Evan's Space

Wonders of Physics


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Fiery Re-entry into Earth’s Atmosphere (updated)

This post was updated following the first astronauts launched by SpaceX returned home safely on 3 Aug 2020.

SpaceX’s Crew Dragon heat shield shown off after first orbital-velocity reentry

How do spacecraft re-enter the Earth? | HowStuffWorks

Why Is It So Difficult For A Returning Spacecraft To Re-Enter Our Atmosphere?

Returning from Space: Re-entry – PDF format

SpaceX In-Flight Abort Test

SpaceX Falcon Heavy- Elon Musk’s Engineering Masterpiece

Shuttle Atlantis STS-132 – Amazing Shuttle Launch Experience

How to Land the Space Shuttle… from Space

First astronauts launched by SpaceX return to earth (3 Aug 2020)


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Parachute Jump – Speed-time graph

In a typical parachute jump, there are various distinct stages/sections as you can see from the graph in the video below.

AB = constant acceleration, free fall, a = 10 ms-2
BC = decreasing acceleration
CD = constant speed, zero acceleration
at D = the time where the parachute is fully opened
DE = constant deceleration
EF = lower constant speed, zero acceleration
FG = constant deceleration

After you have learned Dynamics, you should be able to explain each stage using forces acting on the skydiver, namely the weight and air resistance.

Refer to the video below to understand the motion at various stages and how to explain in terms of forces, esp the part on why the acceleration is decreasing during BC.

You can refer to the detailed explanation in words in the comics below. Hope this post helps you to understand better!

velocity-time graph of parachute jump


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Displacement-time and Velocity-time Graph of Ball Thrown Up / Ball Dropped

1) When a ball is thrown up and it comes down

When the ball leaves the hand, there is no upward force acting on the ball. The only force acting on the ball is its weight. This net force is opposite to the motion of the ball, hence causing the ball to decelerates. Refer to this post for the explanation (comics)

 

2) When the ball is dropped and it re-bounces back (assume no energy lost)

When the ball is released, the only force acting on the ball is its own weight. This net force on the ball is in the direction of the motion, hence causing the ball to accelerate as it falls. Note that the acceleration due to gravity is assumed to be 10 ms-2, where air resistance is negligible.

3) When the ball is dropped and it re-bounces back (in reality with energy lost)

In reality, when the ball hits the ground, there will be some energy converted to heat and sound. So the ball will never return to its original height that it was released. So how will the graphs look like?


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Ticker Tape Timer

Ticker tape timer is a device which punches dots on a tape at specific time interval. The tape can be attached to a moving object. With information of the frequency of the ticker timer and the spacing of the dots, we are able to analyse the motion of the moving object.

View the following videos to understand more about the ticker tape timer and also some typical questions on ticker tape timer.

Explanation of Ticker Tape Timer

Questions on ticker tape timer

Refer to this post for another example of finding acceleration

https://evantoh23.wordpress.com/2010/10/26/20101026ticker-tape-timer-finding-acceleration/


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Distance-time and Speed-time graph of 2 objects in motion

The distance-time (displacement-time) graph on the left is not related to the speed-time (velocity-time) graph. Both are of difference scenarios. But both shows the motion of 2 objects and both graph has an interception point. Do the interceptions point mean the same time in each graph?

The image below shows the basic interpretation of the respective graphs.

motion of 2 objects on distance or speed time graphs

To have examples of the graphs with values and the step-by-step how to find the time it overtakes, refer to the videos below.

Finding the exact time where the overtaking takes place.


Understanding Free-Fall (acceleration due to gravity)

On earth, the gravitational field strength is 10 N/kg and the acceleration due to gravity is 10 ms-2.

That means when you release an object from you hand, the object will fall with increasing speed. The acceleration is a constant 10 ms-2 (acceleration due to gravity). Simply put, it means in 1 second, the speed of the object will increase by 10 m/s.

free fall

In general, if there is no air resistance or air resistance is negligible, the speed-time graph is a straight line with constant gradient and passes through origin, i.e. speed is directly proportional to the time.

If there is air resistance, there is a maximum constant speed if the object continues to fall. Hence the graph is different.

More importantly, remember that any object that you released on earth, whether there is air resistance or no air resistance, the initial acceleration when it is released is always a constant 10 ms-2!

Refer to the video for the explanation and some examples.


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Kinematics Graphs of ball thrown up and come down

When a ball is thrown up vertically from a height and the falls to the ground, the various kinematic graphs below shows the same motion of the ball.

Fixing up as positive, if the ball is going up, it’s displacement and velocity will be in the positive region. When it falls, the direction is opposite, hence it’s in the negative portion.

Recall that the gradient of a velocity-time graph represents the acceleration. Using this concept, it will be easier to understand that the acceleration is a negative constant acceleration, thus – 10 ms-2.

(Thanks Maria for suggesting this post and the addition of acceleration-time graph)

Click here to refer to the post related to this motion.

Velocity-time and Displacement-time Graph for a ball being thrown up

Velocity-time and Displacement-time graph for Ball thrown up and comes down


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Kinematics Videos

1) Speed and Average Speed

Usain Bolt World Record 100 m in 9.58s in Berlin, Germany!
What is his average speed?
Is there portion of the race where his speed is faster than the average speed?

How about his 200 m record?

How about his 4 x 100 m relay?

2) Acceleration

Experience the thrilling acceleration of a launching roller coaster. Calculate the initial acceleration!

Bugatti Chiron 0 – 400 – 0 km/h in 42 seconds.

Ferrari World Abu Dhabi – World’s Fastest Roller-coaster which can reach top speed of 240 km/h in 4.9 s! What is the acceleration in m/s2.

Hyperloop train – next generation of super fast train

3) Free-fall and Weightlessness

Weightlessness in outer space

Zero-gravity plane: You can experience weightlessness on earth!

Jumping From Space! – Red Bull Space Dive

Brian Cox visits the world’s biggest vacuum | Human Universe

Experience G-force in a Centrifuge


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Fiery Re-entry into Earth’s Atmosphere

SPACE CAPSULE simple

SPACE CAPSULE

SpaceX’s Crew Dragon heat shield shown off after first orbital-velocity reentry

How do spacecraft re-enter the Earth? | HowStuffWorks

Why Is It So Difficult For A Returning Spacecraft To Re-Enter Our Atmosphere?

Returning from Space: Re-entry – PDF format

SpaceX In-Flight Abort Test

SpaceX Falcon Heavy- Elon Musk’s Engineering Masterpiece

Shuttle Atlantis STS-132 – Amazing Shuttle Launch Experience

How to Land the Space Shuttle… from Space


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During impact of a free falling ball, the force on ground is greater than the weight of ball

In this post, it shows a free-falling ball from a height of 1.0 m. During the impact, the direction of the force on the ground is downwards and the force on the ground by the ball is greater then the weight.

As the ball is free-falling, the only force acting is its weight downwards. Hence a common misconception is to think that the force on the ground during impact is equal to the weight. This is wrong.

The normal force (force on the ball by the ground = stopping force on the ball by the ground) is greater than the weight.

The force on the ball by the ground is equal and opposite to the force on the ground by the ball. Hence the magnitude of the force on the ground is greater than the weight.

Similar concept can be applied if a man jumps off from a height. But in this case, the man’s leg will exert a stopping force over a short distance. That stopping force, once again, is greater than the weight of the man.


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Man Jumps Vertically Upwards, Pressure On Ground Is Greater During The Jump

This concept is similar to a 2016 O-Level Pure Physics Question P2 Q2, on why the pressure acting on the ground is greater during the jump, compared to when he is standing stationary on the ground.

During the jump, his leg will exert an upward force. This upward force (equivalent to normal force or force on the man by the ground) is greater than the weight of the man. Hence there is a net (resultant force) upwards, causing him to accelerate upwards.

That force on the man by the ground is equal and opposite to the force on the ground by the man. This is an action-reaction pair. Since the force exerted on the ground by the man is greater (greater than weight), the pressure exerted on the floor is greater.

(NOTE: Normal force and Weight is not an action-reaction pair)


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Impressive Bugatti Chiron

This impressive Bugatti Chiron can accelerate from rest to 400 km/h and decelerate to a complete stop in merely 42 seconds! Our normal cars on the expressway travel about 90 km/h and the F1 race yesterday night is about 300 km/h. This Bugatti Chiron is faster than most bullet trains and comparable to the speed of a magnetic levitation train!

Capturehttps://www.bugatti.com/chiron

Before we look at the video, let’s do some calculations:

Capture2

Let’s find the acceleration of the car to reach 400 km/h in 32.6 sec:

Converting 400 km/h to m/s:    400km/1h = 400 000m/3600s =111 m/s

acceleration, a = (v – u)/t = (111 – 0) / 32.6 = 3.4 m/s2

hmmm…. this acceleration doesn’t seem impressive… it is way below free fall acceleration!

But we are not being fair here. To achieve the max speed of 400 km/h is not easy due to the resistive force (air resistance and friction) as speed increases. We should compare fairly the acceleration to reach 100 km/h instead like how we typically compare sports car like Ferrari etc.

Let’s find the acceleration of the car to reach 100 km/h (27.8 m/s) in 2.4 sec:

acceleration, a = (v – u)/t = (27.8 – 0) / 2.4 = 11.6 m/s2

This is greater than acceleration due to gravity (free fall) and much faster than most sports cars in the market like Ferrari or Lamborghini!

Now, let’s find the deceleration of the car when it slows down from 400 km/h to a complete stop in 41.9 – 32.6 = 9.36 s

acceleration, a = (v – u)/t = (0 – 111) / 9.36 = -11.9 m/s2

Take note of the spoiler being activated when it decelerates. This increases the drag (air resistance) to slow down the car, in addition to using the normal brakes. It is the same principle as the aeroplane when it lands and slows down on the runway.

 

 

 


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2013 Nov Sci Phy P2 Q9 – Pressure and Moment

CaptureSolutions:

(a) (i) A bigger force than F can be obtained due to the level system and hydraulic system.
Level system: Applying principle of moments, the anticlockwise moment by the F is equal to the clockwise moment by the force on piston A (note that the handle is pushing the piston down, but the piston A is pushing on the handle upwards – action = reaction). As the perpendicular distance from F to the pivot is greater than the perpendicular distance of the force by piston to the pivot, the force on the piston A is greater than F at handle.
Hydraulic system: As the pressure transmitted in the liquid is the same, pressure at piston A = pressure at piston B. As P = F/A and area of piston A is smaller than area of piston B, a larger force is obtained in piston B. 
Hence these two systems allow the force on piston B to be greater than F at the handle.

(a) (ii) Both liquid and gas molecules are in a continuously random motion. But in liquid, the molecules are closely packed together and able to slide around one another. There is very little empty space between the molecules hence liquid is not compressible. Gas molecules are far apart from one another, hence gas can be easily compressed.

(b) P = F/A = 12000 / 0.060 = 200 000 Pa

(c) Velocity is the vector quantity while speed is a scalar. As the car goes round the bend, the direction of the car changes. Hence velocity is changing even though speed is constant.

Likewise, as the velocity is changing, the car is considered to have an acceleration (not in the sense of increasing speed though).


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What is the speed-time graph of the object when pulling force is removed?

A constant horizontal force F, of magnitude 16 N, is applied to an object at rest on a rough surface. The constant force of friction is 8 N. At X, the force F is removed.

image

Which speed-time graphs correctly shows the motion of the object?

image

Solutions: Option B

Either B or D is correctly. But you need to find the magnitude of the deceleration after X in order to determine if the magnitude (gradient) is greater or smaller than the magnitude of the acceleration before X.

image


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Hammer and Feather in air and in vacuum

In air:
1) hammer will reach the bottom first.
2) feather will reach constant speed (terminal velocity) as its weight = air resistance.
3) hammer will accelerate throughout as the weight > air resistance.

In vacuum:
1) both will reach the bottom at the same time as both have the same acceleration due to gravity (10 m/s2)
2) the speeds of both hammer and feather are the same throughout the drop.
3) the speeds of both hammer and feather just before reaching the bottom are greater than in air (due to no air resistance)


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Austrian Felix Baumgartner has become the first skydiver to go faster than the speed of sound, reaching a maximum velocity of 833.9mph (1,342km/h).

  • Exit altitude: 128,100ft; 39,045m
  • Total jump duration: 9’03”
  • Freefall time: 4’20”
  • Freefall distance 119,846ft; 36,529m
  • Max velocity: 833.9mph; 1,342.8km/h; Mach 1.24

Its representative was the first to greet the skydiver on the ground. GPS data recorded on to a microcard in the Austrian’s chest pack will form the basis for the height and speed claims that are made.

These will be submitted formally through the Aerosport Club of Austria for certification.

There was concern early in the dive that Baumgartner was in trouble. He was supposed to get himself into a delta position – head down, arms swept back – as soon as possible after leaving his capsule. But the video showed him tumbling over and over.

Eventually, however, he was able to use his great experience, from more than 2,500 career dives, to correct his fall and get into a stable configuration.

A

Even before this drama, it was thought the mission might have to be called off. As he went through last-minute checks inside the capsule, it was found that a heater for his visor was not working. This meant the visor fogged up as he exhaled.

“This is very serious, Joe,” he told retired US Air Force Col Joe Kittinger, whose records he was attempting to break, and who was acting as his radio link in mission control at Roswell airport.

The team took a calculated risk to proceed after understanding why the problem existed.

Baumgartner’s efforts have finally toppled records that have stood for more than 50 years.

Kittinger set his marks for the highest, farthest, and longest freefall when he leapt from a helium envelope in 1960. His altitude was 102,800ft (31km). (His record for the longest freefall remains intact – he fell for more than four and a half minutes before deploying his chute; Baumgartner was in freefall for four minutes and 20 seconds).

Kittinger, now an octogenarian, has been an integral part of Baumgartner’s team, and has provided the Austrian with advice and encouragement whenever the younger man has doubted his ability to complete such a daring venture.

“Felix did a great job and it was a great honour to work with this brave guy,” the elder man said.

The 43-year-old adventurer – best known for leaping off skyscrapers – first discussed seriously the possibility of beating Kittinger’s records in 2005.

Since then, he has had to battle technical and budgetary challenges to make it happen.

What he was proposing was extremely dangerous, even for a man used to those skyscraper stunts.

At Sunday’s jump altitude, the air pressure is less than 2% of what it is at sea level, and it is impossible to breathe without an oxygen supply.

Others who have tried to break the records have lost their lives in the process.

Baumgartner’s team built him a special pressurised capsule to protect him on the way up, and for his descent he wore a next generation, full pressure suit made by the same company that prepares the flight suits of astronauts.

Although the jump had the appearance of another Baumgartner stunt, his team stressed its high scientific relevance.

The researchers on the Red Bull Stratos project say it has already provided invaluable data for the development of high-performance, high-altitude parachute systems, and that the lessons learned will inform the development of new ideas for emergency evacuation from vehicles, such as spacecraft, passing through the stratosphere.

Nasa and its spacecraft manufacturers have asked to be kept informed.

“Part of this programme was to show high-altitude egress, passing through Mach and a successful re-entry back [to subsonic speed], because our belief scientifically is that’s going to benefit future private space programmes or high-altitude pilots; and Felix proved that today,” said Art Thompson, the team principal.

In getting to 128,100ft, Baumgartner exceeded the altitude for the highest ever manned balloon flight achieved by Victor Prather and Malcolm Ross, who ascended to 113,720ft (35km) in 1961.

However, the FAI rules, state that to claim an official ballooning record, a balloonist must also bring the envelope down and therefore the Austrian’s altitude will forever remain just an unofficial mark.


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Kinematics – Swimmer in river with current, find distance downstream

Img_3241

Solutions: Option B

Without the current in the river, the time taken for the swimmer to reach Q is 50s. This is based on the vertical constant speed of 1.2 m/s and veritcal (width of river) distance.

With the current, which is constant 1.0 m/s, the swimmer will be drifted downstream. But the time taken to reach R from P is still 50s. So with the same 50s, you can calculate the horizontal distance based on the horizontal current speed of 1.0 m/s.

Capture2


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PP N2010 P1 Q4

P417

Solutions: Option C

When a body is released from rest, the only force acting on the body is its weight due to gravity. Both bodies experience acceleration due to gravity. Hence for both bodies (regardless of mass), will have the same initial acceleration of 10 m/s2.

As the two bodies are of the same size and shape, they will experience the same air resistance for any particular speed. As speed increases, air resistance increases.

For terminal velocity to be reached, air resistance has to be equal to the weight. Since weight is greater for the ball with larger mass, the air resistance has to be bigger. Thus, the ball has to accelerate more (air resistance increases with speed) for the larger air resistance to be equal to the weight. Hence the ball with larger mass will have larger terminal velocity.

[NOTE]
Do not confuse ‘speed of the body is independent of the mass’ as learned in Work Done, Energy and Power. This concept is based on the assumption that there is no air resistance. So not applicable in this question as for terminal velocity to occur, air resistance must be present.

P416


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Ticker-Tape Timer

If the ticker-tape timer has a frequency of 50 Hz, it means it punches 50 holes in 1 second.

Hence the time interval to punch 1 hole is 1/50 = 0.02s. Alternatively, you can use the formula Period, T = 1/f = 1/50 = 0.02 s.

Related posts on ticker tape timer
Finding acceleration from ticker-tape timer

Ticker-tape timer vs oil dropping on the road at regular intervals

 


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Rocket with forward thrust lasting 1.5 s

An experiment rocket model of mass 0.05 kg was launched from the ground, vertically into the air with a propelling force of 4.5 N. The air resistance is assumed to be negligible. The rocket carries propellant only enough for 1.5 s of the upward flight. The rocket then crashed to the ground after some time. (Take g = 10 ms-2)

a) Calculate the weight of the rocket model.   Ans: 0.5 N

b) Calculate the acceleration of the rocket, which is assumed to be uniform, during the first 1.5 s of flight.    Ans: 80 ms-2

c) What is the maximum velocit of the rocket?  Ans: 120 m/s

d) What is the upward acceleration of the rocket after 1.5 s of flight?   Ans: – 10 ms-2

e)Draw a clearly labeled velocity-time and displacement-time graph for the whole flight of the rocket from its take-off to its crashing to the ground.


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Car A and Car B – Find the time one overtakes the other

The speed-time graph of Car A and Car B, along a straight road over 4 seconds is shown below.car A overtakinf car B

a) Calculate the acceleration of Car A and Car B over the 4 seconds.    
Ans: a of Car A = 3 ms-2 and a of Car B =0.75 ms-2

Car A overtakes Car B at time t seconds.

b) Derive two separate expressions for the velocities of Car A and Car B at the point when Car A overtakes Car B, in terms of t.
Ans: Va = 3t    and    Vb = 3 + 0.75t

c) Calculate the time t when Car A overtakes Car B.   
Ans: t = 2.67s

Solutions for (a) and (b)

Solutions for (c)


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Important Concepts on Newton’s 1st and 2nd Law

Basic important concepts on Newton’s Laws which you have to understand.

a) On a rough floor, when the applied force on a moving object is removed, the resultant force acting on object is backwards (frictional force). This resultant force causes the object to decelerate. (Newton’s 2nd Law)

A good example is that of a wakeboarder. When the tension in cable is equal to the opposing force, the wakeboarder moves at constant speed, balanced forces, net force is 0N. But the moment he releases the cable, the only force acting is the opposing force. The resultant force backwards (opposing force) causes him to slow down (decelerate). Hence resultant force backwards does not mean the wakeboarder is moving backwards.

21645029-Capture1

b) On a frictionless floor, when the applied force on a moving object is removed, the resultant force acting on object is 0N, and the object will move with a constant speed in a straight line. (Newton’s 1st law)

View the video tutorial to understand the 2 concepts.


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Basic concepts on Netwon’s 1st and 2nd Law of Motion

Whenever an object is at rest or moving at constant speed in a straight line (constant velocity), Newton’s 1st Law of motion must come to your mind.

Concepts on Newton’s 1st Law are: Net force = resultant force = 0N, a = 0m/s2 , all forces are balanced.

If the object is accelerating or decelerating, that means the forces are unbalanced. Newton’s 2nd Law of Motion must come to your mind.

Concepts on Newton’s 2nd Law are: There is net force or resultant force, unbalanced forces, there is acceleration or deceleration.

This video tutorial gives you an overview basic concepts of Newton’s Laws and understanding of these are vital.


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Vertical Resultant Force of ball rolling off platform

A ball of mass 5 kg rolls along a smooth horizontal surface until it falls off the edge at time = 3s and touches the lower surface at t = 5s. Sketch the graph which represents how the resultant vertical force F acting on the ball varies with time as the ball moves from A to B.

21330272-media_httpevantohfile_HaflH

The question only ask for vertical resultant force, hence there is no need to consider the horizontal forces. Anyway the ball is moving horizontally at constant speed, hence no acceleration, forces are balanced. It will move at constant speed (Newton’s 1st law) When the ball is rolling on the upper/lower platform, the weight is equal to the normal force acting on the ball by the platform, hence net force is 0N.


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Velocity-time and Displacement-time graph for Ball dropped and rebounces back

There is a new updated post on this concepts. Refer to the link below

https://evantoh23.wordpress.com/2020/04/13/displacement-time-and-velocity-time-graph-of-ball-thrown-up-and-comes-down/

When a ball is released from a height, it will accelerate on the way down due to the resultant force (weight) acting downwards. Just before the ball touches the ground, the velocity of the ball is the maximum. When it hits the ground, the speed decreases to 0 m/s instantly. When it rebounces back in the opposite direction, the initial velocity is the maximum. Assume ideal situation (no air resistance, no energy lost to sound or heat). The ball will rebounce back to its original height.

21330268-media_httpevantohfile_cIcvI21330269-media_httpevantohfile_iJcFE

In reality, there is work done against air resistanceenergy converted to heat and sound when ball hits the ground, hence the ball will never reach its original height.

21330270-media_httpevantohfile_EbwHk21330271-media_httpevantohfile_vkxkk

Note that in both situations, conservation of energy always applied. All energy is conserved, just that energy of the ball is converted to other forms like heat and sound.

Click here to view a comic on forces explanations of ball thrown up
Click here to view the post on velocity-time and displacement-time of ball thrown up


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Velocity-time and Displacement-time graph for Ball thrown up and comes down

There is a new updated post on this concept. Refer to the link below

https://evantoh23.wordpress.com/2020/04/13/displacement-time-and-velocity-time-graph-of-ball-thrown-up-and-comes-down/

A ball is thrown vertically upwards from the hand and lands back onto the hand. It is important to note that once the ball leaves the hand, the  resultant force acting on the ball is only its weight! And it is acting downwards throughout the motion. 

Click here for a physics comic on this concept
Click here for velocity-time and displacement-time for ball dropping

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Velocity-Time Graph

Displacement-Time Graph

Key points to note when sketching the v-t graph:
1) Fixing the direction up to be positive (can be down as positive if you want)
2) At t = 0s, the initial velocity of the ball is maximum.
3) As it goes up vertically, due to the weight acting downwards, the ball decreases in speed (decelerates)
4) At the highest point, the ball is momentarily at rest (v =0m/s)
5) On the way down, due to the weight of the ball acting downward, the ball accelerates downwards.


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Ball thrown vertically up from building and falls to ground

A rock is thrown vertically upwards with a velocity of 29.4 m/s from the top of a building 78.4 m high. After how long will the rock reach the ground below? (Gravity = 10 m/s2)

21330265-media_httpevantohfile_kodwn

Many are familiar with calculation when ball is released from rest. This question involves ball being thrown upwards, and then it falls to the ground. To solve this question, just treat the motion as 2 separate parts.
1) Ball going up to max height,

2) Ball going down (similar to ball released from rest from highest point)


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Ball drops above moving train

A stone is dropped from a bridge 2.0 m above the roof of a train moving horizontally below. The train is travelling at a uniform speed of 30.0 m/s. What distant is covered by the train as the stone falls? (Take g = 10 m/s2)train and falling object

The train is moving to the left at a constant speed. We just need to find the time taken for the ball to fall 2 m.  This is normal free fall of ball. Recall: when distance is given, think of area under the graph. Once time is found, the distance travels by the train can be easily found.


If the train is accelerating constantly, would you be able to solve?


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Speed-Time Graph – Motion of 2 Cars and interpreting the graph

At t = 0s, both cars are at Line 1. Car A is moving at constant speed of 20 m/s, while Car B is moving off from rest with a initial constant acceleration and then maintains at a constant speed of 25 m/s after t = 15s. Line 2 is 500 m away from Line 1.2 cars

a) Sketch the motion of both cars on a speed-time graph.

b) What is the time taken for Car A to reach Line 2?

c) What is the time taken for Car B to reach Line 2?

d) If both cars maintain their constant speeds, Car B will overtake Car A eventually. At what time and distance from Line 1 will Car B overtakes Car A?  

Solutions:

a) To solve such questions, even though graph is not required, it is useful to sketch out the speed-time graph for better visualisation.

b) Whenever given distance travelled by car, it is always good to think of area under the graph.

c) Distance covered is still 500 m, which is the area under the graph.

d) As Car B is travelling with a greater constant speed, eventually it will overtake. But when? Definitely after 15s? Or possible to be before 15s?



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N2004P1Q4 – Dynamics + Kinematics

Trolley X and trolley Y are joined by a stretched spring. Trolley X has twice the mass of trolley Y.

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When the trolleys are released, the acceleration of X is   to the right.  What is the initial acceleration of trolley Y to the left?Solutions:Key concept here is Newton’s 3rd Law. Action equals reaction. Both trolleys are connected by the spring, hence the tension force experienced by each trolley is equal and opposite.

Fx    =    Fy
ma       =    ma
2m x 2    =    m x a
a        =   4 m/s2

Another important concept to know. In Fig A, a box of m is pulled with a force of 40 N. Given that the frictional force is 10 N. The resultant force is 30N. Box is accelerating.

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In Fig B, if another identical box m is added on top of the existing box (no change to the floor), the total mass is 2m, the frictional force now will double = 20 N. Hence resultant force now will be 20N. The acceleration of the box will be lower than previously.


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Ticker-tape Timer – Finding acceleration

Click here to learn the basics of a ticker tape timer.
The diagram shows a strip of paper tape that has been pulled under a vibrating arm by a car moving to the left. The arm is vibrating regularly, making 50 dots per second.  What was the acceleration of the car?

ticker tape

Solutions: View the video tutorial to understand where to take time interval.

The arm vibrating 50 dots per second = frequency of 50 Hz   (50 holes are produced in 1 second)

Hence, the period, T = 1/f = 1 / 50 = 0.02s     (every 0.02s, a hole is created on the tape)

To find acceleration, a, we need to find initial velocity, u,and the final velocity, v.

u = dist / time = 0.02m/0.02s = 1 m/s

v = dist / time = 0.04m/0.02s = 2 m/s

To find acceleration, we can use a = (v-u)/t , but the time, t, taken for the increase in velocity is where most students will make a mistake.

Many will take 7 intervals to calculate the t, which is wrong.

But the intervals should be 6, starting from the middle of first interval and middle of last interval.
This will give a more accurate acceleration.

Hence, t = 6 x 0.02s = 0.12s a = (v-u)/t = (2-1)/0.12 = 8.33 m/s2