jc a level h2 physics dynamics notes

Physics SSP 2012 - DYNAMICS

1

A Parachutist is moving downwards at constant

velocity.

Derive an expression of drag force in terms of g.

1. Show proper free-body diagram.

2. Key Points: constant velocity hence

acceleration is zero, which means net force

on parachutist is zero.

3. F = ma

mg – Fdrag = m(0)

Fdrag = mg

B A car is moving at constant speed. Draw the

free-body diagram of the car and include all the

possible forces acting on the car.

1. Free body diagrams to include:

- weight of car

- force of ground on car (acting on

tyres)

- Driving force acting on front wheels

(or both wheels)

- Frictional force acting on wheels

- Drag Force

1.2. Newton’s Second Law

1.2.1. Common problems involving Fnet = ma (15 mins)

A (acceleration of object along a slope)

A box of mass m travels up a slope

with acceleration a.

A constant external force F acting on

the box pushes it up the slope.

A constant frictional force f acts on

the box’s bottom surface.

The slope is angled at o.

Derive an expression of the external

force F in terms of m, f , a and .

F – (mgsin + f) = ma

F = (mgsin + f) + ma

Additional Point: Can discuss power provided by

engine.

i.e. Power = Fv = (mgsin + f)v + mav


2

B (two accelerating objects in contact)

Two boxes A and B of equal mass m

resting on the floor are in contact.

External force F acts on A and as a

result A and B undergoes acceleration

a.

Derive an expression for the

magnitude of force F of B on A.

Body A and B:

F = (2m)a

a = F/2m

Body B only:

F A on B

= (m)a

= m(F/2m)

= F/2

By Newton’s 3rd law, Magnitude of F B on A = F/2

C (atwood pulley)

A rope goes over a fixed pulley and

two masses A and B are secured to

the two free ends of the rope.

Mass A has an acceleration a

upwards.

Mass A is m and Mass B is M.

Derive an expression for the tension

in the rope.

Working:

1. Mass A:

T – mg = ma

Mass B:

Mg – T = Ma

2. Solve for T

D (numerical example)

A car of mass 1000 kg travelling along a straight horizontal road has an acceleration of

1.8 m s–2 when a driving force of 2.6 kN acts in the forward direction.

What is the resistive force acting backwards on the car?

A 0.8 kN B 1.8 kN C 2.6 kN D 4.4 kN F = ma

2600 - Fr = (1000)(1.8)

Fr = 0.8 kN

E The diagram shows a parachutist of mass 82 kg falling towards the Earth. In each case,

determine the net force and the acceleration of the parachutist.

a

b

c


3

For (a) Net force = 500 N downwards

F = ma

500 = 82a

a = 6.1 m s-2 (downwards)

parachutist is speeding up

For (b) Net force = 0

a = 0 m s-2

Constant velocity (unless parachutist changes his body shape)

For (c) Net force = 700 N upwards

700 = 82a

a = 8.5 m s-2 (upwards)

parachutist is slowing down

F

The diagram shows an 80 kg person in a lift.

The normal contact force acting on the person

from the base of the lift is R. Determine the

magnitude of R when the lift:

(a) is travelling upwards at a constant velocity

of 2.0 m s−1

(b) is accelerated upwards at 2.3 m s−2.

(a) R – mg = m(0)

R = mg

(b) R - mg = m(2.3)

R = m(2.3) + mg = 968.8 N

Man feels heavier

1.2.2. Common problems involving Fave = mv / t (10 mins)

A Tennis ball of mass m travelling horizontally at constant

speed v hits a rigid wall and rebounds horizontally with the

same speed. The duration of impact is t.

Derive an expression for the average force Fave of wall on

the ball.

Fave = -2mv/t

Negative means that the

direction of Fave is away

from the wall

B An object with mass m drops on the floor and rebounds with

the same speed.

Derive an expression for the average force Fave of floor on

the object in terms of the object’s weight and rate of change

of momentum of object.

Fave - W = 2mv/t

Fave = 2mv/t + W


4

1.3. Newton’s Third Law (5 mins)

A Describe three examples/situations of Newton’s third law. Draw out the free-body

diagrams for each example/situation.

Emphasise that force is of

same nature

Example: Gravitational

Force of Earth on Man and

Man on Earth

Example: Force of rocket on

fuel/air mixture, force of

fuel/air mixture (thrust) on

rocket

Example: Force of water

on object (upthrust) and

force of object on water

1.4. Impulse (10 mins)

A A moving box initially travelling at constant velocity hits a wall and comes to a stop. Draw

the possible force-time graph for each of the scenarios

Hard Wall :

F-t diagram is hill-shaped with large

maximum force over small duration of time

Wall has a piece of thick sponge

attached:

F-t diagram is hill-shaped with smaller

maximum force over larger duration of

time

Discuss the features of the two graphs.

Similar feature: change of momentum of box is same hence area under F-t graphs are

the same

Derive an expression for the initial velocity of the box in terms of average force of wall on

box, Fave and duration of impact, t.

Fave = (0-mv)/t

v = -(Favet) / m

negative sign indicates that Fave is opposite direction to v


5

2. Conservation of Momentum

2.1. Fill in the blanks (True or False) and any relevant comments. For the last column, state the equations that can be used for problem-solving.

(15 mins)

Total momentum of two

objects before and

after collision are equal

Total momentum of two

objects before, during and

after collision are equal

Total KE of two objects

before and after collision

are equal

Total KE of two objects

before, during and after

collision are equal

Equations to be used

Elastic

Collision

Yes. Always.

Yes. Always. Yes. Always. No. During collision,

Total KE before

collision is converted to

other forms of energy

(eg. EPE)

Total Initial Momentum =

Total Final Momentum

Total Initial KE =

Total Final KE

Rel speed of approach =

Rel speed of separation

Inelastic

collision

Yes. Always.

Yes. Always. No. Part of Total KE

before collision is lost

during the collision

No Total Initial Momentum =


Perfectly

inelastic

collision

Yes. Always.

Yes. Always. No. More or all of Total

KE before collision is

lost during the collision

No Total Initial Momentum =


And common velocities

after collision


6

2.2. Graphical Problem (10 mins)

Scenario: Two masses m and 2m approaches each other with velocities 1 m s-1 and 2 m s-1

respectively. They collide elastically and the impact of duration is 1 second. Sketch the

velocity-time graphs for the two masses.

After drawing graph, check that at every time interval, total momentum is constant.

2.3. Practice Questions (10 mins)

1 A body moving with speed v collides elastically with another body travelling in the

opposite direction with speed 2

v. Which row in the table below correctly gives the relative

velocity of approach and the relative velocity of separation of the two bodies?

Relative velocity

of approach

Relative velocity

of separation

A

2

v

2

v

B

2

v

2

3v

C

2

3v

2

v

D

2

3v

2

3v

Ans: D


7

2 A body of mass m travels with a velocity 3v and collides with another particle of mass 2m

which is initially stationary. After the collision, the two particles move with the same

velocity.

Which row in the table gives the final velocity of the two particles and the loss in kinetic

energy during the collision?

Final velocity Loss in kinetic energy

A v mv2

B v 3mv2

C 1.5 v mv2

D 1.5 v 3mv2

Ans: B

3 In an inelastic collision, which quantities are conserved?

A total kinetic energy and total momentum but not total energy

B total kinetic energy and total energy but not total momentum

C total momentum and total energy but not total kinetic energy

D total kinetic energy, total momentum and total energy

Ans: C

4 What is a correct statement of the principle of conservation of momentum?

A in an inelastic collision the total kinetic energy and momentum are constant

B in any collision the total momentum of an isolated system is constant

C in any isolated system the force on a body equals the rate of change of

momentum

D momentum is constant when mass and velocity are constant

Ans: B

jc a level h2 physics dynamics notes

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