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PHYSICS, M5 2019 VCE 8

A 250 g toy car performs a loop in the apparatus shown in Figure 8.
 

The car starts from rest at point \(\text{A}\) and travels along the track without any air resistance or retarding frictional forces. The radius of the car's path in the loop is 0.20 m. When the car reaches point \(\text{B}\) it is travelling at a speed of 3.0 m s\(^{-1}\).

  1. Calculate the value of \(h\). Show your working.   (3 marks)

  2. Calculate the magnitude of the normal reaction force on the car by the track when it is at point \(\text{B}\). Show your working.   (3 marks)

  3. Explain why the car does not fall from the track at point \(\text{B}\), when it is upside down.   (2 marks)

Show Answers Only

a.    \(0.86\ \text{m}\)

b.    \(8.8\ \text{N}\)

c.    Method 1

→ During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.

→ This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms\(^{-1}\).
 

Method 2

→ The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force. 

\(\dfrac{mv^2}{r}\) \(=mg\)  
\(v\) \(=\sqrt{gr}\)  
  \(=\sqrt{9.8 \times 0.2}\)  
  \(=1.4\ \text{ms}^{-1}\)  

 
→ As 3 ms\(^{-1}\) is greater than the minimum speed required, a normal force downwards will be present at point \(\text{B}\).

→ Therefore the car will not fall from the track at point \(\text{B}\).

Show Worked Solution

a.    Using the law of the conservation of energy:

\(mgh_A\) \(=mgh_B +\dfrac{1}{2}mv^2_b\)  
\(0.25 \times 9.8 \times h_A\) \(=0.25 \times 9.8 \times 0.4 + \dfrac{1}{2} \times 0.25 \times 3^2\)  
\(2.45h_A\) \(=2.105\)  
\(h_A\) \(=\dfrac{2.105}{2.45}\)  
  \(=0.86\ \text{m}\)  

 

b.     \(N + mg\) \(=\dfrac{mv^2}{r}\)
  \(N\) \(=\dfrac{mv^2}{r}-mg\)
    \(=\dfrac{0.25 \times 3^2}{0.2}- 0.25 \times 9.8\) 
    \(=8.8\ \text{N}\)

♦ Mean mark (b) 53%.

c.    Method 1

→ During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.

→ This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms\(^{-1}\).
 

Method 2

→ The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force. 

\(\dfrac{mv^2}{r}\) \(=mg\)  
\(v\) \(=\sqrt{gr}\)  
  \(=\sqrt{9.8 \times 0.2}\)  
  \(=1.4\ \text{ms}^{-1}\)  

 
→ As 3 ms\(^{-1}\) is greater than the minimum speed required, a normal force downwards will be present at point \(\text{B}\).

→ Therefore the car will not fall from the track at point \(\text{B}\).

♦♦♦ Mean mark (c) 23%.
COMMENT: Normal force is poorly understood here..

PHYSICS, M5 2020 VCE 8 MC

A ball is attached to the end of a string and rotated in a circle at a constant speed in a vertical plane, as shown in the diagram below.
 

The arrows in options \(\text{A}\). to \(\text{D}\). below indicate the direction and the size of the forces acting on the ball.

Ignoring air resistance, which one of the following best represents the forces acting on the ball when it is at the bottom of the circular path and moving to the left?
 

Show Answers Only

\(D\)

Show Worked Solution

→ There are only two forces acting on the ball, gravitational force directly down, and the centripetal force (string tension) directly towards the centre of the circular motion.

→ As the ball is following a circular path, the tension in the string is greater than the gravitational force acting on the ball.

\(\Rightarrow D\)

Mean mark 56%.

PHYSICS, M5 2022 VCE 8

A Formula 1 racing car is travelling at a constant speed of 144 km h\(^{-1}\) (40 m s\(^{-1}\)) around a horizontal corner of radius 80.0 m. The combined mass of the driver and the car is 800 kg. Figure 8a shows a front view and Figure 8b shows a top view.
 

  1. Calculate the magnitude of the net force acting on the racing car and driver as they go around the corner.  (2 marks)
  1. On Figure 8b, draw the direction of the net force acting on the racing car using an arrow.  (1 mark)
  1. Explain why the racing car needs a net horizontal force to travel around the corner and state what exerts this horizontal force.  (2 marks)
Show Answers Only

a.    \(1.6 \times 10^4\ \text{N}\)

b.    
       

c.    → A net horizontal force is required for circular motion.

→ The horizontal force is perpendicular to the direction of travel.

→ This is the force which changes the direction of travel.

→ The force of friction from the road acts on the tires provides the centripetal force in this motion.

Show Worked Solution

a.    \(F=\dfrac{mv^2}{r}=\dfrac{800 \times (40)^2}{80}=1.6 \times 10^4\ \text{N}\)

b.   
       

c.    → A net horizontal force is required for circular motion.

→ The horizontal force is perpendicular to the direction of travel.

→ This is the force which changes the direction of travel.

→ The force of friction from the road acts on the tires provides the centripetal force in this motion.

♦ Mean mark 42%.

PHYSICS, M5 EQ-Bank 23

A horizontal disc is rotating clockwise on a table when viewed from above. Two small blocks are attached to the disc at different radii from the centre.

Using the diagram below, draw vector arrows to show the relative linear velocities and centripetal forces for each block as the disc rotates.   (3 marks)
 

Show Answers Only

Show Worked Solution

Points to note on the image

→ Centripetal force and velocity vectors are perpendicular

→ Vector arrows should show relative magnitudes of velocities and centripetal forces.

PHYSICS, M5 EQ-Bank 27

A toy car was placed facing outwards on a rotating turntable. The car was held in place by a force sensor connected to the centre of the turntable. The centre of mass of the car was 0.25 metres from the centre of the turntable. The reading from the force sensor was recorded at varying speeds of rotation. A stopwatch was used to time the rotation of the turntable. The linear velocity was calculated from the period of rotation. The graph shows the force on the car versus the square of the linear velocity of the car.
 


 

  1. Use the graph to determine the mass of the car.   (3 marks)
  1. Identify possible errors in the data and outline how to reduce their effects on the estimation of the mass of the car.   (4 marks)
Show Answers Only

a.   0.034 kg

b.   Errors and reduction strategies

→ If the sensor does not produce accurate readings then systemic errors will result. 

→ Inaccurate readings can be minimised by calibrating the sensor. This can be done against a known force such as the force of gravity on a 1 kg mass.

→ If the stopwatch used to time rotations is manually operated, then error due to human inaccuracy and varying reaction times is introduced. Inaccurate calculations of the motion’s period (i.e. timing of one rotation) makes linear velocity estimations (`v=romega`) less accurate.

→ A strategy for mitigating this inaccuracy is by using the stopwatch to measure several rotations at a time and then dividing the result by the total number of rotations.

Show Worked Solution

a.   Graph passes through (2,0) and (25, 3.1):

`text{Gradient}` `=(3.1-0)/(25-2)`  
  `=0.135`  

 
→ `text{Gradient}\ =(F)/(v^2)=0.135`

→ As the car is undergoing uniform circular motion:

`F_(c)` `=(mv^2)/(r)`  
`(F_(c))/(v^2)` `=(m)/(r)`  
`0.135` `=(m)/(r)`  
`:. m` `=0.135 xx 0.25`  
  `=0.03375`  
  `=0.034\ text{kg}`  

 

b.   Errors and reduction strategies

→ If the sensor does not produce accurate readings then systemic errors will result. 

→ Inaccurate readings can be minimised by calibrating the sensor. This can be done against a known force such as the force of gravity on a 1 kg mass.

→ If the stopwatch used to time rotations is manually operated, then error due to human inaccuracy and varying reaction times is introduced. Inaccurate calculations of the motion’s period (i.e. timing of one rotation) makes linear velocity estimations (`v=romega`) less accurate.

→ A strategy for mitigating this inaccuracy is by using the stopwatch to measure several rotations at a time and then dividing the result by the total number of rotations.

PHYSICS, M5 EQ-Bank 8 MC

A baseball training aid swings a 150 gram baseball around a pole in a horizontal circular path with radius 1.8 m.

The angular velocity of the motion is `5 pi\ text {rad s}^(-1)`.
  


 

What is the magnitude of the centripetal force required to maintain the motion of the ball?

  1. 23 N
  2. 37 N
  3. 67 N
  4. 74 N
Show Answers Only

`C`

Show Worked Solution
`v` `=omegar`  
  `=5pi xx1.8`  
  `=9pi\ text{ms}^(-1)`  

 

`F_(c)` `=(mv^2)/(r)`  
  `=(0.15 xx(9pi)^2)/(1.8)`  
  `~~67\ text{N}`  

 
`=>C`

PHYSICS, M5 EQ-Bank 6 MC

A researcher ties a small lead weight to a string and swings it in a horizontal circle above her head.

She uses the results to evaluate the relationship between centripetal force and speed. 

Which of the following needs to be kept constant to ensure a valid experiment?

  1. The mass of the lead weight and the length of the string
  2. The angular velocity of the lead weight
  3. The mass of the lead weight only
  4. The length of the string only
Show Answers Only

`A`

Show Worked Solution

→ `F_(c)=(mv^2)/(r)`

→ The factors that affect centripetal force are the mass of the lead weight, `m`, the speed,`v`, and the radius of rotation (length of string), `r`.

→ If the student aims to measure the relationship between centripetal force and speed, then to ensure a valid experiment the mass of the lead weight and length of the string must be kept constant to ensure these factors do not impact results.

`=>A`

PHYSICS, M5 2016 HSC 18 MC

A motorcycle travels around a vertical circular path of radius 3.6 m at a constant speed. The combined mass of the rider and motorcycle is 200 kg.
 

What is the minimum speed, in `text{m s}^(-1)`, at which the motorcycle must travel to maintain the circular path?

  1. 0.42
  2. 1.9
  3. 5.9
  4. 35
Show Answers Only

`C`

Show Worked Solution

→ The minimum speed occurs when the centripetal force keeping the motorcycle on its path equals the downwards weight force at the top of the circle.

`F_(c)` `=W`  
`(mv^2)/(r)` `=mg`  
`v` `=sqrt(rg)`  
  `=sqrt(3.6 xx9.8)`  
  `=5.9\ text{m s}^-1`  

 
`=>C`

PHYSICS, M5 2017 HSC 15 MC

A car travelling at a constant speed follows the path shown.
 

An accelerometer that measures acceleration along the `X - X^’` direction is fixed in the car.

Which graph shows the measurements recorded by the accelerometer over the 20-second interval?
 

 

Show Answers Only

`B`

Show Worked Solution

→ The car makes a sharp turn left in a small radius when it enters and exits the roundabout and it turns right in a larger radius around the roundabout.

→ As  `a_(c)=(v^2)/(r)`, this means there will be an acceleration with a large magnitude at the start and end of the interval and an acceleration with lower magnitude and opposite sign during the middle of the journey.

`=>B`


♦ Mean mark 43%.

PHYSICS, M5 2019 HSC 35

The apparatus shown is attached horizontally to the roof inside a stationary car. The plane of the protractor is perpendicular to the sides of the car.
 

The car was then driven at a constant speed `(v)`, on a horizontal surface, causing the string to swing to the right and remain at a constant angle `(theta)` measured with respect to the vertical.

Describe how the apparatus can be used to determine features of the car's motion. In your answer, derive an expression that relates a feature of the car's motion to the angle `theta`.   (4 marks)

Show Answers Only

→ The constant deflection of the mass to the right indicates that the car has a uniform acceleration to the left.

→ Since the car’s speed is constant, it must be travelling in uniform circular motion.

→ The radius of the car’s motion can be found in terms of `theta.`

As there is no vertical acceleration of the mass, the vertical component of its tension will balance its weight:

`T_y=Tcos theta=mg\ \ …\ (1)`

The centripetal force on the mass is given by the horizontal component of its tension:

`T_x=Tsin theta=ma=(mv^2)/r\ \ …\ (2)`

Substitute `T=(mg)/costheta`  from (1) into (2):

`T sin theta` `=(mv^2)/(r)`
`(mg xx sin theta)/(cos theta)` `=(mv^(2))/(r)`
`g xx tan theta` `=(v^(2))/(r)`
`r` `=(v^(2))/(g xx tan theta)`

 
Other expressions could include:

  • `v=sqrt(rg tan theta)`
Show Worked Solution

→ The constant deflection of the mass to the right indicates that the car has a uniform acceleration to the left.

→ Since the car’s speed is constant, it must be travelling in uniform circular motion.

→ The radius of the car’s motion can be found in terms of `theta.`

As there is no vertical acceleration of the mass, the vertical component of its tension will balance its weight:

`T_y=Tcos theta=mg\ \ …\ (1)`

The centripetal force on the mass is given by the horizontal component of its tension:

`T_x=Tsin theta=ma=(mv^2)/r\ \ …\ (2)`

Substitute `T=(mg)/costheta`  from (1) into (2):

`T sin theta` `=(mv^2)/(r)`
`(mg xx sin theta)/(cos theta)` `=(mv^(2))/(r)`
`g xx tan theta` `=(v^(2))/(r)`
`r` `=(v^(2))/(g xx tan theta)`

 
Other expressions could include:

  • `v=sqrt(rg tan theta)`

♦ Mean mark 52%.

PHYSICS M5 2022 HSC 35

A capsule travels around the International Space Station (ISS) in a circular path of radius 200 m as shown.
 


 

Analyse this system to test the hypothesis below.  (5 marks)

The uniform circular motion of the capsule around the ISS can be accounted for in terms of the gravitational force between the capsule and the ISS.

Show Answers Only

Find the gravitational force between the capsule and the ISS:

`F` `=(GMm)/(r^2)`  
  `=(6.67 xx10^-11 xx4.2 xx10^5 xx 1.2 xx10^4)/(200^2)`  
  `=8.4 xx10^(-6)  text{N}`  

 
Find the centripetal force required to keep the capsule in its orbit:

`F_(c)` `=(mv^2)/(r)`  
  `=(1.2 xx10^4 xx 0.233^2)/(200)`  
  `=3.26  text{N}`  

 

→ The gravitational force is not sufficient to provide the necessary centripetal force to keep the capsule in orbit around the ISS.

→ The hypothesis is invalid.

Show Worked Solution

Find the gravitational force between the capsule and the ISS:

`F` `=(GMm)/(r^2)`  
  `=(6.67 xx10^-11 xx4.2 xx10^5 xx 1.2 xx10^4)/(200^2)`  
  `=8.4 xx10^(-6)  text{N}`  

 
Find the centripetal force required to keep the capsule in its orbit:

`F_(c)` `=(mv^2)/(r)`  
  `=(1.2 xx10^4 xx 0.233^2)/(200)`  
  `=3.26  text{N}`  

 

→ The gravitational force is not sufficient to provide the necessary centripetal force to keep the capsule in orbit around the ISS.

→ The hypothesis is invalid.

PHYSICS, M5 2019 HSC 20 MC

In the apparatus shown, a backboard is connected by a rod to a shaft. The shaft is spun by an electric motor causing the backboard to rotate in the horizontal plane around the axis  X `–` X′.

A cube is suspended by a string so that it touches the surface of the backboard.
 

When the angular velocity of the motor is great enough, the string is cut and the position of the cube does not change relative to the backboard.

Which statement correctly describes the forces after the string is cut?

  1. The sum of the forces on the cube is zero.
  2. The horizontal force of the backboard on the cube is equal in magnitude to the horizontal force of the cube on the backboard.
  3. The horizontal force of the backboard on the cube is greater than the horizontal force of the cube on the backboard, resulting in a net centripetal force.
  4. The force of friction between the cube and the backboard is independent of the force of the backboard on the cube because these forces are perpendicular to each other.
Show Answers Only

`B`

Show Worked Solution

Using Newton’s Third Law:

The backboard and the cube are an action-reaction pair.

The force of the backboard on the cube (centripetal force) is equal in magnitude and opposite in direction to the force of the cube on the backboard.

`=>B`


♦ Mean mark 36%.