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NF0603: Physics for Engineers 1 > Motion > Flashcards

Motion Flashcards

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1
Q

What is displacement?

A

Distance in a given direction

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2
Q

What is speed?

A

Change of distance per unit time

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3
Q

What is velocity?

A

Change in displacement per unit time. or speed in a given direction.

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4
Q

Between speed, velocity and displacement, distance which are vectors or scalar?

A

Speed and distance are scalar quantities. Velocity and displacement are vector quantities.

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5
Q

What is the unit for speed and velocity?

A

ms-1

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6
Q

For example, a car travelling at a speed of 30 ms
−1 on a motorway travels a distance of
30 m every second.
How many meter in every minute?
What about in 1 hour?
Convert 30 ms
−1
to kmh−1?

A

1800 m every minute.

In 1 hour, the car would therefore travel a distance of 108000m or 108km

30 ms
−1 = 108 kmh−1
.

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7
Q

What is the equation for speed?

A

For an object which travels distance s in time t at constant speed.
Speed 𝑣 = s/t
Distance travelled 𝑠 = 𝑣t

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8
Q

What is the equation for an object moving at constant speed on a circle of radius r,

A

𝑣 =
2𝜋𝑟/T

where T is the time to move round once and 2𝜋𝑟 is the circumference of the circle

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9
Q

What is the equation for average speed?

A

𝑣 = ∆𝑠/∆t

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10
Q

What is the graph of an object moving at constant speed on a distance-time graph?

A

a straight line
with a constant gradient.

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11
Q

What is the graph of an object moving at changing speed on a distance-time graph?

A

The gradient of the graph changes so the speed of the object is calculated as distance travelled/time taken

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12
Q

How is speed calculated on a distance-tie graph?

A

The gradient of the line i.e distance travelled/ time taken

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13
Q

Can an object moving at constant velocity move at the same speed when changing direction?

A

An object moving at constant velocity moves at the same speed without changing
its direction of motion.
If an object changes its direction of motion or its speed or both, its velocity changes.
For example, the velocity of an object moving on a circular path at constant speed
changes continuously because its direction of motion changes continuously

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14
Q

What is Acceleration?

A

Change of velocity per unit time (ms-2). It is a vector

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15
Q

What is uniform acceleration?

A

The velocity of an object moving along a straight line changes at a constant rate. This is when acceleration is constant.

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16
Q

What is is the equation to calculate acceleration

A

𝑎 = (𝑣 − 𝑢)/t

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17
Q

What is non-uniform acceleration?

A

the direction of motion of an object changes, or its speed changes, at a
varying rate.

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18
Q

What is the gradient of a velocity-time graph used to calculate?

A

Non uniform acceleration.

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19
Q

What are the 4 suvat equations?

A
  1. 𝒗 = 𝒖 + 𝒂𝒕
  2. 𝒔 = ((𝒖+𝒗) 𝒕)/2
  3. 𝒔 = 𝒖𝒕 + 0.5𝒂𝒕^𝟐
  4. 𝒗^𝟐= 𝒖^𝟐 + 𝟐𝒂s
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20
Q

How can you find displacement using a velocity-time graph?

A
  1. If at constant velocity: The area under the line between the start and time t since displacement = v x t

2: If at constant acceleration: since 𝒔 = ((𝒖+𝒗) 𝒕)/2 then displacement is the area under the line between start and time t, calculated as a triangle

  1. Changing acceleration: the area under the line where v represent the velocity at time t and v + 𝛿v represent the velocity a short time
    later at t + 𝛿t . (𝛿 is pronounced ‘delta’.)
    Because the velocity change 𝛿v is small compared with the velocity v, the displacement
    𝛿s in the short time interval 𝛿t is v𝛿𝑡.
    This is represented on the graph by the area of the shaded strip under the line, which
    has a base corresponding to 𝛿t and a height corresponding to v. In other words,𝛿𝑠= v𝛿t
    is represented by the area of this strip
21
Q

What is the acceleration of freefall?

A

The acceleration of freefall is the constant acceleration of a free falling object denoted by g. It is equal to 9.81ms^-2

21
Q

Can suvat be used in freefall situations?

A

Only when air resistance is negligible

21
Q

How should freefall values be entered into suvat equations?

A

+ for items thrown upwards, - for falling downwards.

21
Q

Describe a Displacement time graph
for on object projected upwards

A
  • Immediately after leaving the thrower’s hand, the velocity is
    positive and large, so the gradient is positive and large.
  • As the ball rises, its velocity decreases so the gradient decreases.
  • At maximum height. its velocity is zero, so the gradient is zero.
  • As the ball descends, its velocity becomes increasingly negative,
    corresponding to increasing speed in a downward direction.
    So the gradient becomes increasingly negative It looks like an upside down U
22
Q

A coin was released at rest at the top of a well. It took 1.6s to hit the bottom of the well.
Calculate; (g = 9.8 ms
−2)

a) the distance fallen by the coin,
b) its speed just before impact.

A

a= -12.5m b =-15.7

23
Q

What is the gradient of a Displacement-time graph
for an object projected upwards

A

Velocity

24
Q

What is the gradient of a Distance-time graph for an object
projected upwards

A

Speed

24
Q

Distance-time graph for an object projected upwards

A

The gradient of this line represents the speed.
From projection to maximum height, the shape
is exactly the same as the previous slide.
After maximum height, the distance continues to
increase so the line curves up. It is half an upside down u and then the other half turn upwards like a regular U

25
Q

The difference between a speed-time
graph and a velocity-time graph

A

The object’s velocity decreases from its initial positive (upwards) value to zero at
maximum height and then increases in the negative (downwards) direction as it falls.Speed mainly increases and decreases

26
Q

What is the gradient of a velocity time graph?

A

The gradient of the line represents the object’s acceleration.
This is constant and negative, equal to the acceleration of free fall, g.
The acceleration of the object is the same when it descends as when
it ascends so the gradient of the line is always equal to -9.8 ms
−2

27
Q

What does the area under the line represent on a velocity time graph?

A

The area under the line represents the displacement of the
object from its starting position.
The area between the positive section of the line and the time axis
represents the displacement during the ascent.
The area under the negative section of the line and the time axis
represents the displacement during the descent.
Taking the area for a as positive and the area for b as negative, the
total area is therefore zero. This corresponds to zero for the total
displacement.

28
Q

What does a velocity time graph of an object thrown and caught look like?

A

It looks like this \, The object’s velocity decreases from its initial positive (upwards) value to zero at
maximum height and then increases in the negative (downwards) direction as it falls.

29
Q

A ball released from a height of 1.20 m above a concrete floor rebounds to a height of 0.82m.
a) Calculate i its time of descent, ii the speed of the ball immediately before it hits the floor.
b) Calculate i the speed of the ball immediately after it leaves the floor, ii its time of ascent.
c) Sketch a velocity-time graph for the ball. Assume the contact time is negligible compared with
the time of descent or ascent and that positive velocity is upwards

A

a. t= 0.49, ii. -4.8ms-1 b u = 4.0ms-1
ii. t = 0.41s

30
Q

A space vehicle moving towards a docking station at a velocity of 2.5 ms
−1
is 26 m from the
docking station when its reverse thrust motors are switched on to slow it down and stop it when it
reaches the station. The vehicle decelerates uniformly until it comes to rest at the docking station
when its motors are switched off.
Calculate a its deceleration, b how long it takes to stop

A

initial velocity u = +2.5 ms
−1
, final velocity v = 0, displacements= +26 m.

a. -0.12ms-2
b. 21s

31
Q

What is the link between acceleration in two - stage problems

A

The acceleration in each stage is not the same. The link between the
two stages is that the velocity at the end of the first stage is the same as
the velocity at the start of the second stage.

Consider an object released from rest, falling, and then hitting a bed of sand.
The motion is in two stages:
➢ falling motion due to gravity: acceleration = g (downwards)
➢ deceleration in the sand: initial velocity = velocity of the object just before impact

32
Q

a ball is released from a height of 0.85 m above a bed of sand and creates an impression in the
sand of a depth 0.025 m. Calculate the speed of impact.

A

a. v=-4.1ms-1
stage 2= decellerarion = 336ms-2

33
Q

What is a projectile?

A

A projectile is any object acted upon only by the force of gravity.

34
Q

What are the principles of a projectile?

A
  • The acceleration of the object is always equal to g and is always downwards - g is the only factor affecting vertical motion
  • The horizontal velocity of the object is constant because the acceleration of the object does not have a horizontal component.
  • Horizontal and vertical velocity are independent of each other
35
Q

what are the equations of displacement y and velocity v after time t during vertical projection?

A

𝒗 = 𝒖 − 𝒈𝒕
𝒚 = 𝒖𝒕 − 1/2𝒈𝒕^2

36
Q

Describe the horizontal projection of A stone thrown from a cliff top follows a curved path downwards before it hits the water

A

If its initial projection is horizontal:
* its path through the air becomes steeper and steeper as it drops
* the faster it is projected, the further away it will fall into the sea
* the time taken for it to fall into the sea does not depend on how fast it is projected.

37
Q

Suppose two balls are released at the same time above a level floor such that one ball drops
vertically and the other is projected horizontally. Which one hits the floor first? Why?

A

They both hit the floor simultaneously.
Why?
They are both pulled to the ground by the force of gravity, which gives each ball a downward
acceleration g. The ball that is projected horizontally experiences the same downward
acceleration as the other ball. This downward acceleration does not affect the horizontal motion of
the ball projected horizontally- only the vertical motion is affected.

38
Q

Describe the graph of a projectile path of a ball projected
horizontally

A

An object projected horizontally falls in an arc towards the ground. If its initial velocity is 𝑼, then at
time t after projection:

The horizontal component of its displacement,
𝒙 = 𝑼𝒕
(because it moves horizontally at a constant speed).

The vertical component of its displacement,
𝒚 =𝟏/𝟐𝒈𝒕^𝟐
(because it has no vertical component of its initial velocity).

lts velocity has a horizontal component 𝑣𝑥 = 𝑈, and a vertical component 𝑣𝑦 = −𝑔𝑡.

Note that, at time t, the speed = (𝑣𝑥^2+ 𝑣𝑦^2)1/2

39
Q

An object is projected horizontally at a speed of 15 ms−1from the top of a tower of height 35.0m.
Calculate:
a) how long it takes to fall to the ground
b) how far it travels horizontally
c) its speed just before it hits the ground

A

a. t = 2.67
b x = 40m
c. v = 30.2ms-1

40
Q

What are the equations for vertical projection?

A

v = u -gt
y = ut -1/2gt^2

41
Q

What is the measurement for gravitational field strength?

A

9.81 N kg-1

42
Q

What is Newton’s Second Law of motion?

A

Acceleration of an object is proportional to the resultant force acting on it. The rate of change of momentum of an object is proportional to the resultant force on it. The resultant force is proportional to the change of momentum per second

43
Q

What is Newton’s third law?

A

When two objects interact they exert equal and opposite forces on each other.

44
Q

How are projectiles calculated?

A

Since horizontal and vertical motion is separate, we calculate vertical motion using suvat and horizontal motion with = speed=d/t

NF0603: Physics for Engineers 1 (13 decks)

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