Chapter 2 Force and Motion

2.1Analysing Liner Motion1.Linear motion is motion in a straight line.Distance and Displacement

1.Distance is the _________________________________________________________________by an object.2.Distance is a ______________________________________. It has _______________________________________ but _____________________________________. 3.Displacement is the ___________________________ of its final position from its initial position in ____________________________________________.4.Displacement is a ____________________________________________. It involves both _________________ and _________________________________.5.Both distance and displacement have the same SI unit which is _____________________________.Exercise1. A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.

(i) what is the distance(ii) what is the displacement

2.Ahmad walks to Chongs house which is situated 80 m to the east of Ahmads house. They then walk towards their school which is 60 m to the south of Chongs house.

(a) What is the distance traveled by Ahmad and his displacement from his house?

Speed and Velocity1.Speed is the ________________________________________________________. Speed during the course of a motion is often computed using the following equation:

2.Velocity is______________________________________________________________. Velocity is often computed using the equation:

3.Speed is a ____________________________ whereas velocity is ______________________________________.4.Both speed and velocity have the same SI unit. They are measured in meter per second or m s-1. Other unit may be in cm s-1 or km h-1.5.The average speed of a motion is often computed using the following equation:

6.Meanwhile, the average velocity is often computed using the equation:

Exercise1.A man running in a race covers 60 m in 12 s.(a) What is his speed in,(i) m s-1(ii) km h-1(b) If he takes 40 s to complete the race, what is his distance covered?(c) Another man runs with a speed of 7.5 m s-1, how long did he take to complete the race?

2.

The physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. The entire motion lasted for 24 seconds. Determine the average speed and the average velocity.

Acceleration and Deceleraion1.Acceleration is____________________________________________________________________________________.2.It can be written as;

3.Acceleration is a____________________________.The SI unit for acceleration is m s-2.4.The acceleration is positive if the velocity increases with time. The acceleration is negative if the velocity decreases with time. It is also called deceleration.5.Figure 2.1, shows that the car experiences acceleration, a constant velocity and then a deceleration.

Figure 2.1Exercise1.A vehicle accelerates uniformly from rest to a speed of 25 m s-1 in 100 s along a straight road. It then decelerates uniformly at 0.2 m s-2 for 60 s. Find(a) the initial acceleration(b) the final speed

Analysing of Liner Motion

Ticker-timer

1.In the laboratory, a ticker-timer as shown in figure 2.2, with a trolley is used to study the motion of an object for a short time interval.

Figure 2.22.A ticker-timer consists of a small electrical vibrator which vibrates at the frequency of 50 Hz.3.The time taken to make 50 dots on the ticker tape is 1 s. Hence, the time interval between 2 consecutive dots is = 0.02 s

Figure 2.34.To determine the time interval of motion of the object:Time interval = Number of tick x 0.02 s5.The following shows the different types of motion recorded on the ticker tape.Ticker TapeCharacteristicsInferences

distance between the dots is equally distributed.Consistent distance= ______________________velocity

distance between the dots is equally distributed.Short Distance=_______________________velocity

distance between the dots is equally distributed.Long distance=________________________velocity

the distance between the dots increases uniformly.Increasing distance=________________________velocity=________________________

the distance between the dots dcreases uniformly.decreasing distance=________________________velocity=________________________

Exercise1.A trolley pulled a ticker tape through a ticker timer while moving down an inclined plane. Figure 2.4 shows the ticker tape produced

Figure 2.4Determine the average velocity of the trolley.

2.Figure 2.5 shows a strip of ticker tape pulled through a ticker-timer by a freely falling metal sphere. The ticker-timer vibrates at a frequency of 50 Hz. Determine the acceleration of the sphere.

Activity 2.1

Aim of the activity:To determine displacement, average velocity, acceleration and deceleration.

Apparatus/ Materials:___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Setup:

Procedure:1.Set up the apparatus as shown in the diagram.2.Incline the runway at an angle of 10 with the horizontal.3.Attach the ticker-timer tape to the trolley at the top of the runway.4.Switch on the ticker-timer and release the trolley.5.Mark the tape and cut it into five strips of ten ticks from the start of the tape.6.Repeat the experiment with the angles of inclination of 20 and 30 .7.Using graph paper, build a ticker-timer tape chart for each angle of inclination.

Collecting data:1.The ticker tape can be cut into strip. Prepare a 10-tick strip for each different angle produced. and paste the strip

(a) 1st strip (10)

(b) 2nd strip (20)

(c) 3rd strip (30)

2.Determine the displacement, time interval and velocity

1st strip (10)2nd strip (20)3rd strip (30)

The displacement of 10 ticks

The time interval for 10-ticks strip

Average velocity of 10-ticks strip

3.Determine the acceleration or deceleration

1st strip (10)

The displacement of first 1-tick

The time interval for first 1-tick

The initial velocity, u

The displacement of last 1-tick

The time interval for last 1-tick

The final velocity, v

acceleration, a

2nd strip (20)

The displacement of first 1-tick

The time interval for first 1-tick

The initial velocity, u

The displacement of last 1-tick

The time interval for last 1-tick

The final velocity, v

acceleration, a

3rd strip (30)

The displacement of first 1-tick

The time interval for first 1-tick

The initial velocity, u

The displacement of last 1-tick

The time interval for last 1-tick

The final velocity, v

acceleration, a

4.The ticker tape can be cut into strips of equal time (equal number of ticks) and paste together to form a chart for analysing the motion of a trolley.

Prepare 3 charts for (a) = 10, (b) = 20 (c) = 30 and calculate the acceleration.

3.The following shows the different types of motion recorded on the ticker tape and tape chart.

Ticker tape and chartCharacteristicsInference

The separation between dots stays the same.

The length of the strips of the tape chart is equal.

The distance between the dots increases uniformly.

The length of the strips of the tape increase uniformly

The distance between the dots and the length of strips of the tape decreases uniformly.

4.Change in distance between dots indicates a changing velocity and thus, acceleration. A constant distance between dots represents a ____________________________________________ and _______________________________.ExerciseFigure 2.6 shows a ticker tape chart obtained in an experiment to study the motion of trolley on an inclined plane. Calculate the acceleration of the trolley.

More ExercisesAnalysis of motion on a ticker tape.

Ticker tape or chartType of motionCalculation of velocity / acceleration

Moving with _______________________Velocity=

Moving with ________________________Velocity =

Moving with _______________________Velocity =

________________________ or increasing velocityInitial velocity,u=

Final velocity,v=

Accelerationa=

________________________ or decreasing velocityInitial velocity,u=

Final velocity,v=

Accelerationa=

________________________ or increasing velocityInitial velocity,u=

Final velocity,v=

Accelerationa=

________________________ or decreasing velocityInitial velocity,u=

Final velocity,v=

Accelerationa=

Exercise

Ticker tape or chartCalculation of velocity / acceleration

1

2

3

4

Equations of Liner Motion with Uniform Acceleration

1.For an object in linear motion with uniform acceleration (change in velocity), problems involving the displacement, velocity, acceleration and time of motion can be solve by using the equations of motion.(u = initial velocityv = final velocityt = timea = accelerations = displacement )

EquationVariables involvedVariables not involved

v, u, a, t

v, u, a, s

s, u, t, a

s, u, v, t

2.For motion with constant velocity (zero acceleration), the formula is v =

Exercise1.u (ms-1)v (ms-1)t (s)a (ms-2)s (m)FormulaAnswer

241?

04?5

?4416

3?25

812?1

342?

48?10

?2420

2. A car travelling at 10ms-1 stops after a time of 20s. What is its deceleration?

3.A car starts off at 2 ms-1 and accelerates at 3ms-2 for 5s. What is its final velocity?

4.An object, initially at rest, moves in a straight line with constant acceleration and covers a distance of 250 m in 10s.Calculate(a)Acceleration(b)Final velocity

5.A car travels for 30 minutes and is brought to rest by application of its brakes. The distance travelled before stopping is 9 km.Calculate the(a)Average velocity in the first 30 minutes.(b)initial velocity of the car(c)deceleration of the car

6.A bullet is fired towards a nearby tree with a speed of 200 ms-1. The bullet is shot at a depth of 5 cm. Find the average deceleration of the bullet inside the tree. What is the time taken for the bullet to stop after hitting the tree?

7.Salina is driving her car at a velocity of 10 ms-1. On seeing an obstacle in front, she applies the brakes to stop her car. If the deceleration of the car is 2 ms-2, what is the distance her car travels before it comes to a halt?

8.By applying the brakes, a driver reduce his car's velocity from 20ms-1 to 10 ms-1 after travelling a distance of 30 m. Find the deceleration of the car.

9.Ali is driving a car at velocity of 30 ms-1. On seeing a student crossing the road, Ali steps on his brake to stop the car. The speed of the car decreases uniformly and stops after travelling 150 m.(a)What is the deceleration of the car when the brakes are applied?(b)What is the time interval before the car stops?

2.2Analysing Motion Graphs

Motion Graphs1.There are two main types of linear motion graphs:(a) displacement-time(b) velocity-time

Displacement-Time Graph1.A displacement-time graph is a graph showing how(a)the position of an object from its original position changes with time.(b)the displacement of the object changes with time.

2.A student walks at a constant velocity from position A to reach position B in 200 s. He rests for 100 s at position B and then walks back to position A using the same straight path. He reaches position A after 200 s.

3.The graph below shows the change in the distance and direction with time of the student.

(a).The slope or gradient of a displacement-time graph is equal to the ________________________ of the object.(b).Gradient of the graph in section I,=

(c).On the displacement-time graph, a horizontal line (gradient = 0) shows the object is _______________________________.(d).In section II of the graph, the student remains at position B from 200 s to 300 s.(e).Gradient of the graph in section III,=

(f).The negative sign shows that the direction of motion is opposite to the original direction. (Velocity is a vector quantity)

(g)At t = 500s, the graph intersects at t-axis. The displacement at this moment is zero that is the student has returned to the original position.

4.The various displacement time graphs are shown in the following table.Displacement time graphPosition and motion of the objectDisplacement and velocity

Gradient of graph is _____________.

Hence, the object is at ___________ over a period of timeThe displacement is _____________. Its velocity is zero.

Gradient of graph is _____________.Hence, the object is moving with uniform velocity.The object moves in positives direction from its original position.Its displacement is increasing linearly (at constant rate).

Its velocity is ___________________

and ______________________________.

Gradient of graph is _____________

but _______________________________.Hence, the object is moves in negative direction back to its original positionIts displacement is decreasing linearly (at constant rate).

Its velocity is _____________________

and ________________________________

Gradient of graph varies and is

_____________________________________Hence, the object moves in with

___________________________velocity.

Its displacement is increasing at non-constant rate.

Its velocity is _____________________ with time. The is moving with constant

____________________________________.

Gradient of graph varies and is

____________________________________.Hence, the object moves in with

___________________________velocity.

Its displacement is increasing at non-constant rate.

Its velocity is _____________________ with time. The is moving with constant

____________________________________.

Exercise:

The displacement time graph shows the motion of an object.

(a) Briefly describe the motion of the object represented by AB, BC, CD and DE.(b) Find (i) The displacement after 20s,(ii) The time taken to move 35m from the origin.(c) Calculate the average velocity in each of these time intervals:(i) 0s 5s(iii) 10s 20s(ii) 5s 10s(iv) 20s 28s

Solution(a) AB: The object is at rest ________________m from the ________________________________.

BC: The object moved __________________m forward with _________________________velocity.

CD: The object moved another ____________m forward with ______________________velocity.

DE: The object moved ____________m backwards with _____________________velocity and returned to its starting point.

(b) (i)When t = 20s, s = _________________________________ = _________________________ m

(ii)When s = 35 m, t = _________________________s

(c)(i)object is at __________________________, hence the velocity = _________________ms-1.(ii)velocity = gradient of the graph

v =

(iii)velocity, v =

(iv)velocity, v =

Velocity-Time Graph

1.A car starts from rest and accelerates for 20 s until it reaches a velocity of 60 ms-1. The driver maintains this velocity for 20 seconds. The velocity of the car is then gradually reduced until it stops at t =60 s.

2.The graph below shows how the velocity of the car changes over a certain period of time.

(a)On a velocity time graph, the gradient of the graph is equal to the ______________________ of the object.(b)In section I, the acceleration of the car;

(c)On the velocity time graph, a horizontal line represents a _________________________________.

The gradient of the graph is equal to ____________________.

(d)In section II, the car travels at a constant velocity of _______ms-1 from t = _____s to t =_____s. (e)In section III, gradient of the graph;

(f)The negative sign indicates a __________________________________________.(g)On the velocity time graph, the area under the graph is equal to the ___________________.(h)In section I, the area under the graph

(i)In section II, area under the graph,

(j) In section III, area under the graph,

3.The gradient of a velocity time graph gives the ___________________________________.

4.On the velocity time graph, the area under the graph is equal to the _____________________.5.The various velocity time graphs are shown belowVelocity time graphVelocity and acceleration

(a) Object at rest

The gradient of the graph represents acceleration.

The gradient of the graph is __________________ and hence the acceleration is always zero.

The area under the graph represents ____________________ travelled.

The area under the graph is _______________ and hence the displacement travelled is zero.

(b) Object moving with constant speed

The velocity stays the same. The object is moving with constant velocity.

The gradient of graph is __________________________.

Object is moving with _______________________ velocity.

Its acceleration (= gradient of the graph) is ______________.

(c) object moving with constant acceleration

The velocity is increasing uniformly with time.

The gradient of graph is ___________________________.Hence, The object is moving with

______________________________________.

(d) Object moving with constant deceleration

The velocity is decreasing uniformly with time.The gradient of the graph is constant but

________________________.The object is moving with constant

__________________________.

(e) Object moving with increasing accelerationThe gradient of the graph is increasing.

The acceleration of the object is

________________________________.

(f) Object moving with decreasing acceleration.

The gradient is ______________________ with time.

The object is moving with constant

______________________________.

Exercise

1. The velocity time graph of an object starting from rest and travelling towards the east is as shown in figure.(a)How long does the object travel towards east?(b)How long does the object travel towards the west?(c)Find the average speed and the average velocity.

A comparison between the displacement time and the velocity time graph.Displacement time graphVelocity time graphThe object moves at a constant velocityNon- horizontal straight lineThe object moves at a constant acceleration

Represents the velocity of the objectGradientRepresents the acceleration of the object

The object is stationaryHorizontal lineThe object moves at a constant velocity

The object returns to its original position.Intersection on the time - axisThe object stops

+ve The object moves in a specific direction-ve The object moves in the opposite directionSign of the gradient (positive or negative)+ve Acceleration-ve - Deceleration

No significanceArea under the graphNumerically equals the distance travelled by the object

Summary of Linear Motion GraphDisplacement time graphsVelocity time graphsAcceleration time graphs

Velocity = gradient of the graph Acceleration = gradient of the graphDisplacement = area under the graphVelocity = area under the graph

v = 0(a = 0)

v = constant(a = 0)

va = constant

va = constant

va

va

a increasing rate

a decreasing rate

Try this out, past year SPM questions2007Diagram 1 shows the path travelled by a car from P to S.

Diagram 1What is the displacement of the car? A 5.0 kmC 8.2 kmB 6.8 kmD 9.0 km

2007Diagram 2 is a velocity-time graph showing the motion of an object.

Diagram 2Which of the following describes the motion of the object? OJJKAUniformDecreasingaccelerationacceleration

BIncreasingDecreasingaccelerationacceleration

CIncreasingUniformaccelerationdeceleration

DUniformUniformaccelerationdeceleration

2010Which physical quantity is equal to ?

ASpeedBVelocityCDistanceDAcceleration

2012Diagram 3 shows Ali stands at O. He walks towards A, then moves towards B and stops at B.

What is the displacement of Ali?A2m towards westB5m towards eastC7m towards east

2012Which tape shows a movement with uniform velocity and then deceleration?

2005The acceleration-time graph below shows the movement of an object.

Which velocity-time graph represents the movement of the object?

AC

BD

2008Diagram 4 shows the velocity-time graph of an object. Diagram 4 Which acceleration-time graph represents the same motion as the object?

A

B

C

D

2009Diagram 5 shows a velocity-time graph for motion of an object.

What is the total distance, in 8s?A 18B 24C 32D 64

2010Diagram 6 shows a car moving up a hill. The car decelerates as it moves up the hill and accelerates as it moves down the hill.

Which graph shows the correct relationship between the velocity, v, of the car and the time, t, of the motion?

2011Diagram 7 shows a velocity time graph for a motion of a toy car.

What is the displacement of the car in 6 s?A 0mB 30mC 45mD 75m

2012Diagram 8 shows a velocity time graph of a bus with passengers on board.After 10 minutes, the driver steps on the brake pedal to stop the bus.

What is the velocity of passengers immediately after the brake is applied?A 0 ms-1B 2ms-1C 6ms-1D 72ms-1

2.3Understanding Inertia1.All objects tend to continue with what they are doing.2.Inertia is the tendency of an object to maintain its state of _______________________, or if moving to continue its _____________________________in a straight line.3.Newtons first law of motion (Law of Inertia) states that an object at __________________ tends to stay at rest, and if in _______________________ tends to stay in motion unless acted upon by an ______________________________.3.The following situations show example of inertia:SituationExplanation

When a cardboard is pulled away quickly, the coin drops straight into the glass. The inertia of the coin maintains its state of _____________. When the card is pulled away, the coin falls into the glass due to __________________________.

If a book is pulled out very quickly from the middle of a pile of books, The books above will drop instead of moving along with it. The _______________________ of the books above keeps them in their original ___________________. If the book is pulled out slowly, the books above will move together with the book.

Body moves forward when the car stops suddenly. The passengers were in a state of motion when the car was moving. When the car stopped suddenly, the inertia in the passengers made them maintain their state of motion. Thus when the car stop, the passengers moved ______________________..

Relationship between Mass and Inertia1.The diagram shows two buckets, one bucket filled with sand while the other bucket is empty.2.It is more difficult to move the bucket filled with sand compared to the empty bucket when both are initially at rest.3.When both the buckets are moving, it is also more difficult to stop the bucket filled with sand compared to the empty bucket.4.The bucket filled with sand has a greater tendency to be at __________________________ and also to continue to be in ______________________ compared with the empty bucket.5.The bucket filled with sand has more ___________________________ than the empty bucket. The bucket filled with sand has a bigger ____________________ than the empty bucket. Quantitatively, the inertia of an object is measured by its ________________________.Experiment 2.1: Relationship between Inertia and MassEffects on Inertia1.The positive effect of inertia

Chili sauce in the bottle can be easily poured out if the bottle is moved down fast with a sudden____________________. When the bottle stops suddenly, the sauce continue in its state of motion due to the effect of its ____________________.

A boy runs away from a cow in a zig-zag motion. The cow has a large inertia making it difficult to change direction.

The head of hammer is ________________________ to its handle by knocking the end of the handle, held vertically, on a hard surface. The head of hammer has a __________________ mass and will remain in its ______________________, thus fitting it tighter on the handle.

The drop of water on a wet umbrella will fall when the boy rotates the umbrella. This is because the drop of water on the surface of the umbrella moves simultaneously as the umbrella is rotated. When the umbrella _________________rotating, the inertia of the drop of water will continue to maintain its motion.

2.The negative effects of inertia

If a car crashes while travelling, the driver of the car is still in ____________________ even though the car has stopped. The driver may be ______________________against the windscreen and suffer injuries. This can be prevented if the drive is wearing a ________________________ that will him back from the forward motion.

The tank of a lorry carrying liquid is divided into several _______________________. This will reduce the impact of the ______________________ of the liquid if the lorry stops suddenly. The __________________________________________ between the drivers cabin and the load to stop the initial movement of heavy load towards the driver when the lorry is brought to a halt suddenly.

Furniture carried by a lorry normally is tied up together by _____________________. When the lorry starts to move suddenly, the furniture is more difficult to fall off due to their ________________________ because their combined mass has increased.

3.Ways of reducing the negative effects of inertia.(i) Wearing safety belts when driving.The safety provides the external force that prevents the driver or passengers from being thrown forward.(ii) An air bag is fitted inside the steering wheel.The airbag inflate automatically when a collision occurs. This prevents the driver or passengers crashing into dashboard.(iii) Subdivision of the mass to reduce its inertiaThe oil tank of an oil tanker lorry is usually divided into few smaller compartments so that the effects of inertia can be reduced.(iv) Strong structure behind the drivers cabinIf a loaded lorry stops abruptly, its heavy load, for example, timber logs, will continue to move forwards the drivers cabin because of its massive inertia. A strong iron structure between the drivers cabin and the load ensures the drivers safety.

Experiment 2.1 relationship between inertia and massProblem statement:A bucket filled with sand is more _________________________ to swing compared to the empty bucket.What is the relationship between inertia and mass?Inference:The ________________________________ of an object depends on its____________________________________.Hypothesis:When the mass of an object ______________________, the inertia of the object _______________________Aim:To investigate the relationship between inertia and massVariables:(a) Manipulated variable:____________________________________________________________________________(b) Responding variable:_____________________________________________________________________________(c) Constant variable: ________________________________________________________________________________Materials:_________________________________________________________________________________________________________Apparatus:_________________________________________________________________________________________________________Arrangement of apparatus:

Procedure:1.The apparatus is set up as shown above.2.A 25 g of plasticine is fixed to the free and of the hacksaw blade.3.The time for 20 complete oscillations is measured using a stopwatch and is recorded. The step is repeated twice to calculate the average time taken. Then the period, T is determined.4.Step 3 is repeated with different mass of plasticine, m = 50 g, 75 g, 100 g and 125 g.

Tabulation of data:

Analysis of data: (graph)

Discussion:

Conclusion:

2.4Analysing Momentum

Lets think about it!?

You can catch a fast moving ping-pong ball easily with your bare hand.A softball keeper must wear a glove to catch a hard and fast moving softball.Why is a slow moving softball much easier to catch?

If a loaded lorry and a car are moving at the same speed, it is more difficult for the lorry to stop.This is because the lorry possesses a physical quantity, momentum, more than the car.All moving objects possess momentum.

Activity 2.2To compare the effects of stopping two objects in motion

Apparatus / Materials:One steel ball and one wooden ball of the same diameter, 2 slabs of plasticine.

Arrangement of apparatus:

(a) Two objects of the same mass moving at different velocities.(b) Two objects of different masses moving at the same velocity

Procedure:1 A steel ball is first released from a height of 50cm and then from 100cm above a slab of plasticine as shown in Figure.2 The depths and sizes of the cavities caused by the steel ball on the slab are observed and compared.Procedure:1. A steel ball and a wooden ball of the same diameter are released from a height of 50cm above a slab of plasticine as shown in figure.2. The depths and sizes of the cavities formed are observed and compared.

ObservationsObservations

ConclusionThe moving balls produce an effect on the plasticine which is there to stop the motion.The __________________the mass or velocity of the moving object is, the _________________ is the effect (the depth and size of the cavity), the _________________is the momentum.

Linear Momentum1. The linear momentum, p, of a mass, m, moving at a velocity, v, is defined as the product of mass and velocity.

p = _________________________________2. The unit of momentum is ____________________________. It can be also be written as N s (newton second)3. Momentum is a ________________________ quantity with the same direction as velocity.4. If the direction to the right is denoted as positive, an object moving to the right possesses a positive momentum while an object moving to the left will have a _____________________ momentum.5. Examlple:

A billiard ball A of mass 0.5 kg is moving from left to right with a velocity of 2 m s-1 while another billiard ball B of equal mass is moving from right to left with the same speed. Calculate the momentum for both balls.Solution

Momentum of ball AMomentum of ball B

= mAvA= mBvB

= 0.5 x 2= 0.5 x (-2)

= 1 kg m s-1= -1 kg m s-1

Negative sign shows the object is moving in the opposite direction.

6. The diagram shows a baseball player hitting a ball. The mass of the ball is 200g. What is the momentum of the ball flying with a velocity of 50ms-1?

7. A ball of mass 0.8kg strikes a wall at a velocity of 10ms-1 and rebounds at 6ms-1. What is its momentum(a) Before it strikes the wall, and(b) After the rebound?

8. A bull of mass 250kg is moving at a momentum of 750kg ms-1. Find the velocity.

Conservation of Momentum1. The Principle of conservation of momentum states that:The total momentum of in a closed system is constant, if no external force acts on the system.

Conservation : the total amount of matter / quantity remains the same before and after the occurrence of an event. A closed system : the sum of external forces acting on the system is zero. Example of an external force is friction.2. The principle of conservation of momentum shall be discussed in two situations.A collisionAn explosion

The total momentum of the objects: before a collision = after the collision.The sum of the momentums remains as zero after an explosion.

Collision3. There are two types of collisions.Elastic CollisionsInelastic collisions

Two objects collide and move apart after a collision.Two objects combine and stop, or move together with a common velocity after a collision.

Momentum is conserved Total energy is conserved Kinetic energy is conserved Momentum is conserved Total energy is conserved Kinetic energy is not conserved:the total kinetic energy (after the collision) the total kinetic energy (b4 the collision)

Formula:

Formula:

4. There are two types of collisions. The first is an inelastic collision, which is more common in everyday life. 5. During an inelastic collision, some of the kinetic energy, or movement energy, is lost on impact. This energy is converted into another type of energy, such as sound or heat.6. An elastic collision occurs when the total kinetic energy, or movement energy, of two or more objects is the same after a collision as before the collision.7. Unlike an inelastic collision, no energy is transformed into another type. Completely elastic collisions dont usually happen in the real world, aside from between subatomic particles, but the collision between two billiard balls is a close approximation.

8. Another example of approximate elastic collisions

When the 1st ball is pulled to the side and then released so as fall back and strike the 2nd ball. It is observed that the 1st ball stops, but the last ball swings out to the same height from which the ball P was released. This shows the last ball possesses the same amount of momentum and kinetic energy as 1st ball before it struck the 2nd ball. The total momentum of the ball: before a collision = after the collisionActivity 2.3To verify the principle of conservation of momentum in(a) Elastic collisions, and(b) Inelastic collisionsApparatus / Materials:Ticker-timer, 12 V a.c. power supply, runway, 4 trolleys, wooden block, ticker tape, cellophane tape, and plasticine.(A) Elastic CollisionArrangement of apparatus:

Procedure:(i) The runway is adjusted to compensate the friction.(ii) Trolley A with a spring-loaded piston is placed at the higher end of the runway and trolley B is placed halfway down the runway and stayed at rest.(iii) Two ticker tapes are passed through the ticker timer, one attached to trolley A and another attached to trolley B.(iv) The ticker-timer is switched on and trolley A is given a slight push so that it moves down the runway at a uniform velocity and collides with trolley B which is stationery.(v) After collision, the two trolleys move separately.

Result:Ticker-tape obtained:

(a) Trolley A

(b) Trolley B

Discussion1. The spring-loaded piston acts as a spring buffer in the collision in order to make the trolley bounce off the other one.2. Strictly speaking, this collision is not a perfect elastic collision as part of the kinetic energy of the colliding trolley changes to sound or heat energy during the collision.Conclusion:Total momentum before collision = total momentum after collisionThe principle of conservation of momentum is verified.

(B) Inelastic collisionArrangement of apparatus

Procedure:(i) The runway is adjusted to compensate the friction.(ii) The spring loaded piston of trolley A is removed and some plasticine is pasted onto trolley A and B.(iii) A ticker tape is attached to trolley A only.(iv) The ticker-timer is switched on and trolley A is given a slight push so that it moves down the runway at a uniform velocity and collides with trolley B which is stationery.(v) After collision, the two trolleys are move together.

Result

Conclusion:Total momentum before collision = total momentum after collisionThe principle of conservation of momentum is verified.

Exercise:

1. Block A of mass 5 kg is moving with velocity 2 m s-1 and collides with another stationery block B of unknown mass. After the collision, block A moves with velocity 0.5 m s-1. Given that the collision is elastic. Find the momentum of block B after the collision.

2. A truck travels at a velocity of 15 m s-1 collides head-on with a car that travels at 30 m s-1. The mass of the truck and the car are 6000 kg and 1500 kg respectively. What is the final velocity of the two vehicles after the collision if they stick together?

3. An astronaut of mass 90kg moves at a velocity of 6ms-1 and bumps into a stationary astronaut of mass 100kg. How fast do the two astronauts move together after collision?

4. A 50kg skater is moving due east at a speed of 3ms-1 before colliding into another skater of mass 60kg moving in the opposite direction at a speed of 7 ms-1. After the collision, the two skater hold on to each other. In which direction will they move? What is the speed of the two skaters?

5. A trolley of mass 3 kg moving at a velocity of 2 ms-1 collides with another trolley of mass 0.5 kg which is moving at a velocity of 1 ms-1 in the same direction. If the 0.5 kg trolley moves at a velocity of 2.5 ms-1 in the same direction after collision, what is the velocity of the 3 kg trolley?

6. The diagram shows two trolleys of different masses before and after a collision. Find the velocity of the 3 kg trolley after the collision.

Momentum and Explosions1. Rifle(a) Before explosion(b) After explosion1. When the rifle is fired, the explosion of the gunpowder forces the bullet out of the barrel. A momentum in the forward direction is created.2.The explosion creates a backward momentum on the rifle. This causes the rifle to recoil backwards.Total momentum of the rifle and the bullet is zero as they are stationary.

2. Air escapes from a deflating balloonThe balloon shoots upwards, moving with an upward momentum

Total momentum of the balloon is zero as it is stationary

Air has mass and moves with a velocity. This creates a momentum in the downward direction.

3. A cannon firing a shell

It produces two fragments of masses m (shell) and M (cannon) with velocities u and v respectively.4. An explosion is a closed system which does not involve any external force.The total momentum is conserved in an explosion.

The two momenta have the same magnitude but different direction.ExerciseThe diagram show a boy of 60kg and a girl of 50kg on the roller skates. Initially, they hold each others hand. Then, the boy pushes the girl and both of them release their hands. If the velocity of the girl is 3 ms-1 after being pushed, find the velocity of the boy. (Assume that no frictional force)

1.

Jane and John go ice skating. With their skates on, Jane and John push against each other on level ice. Jane, of mass 50kg, moves away at a velocity of 3ms-1 to the right. What is Johns velocity if he is 75kg?

2.

3. Harfeez alight a boat at a velocity of v ms-1 and lands on the dock. The boat bounces backwards with a speed of v. If the masses of Harfeez and the boat are m and 3m respectively, find the speed of the boat (ignoring friction due to water).

Activity 2.4To verify the principle of conservation of momentum in an explosionApparatus / Materials:4 trolleys, wooden block, 2 wooden blocks, a hammer and a metre rule.NotesThe positions of the wooden blocks are adjusted so that each trolley collides with the corresponding wooden blocks at the same time, t. Substitute in

Arrangement of apparatus:

Procedure:(i) The apparatus is arranged as shown in figure.(ii) Two trolleys A and B of equal mass are placed in contact with each other on a smooth surface. The spring-loaded piston in trolley B is compressed.(iii) The release pin on trolley B is given a light tap to release the spring-loaded piston which then pushes the trolleys apart. The trolleys collide with the wooden blocks.(iv) The experiment is repeated and position of the wooden blocks are adjusted so that both trolleys collide with them at the same time.(v) The distance dA and dB are measured and recorded. Note: dA is positive while dB is negative since d, displacement is a vector quantity.(vi) The experiment is repeated using(a) 1 trolley with 2 stacked trolleys(b) 3 stacked trolleys with 1 trolley

Tabulation of data

Discussion

Conclusion

Application of the Conservation of Momentum1. Rocket54In accordance with the principle of conservation of momentum, the rocket gains a forward momentum and moves forward at high velocity.A backward momentum is createdHot exhaust gases at high velocity321A rocket carries liquid hydrogen and liquid oxygenThe mixture of hydrogen fuel and oxygen burns vigorously in the combustion chamberThe gases formed expand rapidly and are forced to discharge through the exhaust nozzle at a high velocity

Demonstration to show the principle in rocketWater rocket

2. Jet engine321

In the combustion chamber, kerosene fuel burns vigorously with the compressed airAir from the atmosphere is drawn into the engine and compressed by a compressor before it is forced into combustion chamber at high pressure

In accordance with the principle of the conservation of momentum, a forward momentum for the engine is produced. The plane thus flies forwards.6The ejected high speed exhaust gases create a backward momentum565Jet of exhaust gasesMoves forwardsThe hot gases formed expand rapidly and are forced out of the nozzle at high speed through the turbine which rotates the compressor4

3. The principle of the conservation of momentum occurs in nature.A squid uses it to propel itself in the water.The squid moves forward by discharging a jet of water from its body. An equal and opposite momentum created thus propel the squid in the opposite direction.Diagram show how does a squid escape.The squid fills its body up with water which it forces through a tube. This makes the squid move on the opposite direction very quickly. This is called jet propulsion.

4. The shower of burning fragments from an exploding fireworks launched into the sky is governed by the principle of the conservation of momentum. The symmetrical pattern indicates that the total momentum is conserved.

5. The large volume of water that rushes out from a water hose with a very high speed has a large momentum. In accordance with the principle of conservation of momentum, an equal and opposite momentum is created causing the fireman to fall backwards. Thus, several firemen are needed to hold the water hose.

Exercise1. A pigeon of mass 120 g is flying at a velocity of 2 ms-1. What is its momentum?

2. Calculate the momentum of a car of car of mass 2 000 kg moving with a velocity of 30 ms-1.

3. A bull of mass 250 kg is moving at a momentum of 750 kgms-1. Find its velocity.

4. A bullet with mass of 50 g is fired from a 2 kg pistol with a velocity of 120 ms-1. What is the total momentum of the bullet and the pistol after explosion?

5. During training, Othman fires a pistol of 1.5 kg mass.

A bullet with a mass of 30 g is released at a velocity of 300 ms-1. What is recoil velocity of the pistol?

6. A 500 kg car travelling at 30 ms-1 collides with a 3000kg lorry which is at rest. After the collision, both the car and lorry move together. What is the common velocity after the collision?

7. Hizam and his son Jamal are at an ice rink.Jamal with a mass of 20 kg is moving at a velocity of 2 ms-1 while Hizam with a mass of 60 kg, is directly behind Jamal and moving at 6 ms-1. Hizam decides to pick Jamal up and continues moving without stopping. Determine the final velocity of Hizam and Jamal.

8. Trolley A of mass 3 kg moving at 4 ms-1 collides with trolley B of mass 4 kg which is moving in the opposite direction. If both the trolleys move together at 3 ms-1 in the direction of trolley B after the collision, find the initial velocity of trolley B.

9. A butterfly rests on a leaf floating on the surface of a pond. The butterfly then starts moving to the tip of the leaf at a speed of 5 cms-1 relative to the water. The leaf, in accordance with the principle of the conservation of momentum, moves at 3 cms-1 relative to the water in the opposite direction. If the mass of the leaf is 8g, determine the mass of the butterfly.

10. Boat A and boat B are moving at a speed of 2 ms-1 and 1 ms-1 respectively before the two collide head on. The masses of boat A and B (including the passengers) are 150 kg and 250 kg respectively.If boat A bounces back with a velocity of 0.5 ms-1, what is the velocity of boat B?

11. Sau Fei and Siew Ling, each with a mass of 60 kg and 49.5 kg respectively, are standing at rest on an ice rink. Sau Fei throws a ball of mass 0.5 kg towards Siew Ling.

What is the recoil velocity of Sau Fei if the velocity of the ball is 8 ms-1? What is the velocity of Siew Ling after she receives the ball?

2.5 Understanding the Effects of a Force

What is Force?Kicking a ballStretching a chest expanderPulling off the ring of a soft drink tinLifting objectsA force is a push or a pullPressing a switch

1. When you push or pull on an object, you need to know(a) The strength or magnitude of your force, and(b) The direction in which you are pushing and pulling.

2. Force is a _________________________ quantity. Since it has both _____________________ and direction.

3. A spring balance, which is used to measure the weight (or the gravitational pull of object) can be used to measure the magnitude of a force.

The force pulling a cup can be measured as shown in above figure.

The Effect of a Force1. A force can change the shape of an object. (deformation of an object).

Can change the shape of an object A platicine is flattened

Can change the size of an object A can is crushed.

A spring lengthens or compresses when you stretch or compress it.

A plastic ruler can be bent when a force exerted on it.

2. A force can change the original state of motion of an object.(from stationary to move / from moving state to stationary)To move a stationary object. A pushing force is required to move a stalled car

To stop a moving object A goalkeeper stop the ball

Can change the speed of a moving object A ping-pong ball moves faster

Can change the direction of a moving object A ping-pong ball changes its direction

Relationship between Acceleration, Mass and ForceExperiment 2.2Relationship between acceleration and force applied on a constant mass.Situation:

Figure (a) shows car A and car B of the same mass at the same starting line. Car B is a sport car.The engine capacity of sport car B is much bigger than car A. (a car with a bigger car capacity can provide greater engine thrust.)Figure (b) shows that sport car B has built up a higher velocity than car A after 3 seconds.

Can you make an inference about the situation?Inference:______________________________________________depends on _____________________________________________.HypothesisWhen the mass of an object is constant, the greater ______________________________________________, the greater _______________________________________________.Aim:To investigate the relationship between acceleration and the force applied on a constant mass.Variables:(a) Manipulated: _________________________________________________________________________________(b) Responding: __________________________________________________________________________________(c) Constant: _____________________________________________________________________________________

Notes:The force in this experiment is the stretching force in an elastic cord used to pull the trolley. A length of elastic cord attached to the trolley and stretched to a fixed length represents one unit of force acting on the trolley.Apparatus/Materials:Trolley, 3 identical elastic cords, runway, ticker-timer, carbonised ticker-tape, cellophane tape, 12 V a.c. power supply and a wooden block.Arrangements of apparatus:

Procedure:1. A friction-compensated inclined runway is prepared.2. The apparatus is then set up as shown in figure.3. The ticker-timer is switched on and the trolley is pulled down the runway by an elastic cord attached to the hind post of the trolley.4. The elastic cord is stretched until the other end is level with the front end of the trolley. The length is maintained as the trolley run down the runway.5. The ticker tape obtained is cut into strips of 10-tick. A tape chart is constructed and the acceleration, a, is determined.6. The experiment is repeated with 2, and 3 elastic cords to double and triple the pulling force to the same constant extension as when one elastic cord is stretched.Result:(build your own ticker tape chart)Tabulation of data:

Conclusion:

Experiment 2.3Relationship between acceleration and mass of an object under a constant force

Situation:

Figure (a) shows two similar lorries, A and B in front of a traffic light. When the light turns green, both drivers step on the accelerator simultaneously with the same pressure to provide the same engine thrusts, F.Figure (b) shows that within 3 seconds, the empty lorry has built up a higher velocity than the heavy one.Can you make an inference about the situation?Inference:____________________________________________ depends on _______________________________________________HypothesisWhen the force applied on an object is constant, the greater ____________________________________, the ____________________________________________________.Aim:To investigate the relationship between acceleration and the mass of an object under a constant forceVariables:(a) Manipulated: _________________________________________________(b) Responding: __________________________________________________(c) Constant: _____________________________________________________Notes:(a) The mass in this experiment is represented by the number of identical trolley used.(b) The constant force is applied by stretching the elastic cord with the same extension for each repetition of the experiment.Apparatus/Materials:Ticker-timer, 12 V a.c power supply, 3 trolleys, elastic cord, runway, wooden block, ticker-tape and cellophane tape.Arrangements of apparatus:

Procedure:1. A friction-compensated inclined runway is prepared.2. The apparatus is then set up as shown in figure.3. The ticker-timer is attached to the trolley and passed through the ticker-timer.4. The ticker-timer is switched on and the trolley is pulled down the runway by an elastic cord attached to the hind post of the trolley.5. The elastic cord is stretched until the other end is level with the front end of the trolley. The length is maintained as the trolley run down the runway.6. The ticker tape obtained is cut into strips of 10-tick. A tape chart is constructed and the acceleration, a, is determined.7. The experiment is repeated using 2 trolleys (with a second trolley stacked on the first trolley) and 3 trolleys. The elastic cord is stretched to the same fixed length as in the first experiment.Result:(build your ticker tape chart)Tabulation of data

Conclusion:

Newtons Second Law of MotionFrom the experiment 2.2:The acceleration is directly proportional to the force

From the experiment 2.3:The acceleration is inversely proportional to its mass.

The unit of force is Newton, N.In order to make the formula as simple as possible, we make = 1.

Force of 1N is defining as,1 N is the force which gives a mass of 1 kg an acceleration of 1 ms-2

Exercise1. A force of 10 N acts on an object of mass 5 kg on a smooth floor. Find the acceleration.

2. A car of mass 1200 kg travelling at 15 ms-1 comes rest over a distance of 30 m. Find (a) The average retardation, and (b) The average braking force.

Balanced Forces and Unbalanced Forces1. In general, there may be several forces acting on the mass, whether parallel or anti-parallel, or in different directions.2. Thus, the force, F, must be replaced with net or resultant force when there are several forces acting on the mass.

Where a is in the direction of the net or resultant force.3. However, for simplicity, is always used. But F is represented as net force acting on the object (whether a single force or several forces are acting on it.)

Balanced Forces1. When the forces acting on an object are balanced, (net force = 0)2. The object then behave s as if there is no forces acting on it.3. Since Fnet = 0, the acceleration of the object, a =0. Thus, the object remains at rest or moves at________________________________ when there is no net force acting on it.4. This is Newtons first law of motion.5. Example of balanced forces:(a) Balanced forces on a stationary gymnast The weight of the gymnast, W, is balanced by the reaction force, R, from the beam. The two forces are of equal magnitude but opposite in direction. Without the beam (that is, no reaction force), the gymnast will fall to the ground because of her weight.

(b) Balanced forces on a car moving at a constant velocity.

There are 3 horizontal forces acting on a car moving at a constant velocity. The forward thrust, T, provided by the car engine is balanced by the frictional force on the wheels and the air resistance.T = G + Ff

The weight of the car, W, is vertically balanced by the reaction force, R, from the road.W = R

Balanced forces (Fnet = 0, a = 0)Fnet = 0(as no force acting on it)

F1 = F2From Fnet = ma0 = ma (since mass, m, cannot be zero)Object at rest(v = 0 ms-1)Object in motion(v 0, object is moving at constant velocity)

Zamri pushes a heavy cupboard with a force of 200N, but the cupboard does not move.Find the frictional force acting on the cupboard.Linda pushes a book on a table with a force of 5 N. The book moves with a uniform velocity of 2 cm s-1. Find the frictional force acting on the book.

Solution

Using Fnet = maBecause the cupboard does not move

But Fnet = 0, since a = 0

(the frictional force here is known as static friction)

Solution:Because the book move with a uniform velocity

Using Fnet = maBut Fnet = 0, since a = 0

(the frictional force here is known as dynamic friction)

Unbalanced Forces1. When the forces acting on an object are not balanced, the object will accelerate in the direction of the net force.2. The net force is known as the resultant force.

Effect of Balanced Forces and Unbalanced Forces on an Object

Unbalanced forces (Fnet 0, a 0)

When the forces acting on an aircraft do not cancel out each other, a net force known as unbalanced force is acting on the object. Unbalanced forces produce an acceleration to the mass on which force are acting. However, the object will accelerate in the direction of the net force. When an airplane is moving at a constant velocity, if the pilot increases the engine thrust, the forces acting horizontally are no longer balanced. There is a net force and plane will accelerate in the forward direction.

Balanced forces (Fnet = 0, a = 0)

Balanced forces on an aircraft allow it to move at constant velocity at a constant altitude. The engine thrust is balanced by the drag due to air resistance while the weight of the aircraft is balanced by a lift from the wings.

Newtons Third Law of Motion1. When you are practising arm wrestling with your friend, you will feel a force acting on you.2. Your friend will also be feeling a force acting on him.3. This is because while he is applying a force on you, you are also applying a force on him.4. Hence, there is a pair of forces acting and Newtons third law of motion states:If object A exerts a force, F on object B, the object B will exert an equal but opposite force, - F on object A.

5. This law is also known as the action reaction law. Hence, Newtons third law of motion can be quoted as:To every action, there is an equal but opposite reaction.

6. Everyday phenomena that are governed by newtons third law of motion:(a) A swimmer pushes the water backward with his hands and legs with a force, F, while the reaction force of water, F, pushes the swimmer forward.

(b) When a boy presses on the wall with a force, F, the wall presses on his hand with a normal reaction force, F.

(c)

When a man paddles with a backward force, + F (action), the reaction force, F, pushes the boat forwards.

(d)

The principle used in rockets and jet engines can be also be explained by newtons third law of motion. The action that pushes the exhaust gases out through the nozzle results in a forward force (reaction forces) that propels the rocket or jet engine forwards.

Summary of Newtons Law of MotionNewtons Second Law of MotionThe relationship between an object's mass, m, its acceleration, a, and the applied force, F is F = ma.In this law the direction of the force vector is the same as the direction of the acceleration vector.Newtons Third Law of MotionFor every action there is an equal and opposite reaction.Newtons First Law of MotionEvery object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.This is often termed simply the "Law of Inertia"

Exercise1. Find the acceleration of the objects.(a)

(b)

2. Find the value of F.

3. Calculate the value of m.

4. A 10 kg object accelerates from rest to 5ms-1 in 2 s. What is the net force acting on the object?

5. A force of 10 N acts on a 2 kg stationary object. Calculate the distance travelled by the object in 5 s.

6. An object with a mass of 5 kg moves at a constant velocity when it is pulled by a horizontal force of 5 N on a level surface. What is the acceleration of the object if pulled with a force of 25 N

7. Figure shows a small rocket of mass 300 kg at the point of take-off. Find its initial acceleration.

8. John pushes a 12 kg carton with a force of 50 N.

If the carton moves an acceleration of 2 ms-2, what is the frictional force acting on the carton?

Note: When a floor is _____________________, friction force is ____________________.

When an object on a ________________________ moves at uniform velocity with a horizontal force acting on it, friction is ____________ in magnitude to the applied force (but acts in the ______________________ direction.)

9. A shopper pushes a trolley with a force of 20 N. The trolley with a mass of 5 kg, moves at a uniform velocity of 1 ms-1. He then increases his force to accelerate the trolley. What force should he apply in order to give the trolley an acceleration of 2 ms-2?

10. Mr Brown whose mass is 70 kg, performs as a human cannonball at a circus. He is propelled from a 6 m long cannon. He is in the barrel of the canon for 0.9 s. Find the average net force exerted on him.

11. The figure below shows a car of mass 1200 kg moving at an acceleration of 2 ms-2. If the frictional force acting on the car is 750 N, find its engine thrust.Frictional force, 750 Nacceleration

12. What force is needed so that an object with a mass of 3kg has an acceleration of 2ms-2?

13. A wooden box of mass 2kg is placed on a smooth plane. If a force of F is applied to the box, it moves at an acceleration of 3ms-2. Find the acceleration of another box with mass of 6 kg if the same force is exerted on it.

14. Puan Zaitun pushes a trolley of mass m kg with a force of 30 N. The trolley moves with a uniform velocity of 1 ms-1. When she doubles her force, the trolley accelerates at 2 ms-2. Find the mass of the trolley.

15. Faizal rides a bicycle at a constant speed of 16 ms-1. He stops pedaling and the bicycle stops completely after 6 s. Given the total mass of Faizal and his bicycle is 72 kg, find the average opposing force on Faizal and his bicycle.

16. A race car of mass 1200 kg accelerates from rest to a velocity of 72kmh-1 in 8s.(a) Find the acceleration of the car.(b) Find the net force acting on the car.

17. (a) When a 2 kg block is pushed with a force of 12 N to the right, it accelerates with an acceleration of 3 ms-2. In which direction does friction act? Find its magnitude.

(b) An additional force of 20 N to the left then acts on the block. In which direction does friction now act? Find the new acceleration of the block.

18. A car with a mass of 1560 kg decelerates from a velocity of 22.5 ms-1 to a stop in front of a traffic light. The time taken for the deceleration is 4.5 s.

(a) Calculate the deceleration.(b) What is the net force needed to give this deceleration?

2.6 Analysing Impulse and Impulsive Force

Impulse and Impulsive Force When a player kicks a football, his boot is in contact with the ball for a time t During the time t, an average force F acts on the ball which makes the ball fly off with a momentum. Thus, the force F, acting for a period of time t, produces a change in momentum to the ball, since the ball with a mass of m acquires velocity after the time t.

When a badminton racket hits an oncoming shuttle cock, the badminton racket delivers a large force that acts on the shuttle cock for a short time t. The shuttle cock bounces off in the opposite direction. Again, there is a change in momentum. The change in momentum is due to the force F acting on the object for a time t.

Impulsive Force = Impulse is defined as the product of a force, F and the time interval, t during which force acts.Impulse = Force Time = F t

substitute

F

ImpulseChange in momentum

1. Both impulse and impulsive force are vector quantities.Rebound and Impulse1. An object might rebound from a wall, or stick to it without rebounding after striking it.2. In which situation will the wall exert a greater impulse? 3. A tennis ball and a piece of mud with the same mass (0.060 kg) which are moving at 9 ms-1 strike a wall. The mud sticks to the wall while the ball rebounds at 6 ms-1. Find the impulse on each object.

4. Thus, a greater impulse is exerted on an object if it ________________________ after a collision.Exercise:1. Figure shows a 2 kg wooden block initially at rest on a smooth surface. A force of 8 N is applied on the wooden block.

If the force acts for 3 s,(a) What is the impulse on the block?(b) What is the velocity of the wooden block after 3 s?

2. A player spikes an oncoming volleyball moving towards him at a speed of 5 ms-1 to reserve its direction at a speed of 20 ms-1. The mass of the ball is 0.36 kg.

(a) Find the impulse on the ball.(b) If the average force acting on the ball is 300 N, how long is the time of contact between the hand and the ball?

Effects of Time on Impulsive ForceF t small, F large t large, F small

1. From F

If the change in momentum is constant, then: F 2. A student throw a raw egg at a high speed at a wall, and another egg against a towel held by his friends. In which case will the egg break? The egg is stopped by the wall in a very short time interval, resulting in a large force which causes it shatter.The egg thrown against a towel falls on a few towels stacked below and does not break. The movement of the egg is stopped in a longer time interval, resulting in a small force.

In both cases, the eggs are stopped. Thus, the change in momentum is the same.

3. To understand the effect of time in a collision, look at the following example.Herman (mass, m = 50 kg) jumps down from a wall. He lands on a cement ground at a velocity of 6 ms-1.

Herman bends his knees upon landing.The time taken to stop his motion is 1.0 s.

v = 0 ms-1u= 6 ms-1

Herman didnt bend his knees upon landing.The time taken to stop his motion is 0.05 s.

v = 0 ms-1u= 6 ms-1

This example clearly shows that an effective way to reduce the impulsive force is to lengthen the collision time.The effects of time on the magnitude of the impulsive force Increasing the impulsive force by reducing the time of impactt smallF large

1. In all sports shown below, the time of impact with the ball is very small and the impulsive force producedis large.

A tennis player hitting a tennis ball

A golfer driving a golfball with a club

A footballer kicking a football

It should be noted that impulsive forces always exits in pairs. In the figure, the change in momentum of the tennis ball produces a large impulsive force on the racket which reacts to give rise to an equal but opposite impulsive force to the ball this s on accordance with Newtons third Law of motion) Both the racket and the tennis ball are formed temporarily due to the large force being exerted on each other.

F2F1

A massive hammer head moving at a fast speed is bought to rest upon hitting the nail. The large change in momentum within a short time interval produces a large impulsive force which drives the nail into the wood.

2.

An expert in karate can split a thick wooden slab with his bare hand which is moving at a very fast speed. The momentary contact between the fast-moving hand and the wooden slab produces a large impulsive force which split the wooden slab.

3.

4. Food such as chillies and onions can be pounded using a mortar and pestle. The pestle is brought downwards at a fast speed and stopped by a mortar in a very short time. This produces a large impulsive force which cruses the food.

Reducing the impulsive force by increasing the time of impactt largeF small

1. Polystyrene and cardboard egg containers stiff but compressible. They will absorb and reduce impulsive force by lengthening the time of impact.Cardboard egg cartonpolystyrene

In sports, the effects of impulsive forces are reduced to prevent injuries to athletes. Thick mattresses with soft surfaces are used in events such as the high jump and pole-vaulting so that the time of impact on landing is extended, thus reducing the resultant impulsive force.

2.

3. The use of padding in certain sports equipment like baseball gloves, goalkeeping mitts, pole vaulting pits, boxing gloves, and gymnastic mats is to prevent injuries to players by reducing the impulsive force.

In baseball, a player must catch the ball in the direction of the motion of the ball. If the ball is caught by stopping it in its path, the impulsive force acting on the hand will be considerable. Moving his hand backwards when catching the ball prolongs the time for the momentum change to occur so the impulsive force is reduced.

When a boxer sees that his opponents fist is going to hit his head, he will move his head backwards or duck. This will increase the stopping time, hence reducing the average force on his head since the momentum change will be longer.

Playgrounds are covered with a coarse fabric material which prolongs the time of impact when the children fall, thus reducing the impulsive force

4.

Exercise1. A tennis player hits an oncoming 0.06 kg tennis ball with a velocity of 60 ms-1. The ball bounces off in the opposite direction at 90 ms-1.Find the time of impact between the racket and the ball if the impulsive force acting on the ball is 125 N.

2. Tiger Woods hits a golf ball of mass 0.045 kg at a velocity of 30 ms-1.If the time of impact is 0.005 s, what is the average impulsive force applied on the ball by the club?

3. A force F acts on a 6 kg object at rest on a smooth surface. If the velocity of the object increases to 2 ms-1 in 6 s, what is the value of F?

4. An object of mass 2 kg is acted on by a force which causes the velocity of the object to increase from 1 ms-1 to 9 ms-1. What is the impulse on the object?

5. The figure shows a helicopter dropping a box of mass 60 kg which touches a sandy ground at a velocity of 12 ms-1. The box takes 2 seconds to stop after it touches the ground.What is the magnitude of the impulsive force on the box?

6. Beckham kicks a ball with a force of 1500 N. The time of contact of his boot with the ball is 0.008 s. what is the impulse delivered to the ball? If the mass of the ball is 0.5 kg, what is the velocity of the ball?

7. A 0.045 kg golf ball strikes a wall at a speed of 30 ms-1 and rebounds at a speed of 20 ms-1. What is the impulse on the ball? If the force on the ball is 500 N, find the contact time of the ball with the ball.

8. A baseball of mass 0.14 kg moving at 40 ms-1 is stuck by a bat and rebounds at 60 ms-1.Given that the time of contact is 5 10-2 s, find the force exerted on the ball.

9. In a crash test, a car of mass 1500 kg crashes into a wall at 15 ms-1. It rebounds at 2 ms-1.If the collision time is 0.18 s, find the force exerted on the car.

10. A pole-vaulter (mass, m = 50 kg) falls onto a foam mattress which exerts a force of 250 N on him over a time interval of 2.0 sFind his velocity just before landing on the mattress.

11. A hockey player uses a hockey stick to hit a hockey ball, causing the ball to travel with a velocity of 30 ms-1. If the mass of the hockey ball is 100 g and the time of impact between the hockey stick and the ball is 5 10-3 s, calculate the impulsive force exerted by the hockey ball.

12. A boy with a mass of 60 kg jumps over a fence. His velocity just before landing on the ground is 10 ms-1. What is the impulsive force on the boys legs if he (a) Takes 0.5 s to stop,(b) Bends his legs and stops in 2.5 s?

13. A trolley with a mass of 500 g is at rest on a smooth surface. The trolley is then given a horizontal impulsive of 5 Ns. What is the velocity of the trolley after the impact?

14. A horizontal impulse of 500 N is exerted on a stationary trolley with a mass of 2 kg. What is the velocity of the trolley after the impact?

15. A baseball with an initial velocity of 20 ms-1 is hit by a player and sent moving in the opposite direction. After the ball is hit, it moves with a velocity of 36 ms-1. The ball has a mass of 0.16 kg and the time of impact is 8.0 10-3 s.Calculate(a) The impulse applied to the ball(b) The impulsive force exerted on the ball by the bat

16. A 0.6 kg metal lock falls straight down with a velocity of 12 ms-1 and hits the roof of a parked car. The lock bounces back vertically from the roof with a velocity of 5 ms-1. The time of impact is 1.2 10-2 s.Calculate the magnitude of the impulsive force acted by the lock on the roof.

17. A golfer hits a golf ball of mass 50 g and the ball leaves the club with a velocity of 75 ms-1. The contact time between the ball and the club is 0.008 s. find (a) The final momentum of the ball,(b) The average force exerted on the ball by the club.

2.7 Being Aware of the Need for Safety Features in Vehicles.

1. It is important to increase the time interval of collision to reduce the impulsive force in an accident. 2. A good car safety system that can prevent injuries to passengers of an accident is of utmost importance.3. To protect passengers from harm in a car crash, the car must be able to absorb as much forces from the impact as possible.

Safety featureImportance

Bumper / Crumple zoneThe front and rear parts of a vehicle are designed to ________________________ slowly upon collision so that the impact time is _____________________ to _________________________ the ______________________ on the vehicle. Hence a ______________________ force will act on the passenger to minimise the possibility of getting serious injury.

Safety beltsMost vehicles are equipped with safety belts. The safety belts will prevent the passenger from _______________________________ the windscreen or being thrown out of the vehicle and _______________________ the forward movement of the passenger when the vehicle stops suddenly.

Air bagsAir bags will prevent our head from striking the ___________________ or __________________________ in a head-on collision. Air bags will act _________________________ and take less than 0.05 s to inflate. The protection from air bags will reduce the __________________________ acting on the head.

Anti-lock braking system(ABS)ABS allows the driver to ____________________________the vehicle even when the brake are applied. The system ______________________ wheels from locking automatically which can cause the vehicle to skid.

HeadrestIn rear-end collision, our head will snap back due to _________________. All passenger vehicles are equipped with headrests to support the head and prevent injuries on neck when the head ____________________________ strongly.

Windscreen with safety glassThe windscreen glass in specially designed to break into ___________________ instead of shattering. This will prevent the driver and the passengers from being ______________________ by the pieces of broken glass.

Padded dashboardThe dashboard is covered with soft materials to ___________________ the time interval of collision. This can ______________________ injuries on head of passenger because the impulsive force produced during impact has been __________________________.

Side- impact barsSide collision can be _____________________ since it involves the weakest part of a car body. The _________________________ give good protection from side collision. The strong metal bars help to protect the driver and passengers from a ___________________ hit on the side.

2.8 Understanding Gravity

1. The concept of gravitational force was introduced by Sir Isaac Newton who, on seeing an apple falls on his head.2. According to Newton, objects fall because they are pulled towards the Earth by the force of gravity.

3. The pull of gravitational force(a) Keeps things on the earth(b) Brings things down to earth when they are thrown upwards(c) Holds the moon in its orbit round the Earth(d) Captures returning space capsules and pulls them into orbit4. The pull of gravity causes objects to fall with acceleration. This means that objects that fall are moving with increasing velocity.5. The magnitude of the acceleration due to gravity depends on the strength of the gravitational field.

Stroboscopic Photograph

1.A stroboscopic photograph is a photograph that shows the images of an object in motion. The images are taken at regular time intervals.2.Figure below shows a stroboscopic photograph of a feather and an apple falling under gravity. Both feather and apple are dropped simultaneously from the same height.3.The time intervals between two successive images are the same.Inference 1Both feather and apple are falling with an acceleration. The distance between two successive images increases, showing that the two objects are falling with increasing velocity, (= acceleration)Inference 2Both feather and apple are falling with the same acceleration.Thus, a heavy object and a light object fall with the same gravitational acceleration. gravitational acceleration is not depends on mass.

Free Fall

1. A free falling object is an object falling under the force of gravity only.2. A free falling object does not encounter other forces like air resistance or friction that would oppose its motion.

3. Figure shows a free fall in a vacuum cylinder (where air resistance does not exist) of a coin and a piece of paper. Both objects reach the bottom of the cylinder at the same time.4. A piece of tissue paper (fall in atmosphere) does not fall freely because its fall is affected by air resistance.5. A heavier golf ball can be considered to be free falling because the air resistance is small compared to pull of gravity and therefore is negligible.Tips: When an object falls; g = 9.8 ms-2 When an object is thrown upwards; g= -9.8 ms-2 At the highest point, v = 0 ms-1

6. Objects dropped under the influence of the pull of gravity accelerate at a constant rate.7. This acceleration is known as the gravitational acceleration, g8. The value of g is 9.8 ms-2. In calculation, the value of g is often taken to be 10 ms-2 for simplicity.9. The acceleration due to gravity does not depend on the mass and shape of the falling object.10. All objects falling freely with the same acceleration.

Hans on Activity 2.5To determine the value of the gravitational acceleration, g

Apparatus/material:Ticker timer, ticker tape, 12 V a.c electrical power supply, retort stand, weights (50 g- 250 g), G-clamp, cellophane tape and soft board.

Procedure:1. A ticker-timer is clamped to a retort stand with a G-clamp and placed on a tabletop.2. One end of a carbonised ticker tape (approximately 1 m in length) is attached to a weight holder with a total mass of 200 g.3. The other end of the ticker tape is passed through the ticker-timer.4. A soft board is placed on the floor below the weight to stop its fall.5. The ticker-timer is switched on and the weight is released so that it falls squarely onto the soft board.6. Six strips are cut off from the middle section of the ticker tape with each strip containing 2 dot-spaces.7. A tape chart is construed. Form the chart, the acceleration of gravity is calculated.

Show your result, calculation of g, discussion and Conclusion in Experiments Report.

Weight, W and Gravitational Field, g

1. A Gravitational field is a region around the earth in which an object experiences a force towards the centre of the earth. This force is the gravitational attraction between the object and the earth.2. The gravitational field strength is defined as the ratio of the weight to the mass of the object or weight per unit mass.Unit = N kg-1

3. Gravitational field strength = =

Rearranging the formula,

Compare this with the formula g

Gravitational field strength g

Gravitational field strength = g = Gravitational acceleration = 9.8 Nkg-1 = 9.8 ms-24. The gravitational field strength at the surface of the earth is 9.8 N kg-1.5. The weight of an object is defined as the force of gravity which is exerted on it by Earth.6. From the formula,

Since weight, W, is the force of gravity acting on an object of mass, m that makes it fall with an acceleration, g,

Wherem = mass of object.g =gravitational acceleration

Weight of object= Mass of object Acceleration due to gravity

7. The unit of weight is Newton, N.

Mass and WeightThe differences between mass, m and weight, W:Mass, mWeight, W

The mass of an object is the amount of matter in the objectDefinitionThe weight of an object is the force of gravity on the object

The mass of an object is constant everywhereChanging of valueThe weight of an object varies with the magnitude of gravitational field strength, g, of the location

A scalar quantityPhysical quantityA vector quantity

A base quantityType of quantityA derived quantity

Kilogram (kg)SI unitNewton (N)

Equations of motion with constant acceleration:Changing the a to g while solving problem involving free fall, the acceleration, a in the equations has a value of: a = 10 ms-2 (for downward motion) a = - 10 ms-2 (for upward motion)

Exercise:1.Wei is a basketball player. His vertical leap is 0.75 m. What is his take-off speed?

2.A coconut take 1.5 seconds to fall to the ground. What is (a) its speed when it strikes the ground?(b) the height of the coconut tree?

3.After winning a game, a pitcher throws a baseball vertically up with a velocity of 30 ms-1.(a)What is the time taken for the baseball to reach the maximum height?(b)What is the speed of the baseball when it returns to his hand?(c)How long is the ball in the air before it comes back to his hands?

4.An iron ball is dropped from the top of a building and takes 2 s to reach the surface of the earth. What is the height of the building? (g = 9.8 m s-2)

5.A rock has a mass of 20.0 kg and weight of 90.0 N on the surface of a planet.(a) What is the gravitational field strength on the surface of the planet?(b) What are the mass and the weight of the rock on the surface of the Earth where its gravitational field strength is 9.8 N kg-1?

6.Have you ever seen an astronaut walking on the Moon? It is known that the acceleration due to gravity near the surface of Moon is just of that on the surface of earth.(a)Find the weight of a 50 kg man on the surface of Moon.(b)If the 50 kg man can jump to a height of 50 cm on the Earth, find the maximum height reached by him on the surface of moon. (Assume that his initial speed is the same on the earth and on the moon)

Problems Involving F = ma and W =mgLiftreading on the scale shows the normal reaction force, RW = true weight = mgR = normal reaction force exerted on the girl by the platform of the scale

Weighing machine

1.When a girl stands on the platform of a weighing scale, there are two forces acting on her:(a)the girls weight, W (= mg) acting downwards,(b)the upward normal reaction force, R exerted on her feet by the platform of the scale2.The reading of the scale gives the value of the normal reaction force, R.3.Different situation in the lift,(a)Lift at rest or moves up or down at a constant velocity

When v=0 or constant a = 0

(b)Lift moves up at an acceleration of a ms-2

(c)Lift moves down at an acceleration of a ms-2

Help! I am floating!What is the apparent weight of a girl on a weighing scale in a lift if the cable of the lift suddenly breaks?

Exercise1.Subra stands on a weighing machine in a lift. If the mass of Subra is 58 kg, determine the reading of the weighing machine when the lift is.(a)stationary(b)moving upward with a uniform velocity of 2 ms-1(c)moving upwards with a uniform acceleration of 1 ms-2(d)moving downward with a uniform acceleration of 1 ms-2(consider gravitational acceleration, g to be 10 ms-2)

2A box of mass 1.6 kg is suspended from a spring balance hanging from the ceiling of a lift.What is the reading on the spring balance if(a)the lift is stationary?(b)the lift moves upwards at an acceleration of 2 ms-2?(c)the lift moves downwards at an acceleration of 3 ms-2?

Pulley System1.A frictionless pulley serves to change the direction of a force.2.The tension, T that results from pulling at the ends of the string or rope has the same magnitude along its entire length.

(A)A force pulling a mass over a pulleyIn this situation, the tension, T, is equal to the pulling force F, even if the rope is slanting.A boy is pulling a bucket filled a bucket filled with sand. The mass of the bucket with the sand is 3 kg. Find the tension in the rope if the bucket is(i) stationary, or(ii) moving up with a constant velocity of 2 ms-1

The boy increases his force to move the bucket upwards with an acceleration of 2 ms-2. Find the applied force, F

(B)A pulley with two masses1.The heavier mass will accelerate downwards while the lighter one will accelerate upwards with the same magnitude.2.The tension is not equal to the weight of either mass.

Exercise1.Two masses of 5 kg and 3 kg are connected to a rope which passes over a frictionless pulley.

Find the tension in the rope and the acceleration of the 3 kg mass when the 5 kg mass is released.

2.A 1.5 kg trolley is being pulled by a 0.5 kg weight with the help of a pulley system.

Calculate the acceleration of the trolley.(Take g = 10 ms-2)

3.Figure shows a pulley system with two loads, A and B, connected by a non-elastic rope that passes over a frictionless pulley. (Take g = 10 ms-2)(a)Determine the resultant force, F, of the system and state the direction of the movement of each load.(b)Calculate the magnitude of the acceleration of load A.

4.A 2 kg weight is connected by a rope to a 3 kg wooden block. The rope passes over a smooth pulley as shown in figure.

The weight is then released. Find the tension in the rope if a friction of 5 N acts against the wooden block.

2.9 Forces in Equilibrium

1. When forces act upon an object and it remains stationary or moves at a constant velocity, the object is said to be in a state of equilibrium.2. When equilibrium is reached, the resultant force acting on the object is zero. i.e, there is no net force acting upon it.3. Newtons Third Law of Motion states that if a force acts upon an object, then there will be an equal and opposite reaction acting upon the same object.

4. Based on the above figure, the airplane will move at a constant velocity if,Lift Force=Gravity ForceThrust Force =Drag Force

5.Another figure shows a stationery block of wood resting on a table. The forces acting on the block of wood are:(a)The weight, W is acting downwards.(b)The normal reaction, R is acting upwards.

6.The weight, W, is balanced by the normal reaction, R.7.Hence, the block of wood is in a state of equilibrium8.Figure below shows a weight hanger attached to a string and which is in a stationary state. The force acting on the weight hanger are:(a)The weight, W, which is acts downwards(b)The tension, T which acts upwards.

9.In this section, we shall study more about the equilibrium of forces, including an object at rest on an inclined plane where three forces are in equilibrium. The resultant force on the object is zero.

10.The cat resting on an inclined plane as shown in above figure is also in equilibrium. The three forces acting on the cat cancel out each other so that the resultant force is zero.11.A tilted surface is called an inclined plane.12.To understand better how three forces work in equilibrium, we need to understand(a)the resultant force of two forces, and(b)the resolution of a force.

Resultant Force1.A resultant force is a single force that represents the combined effect of two or more forces by taking into account both magnitude and the direction of the forces.

To find the Resultant Force

(i)Two parallel ForcesThe resultant force is obtained by simple arithmetic.Forces acting in the same directionForces acting in opposite directions

Add the magnitudes of two forces.The resultant force is in the same direction as the two forces.Resultant force, F = F1 + F2

Subtract the magnitude of the smaller force by the larger one (to find the difference in magnitude between the two forces).The resultant force is in the direction of the larger force.Resultant force, F = F2 F1

(ii)Two Non-parallel Forces (forces at an angle to each other acting at a point)1.Simple arithmetic cannot be applied to find the resultant force of two non-parallel forces.2.Determine the resultant force by drawing scaled diagrams using Two method below.

Method (I) - The Triangle Method (Tail - to - Tip Method)A scaled diagram of triangle of forces constructed to determine the resultant force of the two forces, F1 and F2, acting at an angle to each other. choose a suitable scale for the two forces, for example, 1 cm = 20 kN. Follow the steps below to determine the resultant force.

Method (II) The Parallelogram Rule (Parallelogram of Forces)A scaled diagram of the parallelogram of forces constructed to determine the resultant force of two forces, F1 and F2, acting at an angle to each other choose a suitable scale for the two forces, for example, 1 cm = 50 N. Follow the steps below to determine the resultant force.Note that the tails of both forces, F1 and F2 and the tail of the resultant force, FR are all at the same point, O and FR is in between F1 and F2

(iii)Two Perpendicular ForcesChapter 2 Force and Motion

This is a special situation for two non-parallel forces acting on an object at a right angle to each other.Resultant force, FR = tan =

Two forces (F1 and F2) together with the resultant force, FR using parallelogram rule. The resultant force is obtained using Pythagoras theorem.

Resolution of Forces1.A single force can be resolved into two perpendicular components.2.Figure above shows how a force, F can be resolved into two perpendicular components, the horizontal force, Fx and the vertical force, Fy

3.The magnitudes of the vertical and horizontal components can be determined with knowledge of simple trigonometry.For

For

The Effective Components of a Weight on an Inclined Plane

1.Consider a box resting on an inclined plane at an angle to the horizontal.2.The weight of the box is W = mg. The weight can be resolved into two perpendicular components:(a)the component vertical or perpendicular to the plane = mg cos(b)the component down or parallel to the plane = mg sin

To solve problems involving inclined planes, weight is replaced with its two perpendicular components.Object in equilibrium on a rough inclined planeAcceleration of an object on a smooth inclined plane

The figure below shows a box at rest on an inclined plane

Since the box is at rest:The net force perpendicular to the plane = 0

The net force parallel to the plane = 0

Therefore:

The figure below shows a box on a smooth inclined plane.

For an object on a smooth inclined plane, the vertical components of the forces are balanced. However, the force down the plane is not balanced.

Applying

If If

Example:A workman pushes a carton of mass 50 kg

up an inclined plane into a lorry. The inclinedplane makes an angle of 45 with the horizontal floor and the frictional forcebetween the inclined plane and the carton is135 N. If the workman pushes the cartonwith a force of 500 N(a)Can the carton move up the inclined plane?(