Tag Archives: Topic Kinematics
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
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.
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)
Heavy and light shuttlecocks – which will reach the floor first when released?
Refer to this post on falling of basketball and bowling ball
Refer to this post on a video of falling feather and bowling ball
Basketball vs bowling ball – which will reach the floor first when dropped?
Refer to this post on terminal velocity of shuttlecocks
Refer to this post on a video of falling feather and bowling ball
Kinematics Summary
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.
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.
Free-fall + breaking sound barrier
Kinematics – Swimmer in river with current, find distance downstream
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.
Kinematics – Two balls free-fall with different mass and height
Solutions: Option B
Key concepts: Both balls are free-falling, hence both will experience the same acceleration due to gravity 10 m/s2.
PP N2010 P1 Q4 Two objects same size and shape but one is heavier. Motion when both released
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.
Kinematics Question – Using gradient and area under graph
Solutions: Option B
Object falling at freefall
A particle released from rest at O falls freely under gravity passes through P and Q as shown.
If the particle takes 4 s to move from P to Q where PQ = 120 m, how long does it take to fall from O to P?
A) 1s B) 2s C) 4s D) 8s
Solutions: Option A
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
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.
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.
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)
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.
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.
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.
In reality, there is work done against air resistance, energy converted to heat and sound when ball hits the ground, hence the ball will never reach its original height.
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
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.
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.
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)
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)
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)
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?
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.
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?
Conversion of m/s to km/h and vice versa – Usain Bolt’s world record for 100m
Usain Bolt smashed the 100m and 200m World record with a time of 9.58 s and 19.19s respectively in Berlin 2009! Can’t embed the 100m clip here. So just how fast is he? Is he much faster than your dad’s car on the expressway 90 km/h???
100m in 9.58s means approx. 10.44 m/s = 37.58 km/h. So well, 2 legs are slower than 4 wheels =) How do you convert?
View the step by step conversion from m/s to km/h. (in short, multiply by 3.6)
Below is another record set by him. 200m in 19.19s. You can try to do the conversion!
N2004P1Q4 – Dynamics + Kinematics
Trolley X and trolley Y are joined by a stretched spring. Trolley X has twice the mass of trolley Y.
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.
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.
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?
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
Kinematics – Ticker Tape Timer
Many are familiar with ticker-tape timer. The tape is attached to Car B. As Car B moves to the left, it pulls the tape with it. The tape will go through the ticker-tape timer machine. The machine is stationary and it just punches holes on the tape as the tape is pulled to the left. It is clear that Car A is accelerating as the distance interval of the holes is getting wider.
For Car B, the scenario is different. Oil is leaking from the car onto the road at fixed interval or frequency. Hence Car B is decelerating as the distance interval of oil drops becomes smaller.
Therefore, be careful with the two scenarios and do not be confused.
Dynamics and kinematics: Force-time graph, find frictional force
A 10 kg box is initially at rest on a rough horizontal surface and a horizontal force is then applied to it. The force varies as shown in the diagram.
a) If the box moves with an acceleration of 2m/s2 in the first 15 s, what is the frictional force between the surfaces?
Resultant force on box = ma
Applied force – frictional force = ma
40 – frictional force = 10 x 2
frictional force = 20 N
b) How does the velocity change during the next 5 s?
Next 5 s after time 15 s, there is no applied force. Hence the resultant force acting on box isonly frictional force 20N (in direction opposite to motion of box).
F = ma
-20 = 10 x a
a = – 2 m/s2
The velocity of box at time 15s is required.
Hence when t = 0s, box accelerates at 2m/s2 from rest,
a = (v – u) / t
2 = (v – 0) /15
v = 30 m/s
(Hence the final velocity of the box at time 15s is 30 m/s, just before applied force of 40 N is released)
Find velocity of box at end of next 5s (time = 20s):
a = (v – u) /t
– 2 = (v – 30) / 5
v = 20 m/s.
Therefore, velocity decreases at a constant rate of -2 m/s2 from 30 m/s to 20 m/s in next 5s.
c) Plot a velocity-time graph and calculate the distance travelled by the box in the 30 s.
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