C5 Flashcards
During the car’s acceleration, the driver of the car noticed she felt pushed back in her seat. Use one of Newton’s laws to explain this phenomenon.
N1
According to N1, the driver will
- MAINTAIN her CURRENT motion
- until an unbalanced force acts on her.
- During ACCELERATION the car seat applies an
UNBALANCED force forward which feels like she is
being pushed back.
A rock is thrown vertically into the air from a height of 1.60 m above the ground with a velocity of 24.8 m/s upwards.
Calculate the time until the rock impacts the ground after being thrown into the air.
- Calculate time to travel up
v = u + at
t = 2.53 sec - Calculate time to travel down
s = ut + ½ at^2
∴ t = 2.59 sec
Total time = 2.59 + 2.53 = 5.12 sec
What is meant by terminal velocity?
When Fg = -Ff
- Zero acceleration
- when the accelerating force is
BALANCED by frictional force.
- no net force
In a similar experiment, the rock is replaced with a light ball of the same size. The ball rises a smaller distance before falling and quickly reaching its terminal velocity.
In this situation, why does the ball reach terminal velocity so rapidly, yet the rock appears to
be affected by gravity alone?
The gravitational force (Fg) of the light ball is small and so it requires a small frictional force to balance it. This small frictional force is achieved at a slow speed.
Or
The light ball has a large surface area to mass ratio
meaning it is more affected by air resistance at low
speed.
A netball player throws a ball horizontally towards a teammate standing 12 m away. The ball is thrown from a height of 2.1 m. The teammate catches the ball at a waist height of 0.90 m.
ball was in the air for around 0.495s
horizontal initial velocity: 24.24 (800/33
c) What final velocity did the ball have when caught by the teammate?
v = u + at
v = 0 + 9.81 x 0.495
∴ v = 4.85 m/s
c2 = a2 + b2
∴ v = 24.8 m/s
θ = arctan (4.85/24.3) = 11.3 degrees
Velocity = 24.8 m/s 11.3 degrees below horizontal
When the boat reaches 6.0 m/s, the wind stops and hence no longer exerts a force on the boat. Describe and explain the motion of the boat over the next few seconds
using Newton’s law(s).
The boat will continue moving at 6.0 m/s, as stated in Newton’s first law.
As the water exerts an unbalanced frictional force on the boat, the boat will experience a negative acceleration, as stated in Newton’s second law, and will begin to slow down.
Two skydivers with parachutes jumped from a plane.
Both skydivers reached terminal velocity before they opened their parachutes.
Define the term terminal velocity.
Terminal velocity is the constant speed which is
reached by a falling object when the air resistance
force is equal and opposite to the force due to gravity.
Two skydivers with parachutes jumped from a plane.
Both skydivers reached terminal velocity before they opened their parachutes.
Both skydivers present the same surface area as they fall but the second has a greater mass.
Would the heavier skydiver reach a terminal velocity which is lower, higher or the same as the other skydiver? Explain your reasoning.
The heavier skydiver would reach a ………………………………………………….. terminal velocity.
Explanation:
The heavier skydiver will reach a higher terminal
velocity.
As their weight force is higher, it takes longer for the
air resistance force to increase to the point that it is
equal to the weight force, and thus the skydiver will be
at a higher maximum velocity.
As shown in the diagram on the next page (page 9), the archery target is supported by a diagonal brace which provides support in the horizontal direction.
With reference to Newton’s laws, explain why the target needs horizontal support.
When the arrow hits the target, the target exerts a
horizontal force to stop the arrow and the arrow
exerts a horizontal force on the target in accordance with Newton’s THIRD law.
In order to stop the target from accelerating away from the arrow in accordance with Newton’s SECOND law, the horizontal brace provides a resisting horizontal force, which means the net force is 0 and the target doesn’t fall, in accordance with Newton’s FIRST law.
In April 2021 the Mars Ingenuity helicopter conducted
the first controlled powered flight on another planet. It
has a mass of 1.80 kg and has been designed with
oversized rotor blades to give sufficient lift in Mars’ thin
atmosphere.
On Mars, the acceleration due to gravity is 3.72 m s–2
.
However, the atmosphere is so thin that the blades only
provide a maximum lift force of 8.00 N when the
helicopter is operating at full power.
e) Engineers needed to determine how the helicopter would deal with various flight scenarios on Mars. A Mars flight scenario they analysed was as follows:
▪ Initially, the helicopter is hovering stationary 0.500 m above the ground.
▪ The helicopter then accelerates vertically upwards for 2.50 m at full power.
▪ The engine fails and the helicopter falls.
*Weight of helicopter on Mars: 6.696 N
*Net force on helicopter when operating at full power: 1.30 N
*Maximum acceleration of the helicopter: 0.724 m/s^2 upwards
i. Calculate the expected velocity reached by helicopter just before the engine fails.
v2 = u2 + 2as
= (0) + 2(0.724)(2.5)
v2 = 3.62 m/s
v = √3.62
v = 1.90 m/s (up)
In April 2021 the Mars Ingenuity helicopter conducted
the first controlled powered flight on another planet. It
has a mass of 1.80 kg and has been designed with
oversized rotor blades to give sufficient lift in Mars’ thin
atmosphere.
On Mars, the acceleration due to gravity is 3.72 m s–2
.
However, the atmosphere is so thin that the blades only
provide a maximum lift force of 8.00 N when the
helicopter is operating at full power.
e) Engineers needed to determine how the helicopter would deal with various flight scenarios on Mars. A Mars flight scenario they analysed was as follows:
▪ Initially, the helicopter is hovering stationary 0.500 m above the ground.
▪ The helicopter then accelerates vertically upwards for 2.50 m at full power.
▪ The engine fails and the helicopter falls.
*Weight of helicopter on Mars: 6.696 N
*Net force on helicopter when operating at full power: 1.30 N
*Maximum acceleration of the helicopter: 0.724 m/s^2 upwards
*Expected velocity reached by the helicopter just before the engine fails: 1.90 m/s
ii. The engineers knew that the helicopter could survive a fall to Mars’ surface if the impact speed was less than 4.8 m/s. Determine whether the helicopter could survive the fall in
this scenario.
v^2 = u^2 + 2as
= (−1.90)^2 + 2(3.72)(3.0) {up- , down +}
= 5.09 m/s
As 5.09 > 4.8, the helicopter would not survive.
(g) In order to prevent the rocket being damaged a thick airbag was placed on the ground for
the rocket to fall onto. Explain how this will help protect the rocket. (2 marks)
- f=ma, f=m(v-u)/t
- stopping force made much smaller (less damage)
- by making stopping time longer
- airbag increases t
- increasing t from 0.01 sec to 1 sec results in 100x less net force
Cube A (lighter) is placed on top of cube B.
If a horizontal force, FH, is applied to cube B, state and explain what happens to each
cube when the force is applied, assuming there is no friction between the horizontal
surfaces. Refer to Newton’s Laws to support your answers.
Cube A:
Cube B:
- Cube A in accordance with NI is not being acted upon by any horizontal forces
- therefore remains stationary until B moved away then A falls onto table
- Cube B is being acted upon by an unbalanced force of fH therefore cube B accelerates to the right
(NT 3 for cube B?
While the plane is accelerating along the runway, a passenger has the sensation of being pushed back into their seat. Explain.
- passenger accelerating too so there is a net unbalanced force exerted on them through their seat (NT2)
- passenger tends to remain at a constant velocity (NT1) in a fixed position relative to plane
- therefore plane whilst accelerating tends to move ahead of the comparatively ‘stationary’ passenger
What is the formula for change of momentum?
ΔP= mv-mu
