March 2013

3.1 50N

3.2.1

of 3,43 m.s

3.2.2

USE THE GROUND AS ZERO REFERENCE.

Indicate the following on the graph:

• Height from which the ball is dropped

• Height reached by the ball after the bounce

• Time at which the ball bounces off the floor (5)

3.3

OPTION 1

Ground as zero reference and downward negative:

OPTION 2

Ground as zero reference and downward positive:

OPTION 1

Ground as zero reference and downward negative:

OPTION 2

Ground as zero reference and downward positive:

March 2012

3.1

3.2.1

3.2.2

3.3

OR

OR

3.4

Calculate the initial velocity at which it reached the top of the window.

Use this velocity as the final velocity from the highest point to this point.

Calculate the initial velocity at which it reached the top of the window.

Use this velocity as the final velocity from the highest point to this point.

Nov 2012

3.1 Downward

3.2.1

3.2.2

The object bounces off the balcony at a velocity of 27,13 m.s

Indicate the following velocity and time values on the graph:

• The initial velocity at which the object was projected from the roof of the building

• The velocity at which the object strikes the balcony

• The time when the object strikes the balcony

• The velocity at which the object bounces off the balcony

• The time when the object strikes the ground (6)

3.3

March 2011

3.1 Gradient of the graph is constant.

3.2 At t = 1s and t = 3s

3.3

3.4

The reason why 10m.s^{-2} OR 9,8m.s^{-2} was allowed, is that the acceleration is 10m.s^{-2} from the graph.

The reason why 10m.s

3.5

Displacement

OR

Change in position

Displacement

OR

Change in position

3.6

Nov 2011

3.1 The initial velocity (speed) of the camera is the same as that of the balloon. (upwards)

3.2

The balloon speed would be the initial velocity of the camera.

For the camera, use the the dispacment of 92,4m but the full time of 6 seconds. The initial velocity thus calculated would be the answer since this accomodates the full 6 seconds of up and down, and NOT the time to actually fall the 92,4m.

The balloon speed would be the initial velocity of the camera.

For the camera, use the the dispacment of 92,4m but the full time of 6 seconds. The initial velocity thus calculated would be the answer since this accomodates the full 6 seconds of up and down, and NOT the time to actually fall the 92,4m.

Indicate the following on the graph:

• Initial velocity

• Time at which it reaches the ground (4)

3.3

Use a calculation to show how you arrived at the answer. (5)

3.4

It is understodd that the jogger will get there first, and wait to catch the camera.

It is understodd that the jogger will get there first, and wait to catch the camera.

March 2010

Ignore the effects of friction and take the downwards motion as positive.

5.1

Velocity after a further 18,2m:

He will not be struck

His reaction time is shorter than the time for the brick to reach his head.

Reaction Time is the delay you experience from when you saw something, till you actually react.

He will not be struck

His reaction time is shorter than the time for the brick to reach his head.

Reaction Time is the delay you experience from when you saw something, till you actually react.

Nov 2010

3.1 3 seconds

3.2 Area between graph and time axis

USE THE EDGE OF THE CLIFF AS ZERO OF POSITION.

Indicate the following on the graph:

• The time when projectile

• The time when projectile

3.3

The first projectile,

(Ignore the effects of friction.)

Calculate the following:

3.4.1

3.4.2

March 2009

5.1

5.2 Consider upward motion as positive:

Then

Then

5.3 Take upward as positive:

5.4

Will the ball reach the SAME, GREATER or LESSER height compared to the previous ball? Use principles of physics to explain your answer. (3)

5.5 Smaller

Contact time for softer ball is longer than for rigid ball.

According to F_{net}Δt = Δp, the force exerted by floor on softer ball is
smaller than on the rigid ball.

Contact time for softer ball is longer than for rigid ball.

According to F

Nov 2009 Unused

from t = 0s to t = 0,4s. (4)

4.1

at t = 0s:

ball starts from rest (0 m.s^{-1})

from t = 0s - 0,4s:

• falls at constant acceleration

• constant increase in velocity

at t = 0,4s:

• reaches the floor at 4m.s^{-1} (or 4m.s^{-1} downwards)

at t = 0,4s:

• bounces back at -3m.s^{-1} (or 3m.s^{-1} upwards)

at t = 0s:

ball starts from rest (0 m.s

from t = 0s - 0,4s:

• falls at constant acceleration

• constant increase in velocity

at t = 0,4s:

• reaches the floor at 4m.s

at t = 0,4s:

• bounces back at -3m.s

4.2

To find the distance from a v-t graph, calculate the area under the graph.

To find the distance from a v-t graph, calculate the area under the graph.

Use the given velocity versus time graph for the motion of the ball to sketch the corresponding position-time graph for the time interval 0s to 0,7s. (3)

4.3

4.4 Inelastic

Decrease / change in speed (from 4m.s^{-1} to 3m.s^{-1})

OR

Decrease/change in kinetic energy during collision

Decrease / change in speed (from 4m.s

OR

Decrease/change in kinetic energy during collision

Nov 2009

Two construction workers, Alex and Pete, were arguing about whether a smaller brick would hit the ground quicker than a larger brick when both are released from the same height.

Alex said that the larger brick should hit the ground first. Pete argued that the smaller brick would hit the ground first.

4.1

**Option 1**

Statements not correct

The bricks will experience the same gravitational acceleration of free fall and thus reach the ground at the same time.

**Option 2**

Pete is correct or Alex is wrong.

The smaller brick experiences less air resistance, thus larger acceleration and reaches the ground first.

**Option 3**

Alex is correct or Pete is wrong.

In the presence of air resistance, the larger brick, with larger mass, experiences a larger net force downwards, thus largest acceleration and reaches the ground first.

**Option 4**

Both are correct.

Pete correct: The smaller brick experiences less air resistance, thus larger acceleration and reaches the ground first.

Alex correct: In the presence of air resistance, the larger brick, with larger mass, experiences a larger net force downwards, thus largest acceleration and reaches the ground first.

Option 1 deals with a free fall situation. Options 2, 3, and 4 deals with air resistance present. Option 1 is the best answer with what this question intended.

Statements not correct

The bricks will experience the same gravitational acceleration of free fall and thus reach the ground at the same time.

Pete is correct or Alex is wrong.

The smaller brick experiences less air resistance, thus larger acceleration and reaches the ground first.

Alex is correct or Pete is wrong.

In the presence of air resistance, the larger brick, with larger mass, experiences a larger net force downwards, thus largest acceleration and reaches the ground first.

Both are correct.

Pete correct: The smaller brick experiences less air resistance, thus larger acceleration and reaches the ground first.

Alex correct: In the presence of air resistance, the larger brick, with larger mass, experiences a larger net force downwards, thus largest acceleration and reaches the ground first.

Option 1 deals with a free fall situation. Options 2, 3, and 4 deals with air resistance present. Option 1 is the best answer with what this question intended.

4.2.1 Any two

• Ensure that both bricks are dropped from same height

• Ensure that both bricks are dropped at the same time

• Ensure that the stopwatch starts at instant that each brick is released and stopped at the instant that each brick reaches the ground

• Repeat the experiment several times and use the average of the results

• Make sure that v_{i} = 0 for both bricks

• Make sure that there is no strong wind

• Use bricks made of the same material / of same density

• Ensure that both bricks are dropped from same height

• Ensure that both bricks are dropped at the same time

• Ensure that the stopwatch starts at instant that each brick is released and stopped at the instant that each brick reaches the ground

• Repeat the experiment several times and use the average of the results

• Make sure that v

• Make sure that there is no strong wind

• Use bricks made of the same material / of same density

4.2.2 External force(s) may be present e.g. friction / air resistance / strong wind blowing

Ignore the effects of friction and calculate the speed at which brick B was thrown. (7)

4.3

Nov 2008

Use the information on the graph to answer the following questions:

6.1

6.2 0,5s and 1,5s

6.3 1s

6.4 Difference between areas of two triangles

6.5

Prep Paper 2008

5.1 Consider upward motion as positive

5.2 Consider upward motion as positive:

5.3

up as positive

up as positive

Exemplar Paper 2008

Use information from the graph to answer the following questions:

5.1

Considering upward motion only

Upward motion positive

OR

For complete motion of stone

Upward motion positive

Considering upward motion only

Upward motion positive

OR

For complete motion of stone

Upward motion positive

5.2 Upward motion as positive:

Additional Exemplar Paper 2008

5.1

• Release a stone from the top of the well and let it fall straight down into the well.

• Take the time from it was released until it splashes in the water.

• Use the equation:

with v_{i} = 0 to calculate the depth of the water level.

• Release a stone from the top of the well and let it fall straight down into the well.

• Take the time from it was released until it splashes in the water.

• Use the equation:

with v

5.2 Due to air friction, gravity is not the only force acting on the object.

5.3.1

5.3.2 Velocities will be the same.

Both X and Y experience the same displacement and same acceleration. On its downward flight X has same velocity as Y at a height of h.

Using:

will thus give the same final velocity for both.

Both X and Y experience the same displacement and same acceleration. On its downward flight X has same velocity as Y at a height of h.

Using:

will thus give the same final velocity for both.

Calculate the initial speed at which she threw the second stone. Ignore the effects of friction.

5.4

For X - downwards as positive

For Y - downwards as positive

For X - downwards as positive

For Y - downwards as positive

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