Module 3: C3 - Motion

Question 1

Equation for Speed

Answer 1

Speed (ms^-1) = Distance (m) / Time (s)

Question 2

Speed, Velocity, Instantaneous Speed Definitions

Answer 2

Speed:
Is the distance travelled per unit of time

Velocity:
Is the displacement travelled per unit of time

Instantaneous Speed:
Is the speed at a given instant of time

Question 3

Equation for Velocity

Answer 3

Velocity (ms^-1) = Displacement (m) / Time (s)

Question 4

Acceleration Definition

Answer 4

Acceleration tells us how rapidly something is changing velocity - for instance, the change in velocity in unit time. Deceleration is the same thing, but has a negative sign as the velocity is decreasing.

Question 5

What does the area underneath a Velocity-Time Graph give

Answer 5

It gives the displacement

Question 6

Equation for Acceleration

Answer 6

Acceleration (ms^-2) = Velocity (ms^-1) / Time (s)

Question 7

Parabola Definition

Answer 7

The characteristic path followed by an object moving under gravity in a uniform gravitational field.

Question 8

Projectile Definition

Answer 8

An object is given an initial force and then allowed to move freely under gravity.

Question 9

Range Definition

Answer 9

The horizontal distance from its starting position at which a projectile will land. This can be predicted from the initial conditions of the flight.

Question 10

Trajectory Definition

Answer 10

The path that an object follows through space, due to the forces that act on it. Under the influence of gravity, projectiles in free fall will follow a parabolic trajectory.

Question 11

The 4 SUVAT Equations

Answer 11

v = u + at

s = ut + 1/2at^2

v^2 = u^2 + 2as

s = 1/2(u + v) x t

Question 12

How do you get the Acceleration from a Velocity-Time Graph

Answer 12

You find the gradient. This is done by:

Change in V / Change in t

Question 13

How can we measure acceleration?

Answer 13

1. Set up time gates to measure the acceleration of a truck
2. Draw a suitable table to take results
3. Measure acceleration over the length of the ramp
4. Measure acceleration 10cm from the top of the ramp
5. Measure acceleration 10cm from the top of the ramp

Question 14

How do you derive the SUVAT equation without s

Answer 14

From a velocity-time graph:
a = Δv/Δt = v-u/t

This can be rearranged to give:
v = u + at

Question 15

How do you derive the SUVAT equation without v

Answer 15

The area under the velocity-time graph is equal to the displacement s.

• the rectangular area = ut
• the triangular area = 1/2 x (v-u) x t

From equation 1, (v-u) = at. If you substitute this into the expression for the area of the triangle, you get 1/2 x at x t. With ut for the area of the triangle, this gives the total area s.

s = ut + 1/2at^2

Question 16

How do you derive the SUVAT equation without a

Answer 16

If you treat the area under a velocity time graph as the area of a trapezium (with u and v as the parallel sides , and t as the perpendicular length), this becomes:

s = 1/2 (u+v)t

In other words, the displacement s is the average velocity (u+v/2) multiplied by the time t.

Question 17

How do you derive the SUVAT equation without t

Answer 17

According to Equation 1 (a = (v-u) / a):
t = (v-u) / a

This equation can be substituted into Equation 3 (s = 1/2 (u+v)t):
s = 1/2 (u+v) x (v-u)/a

Rearranging this gives:

(u+v)(v-u) = 2as
v^2-u^2 = 2as
v^2 = u^2 + 2as

Question 18

How does a Skydiver reach their Terminal Velocity

Answer 18

1. At the start of his jump, the air resistance is zero so he accelerates downwards.
2. As his speed increases his air resistance will increase.
3. Eventually the air resistance will be big enough to balance the skydiver’s weight. At this point the forces are balanced so his speed becomes constant - this is called Terminal Velocity.
4. When he opens his parachute the air resistance suddenly increase, causing him to start slowing down.
5. Because he is slowing down his air resistance will decrease again until it balances his weight. The skydiver has now reached a new, lower terminal velocity.

Question 19

What value is gravity in Physics

Answer 19

In Physics, gravity is 9.81ms^-2

(However, in Maths it is 9.8ms^-1)

Question 20

What is Free Fall

Answer 20

When an object is accelerating under gravity, with no other force acting on it, it is said to be in free fall. The acceleration of free fall is denoted by the label g. Since g is an acceleration, it has the unit ms^-2

Question 21

What are the 3 main methods that can be used to determine g

Answer 21

• Electromagnet and Trapdoor
• Light Gates
• Taking Pictures

Question 22

How can you find g by using the Electromagnet and Trapdoor experiment?

Answer 22

An electromagnet holds a small steel ball above a trapdoor. When the current is switched off, a timer is triggered, the electromagnet demagnetises, and the ball falls. When it hits the trapdoor, the electrical contact is broken and the timer stops. The value of g is calculated from the height of the ball and the time taken.

Question 23

How can you find g by using Taking Pictures?

Answer 23

A small, metal ball is dropped from rest next to a meter rule, and it’s fall is recorded on video or with a camera in rapid-fire repeating mode. Alternatively, a stroboscope illuminates the scene with rapid multiple images of the falling ball. The position of the ball at regular intervals is then determined by examining the recording.

Question 24

How can you find g by using Light Gates?

Answer 24

The electromagnet and trapdoor introduce tiny delays into the timing. Instead we can use ‘light gates’, two beams, one above the other, with detectors connected to a timer. When the ball falls through the first beam it interrupts the light and the timer starts. When the ball falls through the second beam, a known distance further down, the timer stops.

Question 25

Stopping Distance Defintion

Answer 25

The stopping distance is the total distance travelled from when the driver first sees a reason to stop, to when the vehicle stops. It has is the sum of both the thinking distance and the braking distance.

Question 26

Thinking Distance Definition

Answer 26

Thinking Distance is the distance travelled between the moment when you first see a reason to stop, to the moment when you use the brake

Question 27

Braking Distance Definition

Answer 27

Braking Distance is the distance travelled from the time the brake is applied until the vehicle stops.

Question 28

How long is the Thinking Distance (+Equation for it)

Answer 28

For a vehicle moving at constant speed:

Thinking Distance = Speed x Reaction Time

Question 29

What Factors Affect Thinking Distance

Answer 29

• Level of Alertness
• Drugs/Alcohol in the system
• Distractions
• Speed of Vehicle

Question 30

What Factors Affect Braking Distance

Answer 30

• Mass of the vehicle
• Speed of the vehicle
• State of the tyres
• Road Condition (wet, icy, surface)

Question 31

What is Thinking Distance proportional to and what is Braking Distance proportional to

Answer 31

Thinking Distance ∝ Speed

Braking Distance ∝ Speed^2

Question 32

How is Braking Distance affect when the road is icy

Answer 32

If the road surface is icy, the braking distance will be 10 times longer than it would usually be.

Question 33

A car is travelling at a constant velocity of 15ms^-1 on a level road. The driver sees a child stepping onto the road, 50m ahead. The driver takes 0.50s to react before applying the brakes. The brakes decelerate the car at 6.0ms^-2. Calculate how far the car stops from where the child stepped onto the road.

Answer 33

0.5 x 15 = 7.5m thinking distance

v^2 = u^2 +2as

0 = 15^2 + 2x-6xs
0 = 225 - 12s
12s = 225
s = 18.7

18.75 + 7.5 = 26.25

50 - 26.25 = 23.75m

Question 34

Method for Solving Questions involving projectiles with vertical and horizontal motion

Answer 34

1. Draw a diagram
2. Define upwards as the positive direction
3. Write down all the equations
4. List what you have in the horizontal direction
5. List what you have in the vertical direction
6. Gravity, a = -9.81m/s2, only in the vertical direction
7. a = 0, in the horizontal direction (no air resistance)
8. Highlight what you need
9. Pick the equation that you need

Question 35

Example Question:

A ball is kicked off a cliff 100m above the sea with an initial horizontal velocity of 10ms-1. Assuming that there is no air resistance, calculate the time it takes for the ball to reach the sea and the horizontal distance it travels.

Answer 35

Vertical:
s = 100
u = 0
v =
a = -9.8
t = ?

s = ut + 1/2at^2
s = 0 + 1/2at^2
t^2 = 2s/a
t = √2s/a
t = 4.52s

Horizontal:
s = ?
u = 10
v
a = 0
t = 4.52s

s = ut+1/2at^2
s = (10x4.52) + 0
s = 45.2m

Question 36

Example Question:

A projectile is launched from the ground at an angle of 30° to the horizontal with a velocity of 20m/s. Calculate the horizontal distance it travels before it hits the ground again.

Answer 36

Vertical: sin(30) x 20 = 10ms^-1
Horizontal: cos(30) x 20 = 17.32ms^-1

Vertical:
s
u = 10
v = 0
a = -9.81
t

v = u+at
0 = 10-9.81t
9.81t = 10
t = 2.04s

Horizontal:
s
u = 17.3
v = 17.3
a = 0
t = 2.04

d = sxt
d = 2.04 x 17.3
d = 35.3m

Question 37

Example Question:

An object is projected horizontally at a speed of 16ms^-1 into the sea from a cliff top of height 45.0m.

Calculate:

a) How long it takes to reach the sea
b) How far it travels horizontally
c) It’s impact vertical velocity

Answer 37

a)
Vertical:
s = 45
u = 0
v
a = 9.81
t

s = ut+1/2at^2
45 = 4.905t^2
9.17 = t^2
3.03 = t

b)
s = ut + 1/2at^2
s = 16 x 3.03
s = 48.5m

c)
v^2 = u^2 + 2as
v^2 = 2 x 9.81 x 45
v^2 = 882.9
v = 29.7ms^-1

Question 37

Example Question:

An object is projected horizontally at a speed of 16ms^-1 into the sea from a cliff top of height 45.0m.

Calculate:

a) How long it takes to reach the sea
b) How far it travels horizontally
c) It’s impact vertical velocity

Answer 37

a)
Vertical:
s = 45
u = 0
v
a = 9.81
t

s = ut+1/2at^2
45 = 4.905t^2
9.17 = t^2
3.03 = t

b)
s = ut + 1/2at^2
s = 16 x 3.03
s = 48.5m

c)
v^2 = u^2 + 2as
v^2 = 2 x 9.81 x 45
v^2 = 882.9
v = 29.7ms^-1

Question 38

How does Horizontal Velocity change and why

Answer 38

Assuming there is no air resistance:

• Horizontal velocity remains constant

Question 39

How does Vertical Velocity change and why

Answer 39

Assuming there is no air resistance:

• The vertical velocity changes due to acceleration of free fall
• The vertical displacement and time of flight can be calculated using equations of motion

Question 40

Why does the horizontal velocity of the projectile remain constant?

Answer 40

Acceleration and velocity are vectors.

The acceleration of free fall is vertically downwards. The component of this acceleration of free fall is vertically downwards. The component of this acceleration in the horizontal direction is zero.

Horizontal acceleration = g cos(90) = 0

The horizontal velocity is therefore unaffected by the fall

Question 41

Worked Example: Cannonball

A cannonball is fired horizontally from a cliff top 44.1m above the sea. The initial horizontal velocity of the cannonball is 304ms^-1.

Calculate:

a) the time of flight
b) the horizontal distance it travels

Answer 41

a) Vertical Direction
Step 1:

Identify the equation needed and list the known values .

We can use the equation s = ut + 1/2at^2 to calculate the time t.

The initial vertical velocity u = 0 (initial vertical velocity u = 304 x cos(90) = 0).

s = 44.1m, u = 0, a = 9.81ms^-2, t = ?

Step 2:

Substitute the values into the equation and calculate the answer.

s = 1/2at^2
t^2 = 2s/a = 2x44.1/9.81 = 8.991s^2
t = 3.00s

b) Horizontal Direction
Step 1:

Identify the equation needed.

No acceleration, therefore horizontal distance = horizontal velocity x time

Step 2:

Substitute the values into the equation and calculate the answer.

Horizontal distance = 304 x 3.00 = 912m

The horizontal range of the cannonball is 912m

Question 42

Worked Example: Arrows

An arrow is fired horizontally at 49.2ms^-1 from a 29.4m high castle. Calculate its velocity when it hits the ground.

Answer 42

Vertical Motion

Step 1:
Identify the equation needed and list the known values.

v^2 = u^2 + 2as

The initial vertical velocity u = 0, a = 9.81ms^-2, s = 29.4m

Step 2:
Substitute the values into the equation and calculate the answer.

v^2 = 0^2 + (2x9.81x29.4) = 576.8m^2s^-2

Final vertical velocity Vy = 24.0ms^-1

Horizontal Motion

Step 1:
Identify the method needed and list the known values.
Use Pythagoras’ theorem
The horizontal component of the velocity Vx = 49.2ms^-1

Step 2:
Substitute the values into the equation and calculate the answer.

Velocity of arrow = √24.0^2 + 49.2^2 = 54.7ms^-1
= tan^-1 24.0/49.2 = 26.0° to the horizontal.

Question 43

Worked Example: Kicking a ball

A ball is kicked with an initial velocity of 15ms^-1 at an angle of 30° to the horizontal. Calculate the range R of this ball.

Answer 43

Step 1: Identify the method needed.

Work out the time of flight from the vertical motion, then the horizontal range from the time.

Step 2: List the values and the equation for the first part

Vertical Motion
The time of the ball in flight can be calculated from its vertical motion. Note that the velocity of the ball when it hits the ground has the same magnitude as, but the opposite sign to its upward velocity when it left the ground.

u = 15sin(30) = 7.5ms^-1
v = -7.5ms^-1
a = -9.81ms^-2
t = ?

v = u + at

Step 2: Substitute the values into the equation and calculate the answer.

v = u + at
t = v-u/a = -7.5-7.5/-9.81 = 1.53s

Step 3: Use the answer for vertical motion to calculate the answer for the range.

Horizontal Motion
The horizontal component of the velocity is 15cos(30) = 13ms^-1. This remains constant throughout the flight. Therefore range R = (15cos30) x 1.53 = 20m