>B$

  • (third-law force pair: equal & opposite)

  • $\therefore F<em>R \Delta t = - F</em>B \Delta t \Rightarrow \Delta p<em>R = - \Delta p</em>B$

  • i.e. ∆ptotal = 0 (provided no net external forces)

Colliding Objects

• Two broad types of collision

  • Inelastic Collisions

    • Momentum is conserved (if no net external forces),

    • Kinetic energy is not conserved.

      • (Some of it's converted: heat energy, sound…)

  • Elastic Collisions

    • Momentum is conserved (if no net external forces).

    • Kinetic energy is conserved

Mass (kg) Collision Duration, ∆t (ms)

Golf ball (collision with club) 0.047 1.0

Cricket ball (with bat) 0.156 2.0

Tennis ball (with racquet) 0.058 4.0

Soccer ball (with boot) 0.425 8.0

Basketball (with floor) 0.55 20

Solving Collision Problems

• Choose a boundary to define the system of interest

  • Which forces are external; which are internal?

  • Try to make complicated forces between objects internal forces

  • If only one object in the system, then all forces are external

• Always draw the system before and after the collision

• Decide on a coordinate origin and determine which direction is positive. (If more than 1D, break into vector components.)

• Is the collision elastic or inelastic?

  • Do the objects stick together afterwards?

Sticky Inelastic Collisions

• A collision is often called “totally” inelastic if the objects stick together after they have collided

• Define `the system’ to include both objects and nothing else, so that there are no external forces.

• Hence Linear momentum is conserved (but inelastic, so Kinetic Energy is not conserved)

• $p<em>i = m</em>1 v<em>{1i} + m</em>2 v_{2i}$

• $p<em>f = m</em>1 v<em>{1f} + m</em>2 v_{2f}$

• or better…

• $p<em>f = (m</em>1 + m<em>2) v</em>f$

• Example 1

  • Before collision$p_{xi} = 3 x 9 + 2 x 4 = 35 ( in \text{kg m s}^{-1})$

  • After collision$p<em>{xf} = (3 + 2)v</em>f = 5v_f$

  • but $p<em>{xf} = p</em>{xi}$

  • $\therefore 5v<em>f = 35 \Rightarrow v</em>f = 7 ( in \text{m s}^{-1})$

Inelastic collision, Example 2

• PHSI191 Final Exam 2019 A 35 kg child and her 70 kg father are on an ice skating rink.

  • The father is stationary when the child skates into him at a speed of $8.0 \text{ m s}^{-1}$.

  • The father and child hold on to each other. What is the speed of the father and child after the collision (in $\text{m s}^{-1}$)?

    • (A) 2.7 (B) 3.5 (C) 5.0 (D) 7.5 (E) 11

Elastic Collisions

• Kinematics: Describes motion using time, distance, speed, and acceleration (both average/instantaneous and vector quantities like displacement, velocity, acceleration).

• Key Equations:

  • Speed: $v = \frac{\Delta x}{\Delta t}$

  • Average speed: $v_{av} = \frac{\Delta x}{\Delta t} = \frac{\text{Total distance travelled}}{\text{Total time it takes}}$

    • Also, at constant acceleration: $v<em>{av} = \frac{v</em>f + v_i}{2}$

  • Acceleration: $a = \frac{\Delta v}{\Delta t}$

  • Uniform acceleration: $\Delta x = v_i \Delta t + \frac{1}{2} a \Delta t^2$

• Adding Velocities: Use vector addition to find resultant velocities (e.g., airplane and wind).

• Gravity: Acceleration due to gravity $g = 9.8 \text{ m s}^{-2} \approx 10 \text{ m s}^{-2}$. Equations: $\Delta v = g \Delta t$, $\Delta x = \frac{1}{2} g \Delta t^2$

• Independence of Motion: Vertical and horizontal motions are independent.

• Relative Motion: Problems involving relative speeds (e.g., bear chasing a man).

• Hooke’s Law: $F = -kx$ (restoring force proportional to displacement).

• Energy in Hooke's Law: Potential energy stored in a spring: $PE = \frac{1}{2} kx^2$

• Simple Harmonic Motion (SHM): Oscillations with period $T$ and frequency $f = \frac{1}{T}$. Total energy: $E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} kA^2 = \frac{1}{2} mv_{max}^2$

  • Period of SHM: $T = 2 \pi \sqrt{\frac{m}{k}}$. Frequency: $f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$

  • Simple Pendulum: $T = 2 \pi \sqrt{\frac{L}{g}}$

• Wave Motion: Velocity $v = f \lambda$. Transverse and longitudinal waves, superposition, and interference.

• Newton’s Laws of Motion: 1) Inertia, 2) $F = ma$, 3) Action-reaction.

• Weight: $W = mg$

• Forces as Vectors: Components: $F<em>x = F \cos{\theta}$, $F</em>y = F \sin{\theta}$

• Normal Force: Force acting on surfaces.

• Circular Motion: Speed $v = \frac{2 \pi r}{T}$. Centripetal acceleration $a = \frac{v^2}{r} = \frac{4 \pi^2 r}{T^2}$. Centripetal force $F = m \frac{v^2}{r}$

• Momentum: $p = mv$. Newton’s 2nd law: $F = \frac{\Delta p}{\Delta t}$

• Collisions: Inelastic (momentum conserved, KE not conserved) and elastic (both conserved).

  • Impulse: $F\Delta t = \Delta p$

• Problem Solving: Define the system, draw diagrams, choose a coordinate origin, and determine if the collision is elastic or inelastic.