SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

PHYSICS, M2 2023 VCE 8

Maia is at a skatepark. She stands on her skateboard as it rolls in a straight line down a gentle slope at a constant speed of 3.0 m s\(^{-1}\), as shown in Figure 8. The slope is 5° to the horizontal.

The combined mass of Maia and the skateboard is 65 kg.
  

  1. In Figure 9, the combined system of Maia and the skateboard is modelled as a small box with point \(\text{M}\) at the centre of mass.
  2. Draw and label arrows to represent each of the forces acting on the system - that is, Maia and skateboard as they roll down the slope.  (3 marks)
     


  1. Calculate the magnitude of the total frictional forces acting on Maia and the skateboard.  (2 marks)

Near the bottom of the ramp, Maia takes hold of a large pole and comes to a complete rest while still standing on the skateboard. Maia and the skateboard now have no momentum or kinetic energy.

  1. Explain what happened to both the momentum and the kinetic energy of Maia and the skateboard. No calculations are required.  (2 marks)

Show Answers Only

a.
       

b.     55.5 N

c.    Momentum:

→ Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible.

Kinetic energy:

→ The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down.

Show Worked Solution

a.
       

b.     Total frictional forces:

→ Constant speed means that the force down the slope of the incline is equal to the sum of the frictional forces acting on Maia and the skateboard.

\(F_{\text{down slope}}\) \(=F_f\) \(=mg\, \sin\, \theta\)
    \(=65 \times 9.8 \times \sin\,5^{\circ}\)
    \(=55.5\ \text{N}\)

 

♦ Mean mark (b) 50%.

c.    Momentum:

→ Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible.

Kinetic energy:

→ The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down.

♦♦♦ Mean mark (c) 26%.

PHYSICS, M2 2018 VCE 8-9 MC

A railway truck \(\text{X}\) of mass 10 tonnes, moving at 6.0 m s\(^{-1}\), collides with a stationary railway truck \(\text{Y}\) of mass 5.0 tonnes. After the collision the trucks are joined together and move off as one. The situation is shown below.
 

   

\(\text{Question 8}\)

The final speed of the joined railway trucks after the collision is closest to

  1. \( 2.0 \text{ m s}^{-1}\)
  2. \( 3.0 \text{ m s} ^{-1}\)
  3. \( 4.0 \text{ m s} ^{-1}\)
  4. \( 6.0 \text{ m s} ^{-1}\)
     

\(\text{Question 9}\)

The collision of the railway trucks is best described as one where

  1. kinetic energy is conserved but momentum is not conserved.
  2. kinetic energy is not conserved but momentum is conserved.
  3. neither kinetic energy nor momentum is conserved.
  4. both kinetic energy and momentum are conserved.
Show Answers Only

\(\text{Question 8:}\ C\)

\(\text{Question 9:}\ B\)

Show Worked Solution

\(\text{Question 8}\)

→ By the law of conservation of momentum:

\(m_Xu_X+m_Yu_Y\) \(=m_Xv_X+m_Yv_Y\)  
  \(=v(m_X + m_Y)\ \ \ (v_X=v_Y) \)  
\(10\ 000 \times 6 +0\) \(= v(10\ 000 + 5000)\)  
\(60\ 000\) \(=15\ 000v\)  
\(v\) \(=4\ \text{ms}^{-1}\)   

 
\( \Rightarrow C\)
 

\(\text{Question 9}\)

→ By the law of conservation of momentum, momentum in the collision is conserved.

→ Kinetic energy conservation:

       \(KE_{\text{init}}\) \(=\dfrac{1}{2}m_Xu_X^2+\dfrac{1}{2}m_Yu_Y^2\)
   

\(=\dfrac{1}{2} \times 10\ 000 \times 6^2 + \dfrac{1}{2} \times 5000 \times 0^2\)
    \(=180\ 000\ \text{J}\)

 

\(KE_{\text{final}}\) \(=\dfrac{1}{2}(m_X+m_Y)v^2\)  
  \(=\dfrac{1}{2} \times 15\ 000 \times 4^2\)   
  \(=120\ 000\ \text{J}\)  

 
→ The kinetic energy of the system decreases after the collision and so is not conserved.

\(\Rightarrow B\)

PHYSICS, M2 2020 VCE 10

Jacinda designs a computer simulation program as part of her practical investigation into the physics of vehicle collisions. She simulates colliding a car of mass 1200 kg, moving at 10 ms\(^{-1}\), into a stationary van of mass 2200 kg. After the collision, the van moves to the right at 6.5 ms\(^{-1}\). This situation is shown in Figure 10.
 

  1. Calculate the speed of the car after the collision and indicate the direction it would be travelling in. Show your working.  (4 marks)

  2. Explain, using appropriate physics, why this collision represents an example of either an elastic or an inelastic collision.  (3 marks)

  3. The collision between the car and the van takes 40 × 10\(^{-3}\) seconds.
    1. Calculate the magnitude and indicate the direction of the average force on the van by the car.  (3 marks)

    2. Calculate the magnitude and indicate the direction of the average force on the car by the van.  (2 marks)

Show Answers Only

a.    \(v_{\text{car}} =1.92\ \text{ms}^{-1}\) to the left.

b.    The collision is an inelastic collision as the kinetic energy decreases after the collision.

c.i   The average force of \(358\ \text{kN}\) is to right as the change in the momentum is also to the right.

c.ii    The average force of \(358\ \text{kN}\) is to the left as the change in momentum is also to the left.

Show Worked Solution

a.    Using the Conservation of Momentum:

\(m_{\text{car}}u_{\text{car}}+m_{\text{van}}u_{\text{van}}\) \(=m_{\text{car}}v_{\text{car}}+m_{\text{van}}v_{\text{van}}\)  
\(1200 \times 10 + 2200 \times 0\) \(=1200 \times v_{\text{car}} + 2200 \times 6.5\)  
\(1200v_{\text{car}}\) \(=12\ 000-14\ 300\)  
  \(=-2300\)  
\(v_{\text{car}}\) \(=-1.92\ \text{ms}^{-1}\)  
  \(=1.92\ \text{ms}^{-1}\ \text{to the left}\)  

  

b.     \(KE_{\text{init}}\) \(=\dfrac{1}{2}m_{\text{car}}u_{\text{car}}^2+\dfrac{1}{2}m_{\text{van}}u_{\text{van}}^2\)
   

\(=\dfrac{1}{2} \times 1200 \times 10^2 + \dfrac{1}{2} \times 2200 \times 0^2\)
    \(=60\ 000\ \text{J}\)

\(KE_{\text{final}}\) \(=\dfrac{1}{2}m_{\text{car}}v_{\text{car}}^2+\dfrac{1}{2}m_{\text{van}}v_{\text{van}}^2\)  
  \(=\dfrac{1}{2} \times 1200 \times 1.92^2 + \dfrac{1}{2} \times 2200 \times 6.5^2\)  
  \(=48\ 687\ \text{J}\)  

 

→ As the kinetic energy of the system decreases after the collision, the collision is inelastic.
 

c.i.   \(\Delta p\) \(=F_{net}\Delta t\)  
\(F_{net}\) \(=\dfrac{\Delta p}{ \Delta t}\)  
  \(= \dfrac{m\Delta v}{\Delta t}\)  
  \(= \dfrac{2200 \times (6.5-0)}{40 \times 10^{-3}}\)  
  \(=357\ 500\ \text{N}\)  
  \(=358\ \text{kN to the right}\)  
 
→ The average force of \(358\ \text{kN}\) is to right as the change in the momentum is also to the right.
 
  
c.ii.   \(F_{net}\) \(=\dfrac{\Delta p}{ \Delta t}\)
  \(= \dfrac{m\Delta v}{\Delta t}\)
  \(= \dfrac{1200 \times (-1.92-10)}{40 \times 10^{-3}}\)
  \(=-357\ 600\ \text{N}\)
  \(=-358\ \text{kN}\)

  

→ The average force of \(358\ \text{kN}\) is to the left as the change in momentum is also to the left.

→ Note: no calculation was required for this question as it is an example of Newton’s third law of motion. Simply stating that the force would be equal in magnitude but opposite in direction earned full marks.

PHYSICS, M2 2022 VCE 7

Kym and Kelly are experimenting with trolleys on a ramp inclined at 25°, as shown in Figure 7. They release a trolley with a mass of 2.0 kg from the top of the ramp. The trolley moves down the ramp, through two light gates and onto a horizontal, frictionless surface. Kym and Kelly calculate the acceleration of the trolley to be 3.2 m s\(^{-2}\) using the information from the light gates.
 
 

  1.  i. Show that the component of the gravitational force of the trolley down the slope is \(8.3 \text{ N}\). Use \(g=9.8 \text{ m s}^{-2}\).  (2 marks)
  1. ii. Assume that on the ramp there is a constant frictional force acting on the trolley and opposing its motion.
  2.      Calculate the magnitude of the constant frictional force acting on the trolley.  (2 marks)
  1. When it reaches the bottom of the ramp, the trolley travels along the horizontal, frictionless surface at a speed of 4.0 m s\(^{-1}\) until it collides with a stationary identical trolley. The two trolleys stick together and continue in the same direction as the first trolley.
    1. Calculate the speed of the two trolleys after the collision. Show your working and clearly state the physics principle that you have used.  (3 marks)
    1. Determine, with calculations, whether this collision is an elastic or inelastic collision. Show your working.  (3 marks)
Show Answers Only

a.i.   See Worked Solutions

a.ii.  \(F_f=1.9\ \text{N}\)

b.i.   \(2.0\ \text{ms}^{-1}\)

 b.ii.  For the collision to be elastic, the kinetic energy must be conserved.

\(KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}\)

\(KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}\)

→ As the kinetic energy of the system decreases after the collision, it is not an elastic collision.

Show Worked Solution

a.i.  The gravitational force down the slope:

\(F\) \(=mg\, \sin \theta\)  
  \(=2.0 \times 9.8 \times \sin 25\)  
  \(=8.3\ \text{N}\)   

 

a.ii.    \(F_{net}\) \(=ma\)
    \(=2.0 \times 3.2\)
    \(=6.4\)

 

\(6.4\) \(=F-F_f\)  
\(F_f\) \(=8.3-6.4\)  
  \(=1.9\ \text{N}\)  
♦ Mean mark (a.ii) 53%.

 
b.i.   By the conservation of momentum:

\(m_1u_1+m_2u_2\) \(=v(m_1+m_2)\)  
\(2 \times 4 + 2 \times 0\) \(=v(2 +2)\)  
\(8\) \(=4v\)  
\(v\) \(=2\ \text{ms}^{-1}\)  
 

 b.ii.  For the collision to be elastic, the kinetic energy must be conserved.

\(KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}\)

\(KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}\)

→ As the kinetic energy of the system decreases after the collision, it is not an elastic collision.

PHYSICS, M2 2022 VCE 6-7 MC

A railway truck \(\text{(X)}\) of mass 10 tonnes, moving at 3.0 m s\(^{-1}\), collides with a stationary railway truck \(\text{(Y)}\), as shown in the diagram below.

After the collision, they are joined together and move off at speed  \(v= 2.0\ \text{m s}^{-1}\).
 


 

\(\text{Question 6}\)

Which one of the following is closest to the mass of railway truck \(\text{Y}\)?

  1. 3 tonnes
  2. 5 tonnes
  3. 6.7 tonnes
  4. 15 tonnes

 
\(\text{Question 7}\)

Which one of the following best describes the force exerted by the railway truck \(\text{X}\) on the railway truck \(\text{Y} \left(F_{\text { X on Y}}\right)\) and the force exerted by the railway truck \(\text{Y}\) on the railway truck \(\text{X} \left(F_{\text {Y on X}}\right)\) at the instant of collision?

  1. \(F_{  \text { X on Y }  }<F_{ \text {Y on X}  }\)
  2. \(F_{ \text { X on Y }  }=F_{ \text { Y on X}  }\)
  3. \(F_{ \text { X on Y }  }=-F_{  \text { Y on X}  }\)
  4. \(F_{  \text {X on Y } }>F_{ \text {Y on X}  }\)
Show Answers Only

\(\text{Question 6:}\ B\)

\(\text{Question 7:}\ C\)

Show Worked Solution

\(\text{Question 6}\)

→ By the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision.

\(m_Xu_X+m_Yu_Y\) \(=m_Xv_X+m_Yv_Y\)  
  \(=v(m_X + m_Y)\ \ \ (v_X=v_Y) \)  
\(10000 \times 3 +0\) \(= 2(10000 + m_Y)\)  
\(15000\) \(=10000 +m_Y\)  
\(m_Y\) \(=5000\ \text{kg}\)   
  \(=5\ \text{tonnes}\)   

\( \Rightarrow B\)

 
\(\text{Question 7}\)

→ By Newton’s 3rd law of motion, each action has an equal and opposite reaction. 

→ Hence, the force of \(F_{ \text { X on Y }  }\) is equal in magnitude to \(F_{  \text { Y on X}  }\) but opposite in direction which is indicated by the negative sign (–).

\( \Rightarrow C\)

♦ Mean mark (Q7) 47%.

PHYSICS, M7 2016 HSC 28

The following diagram shows the acceleration of a rocket during the first stage of its launch.
 

Explain the acceleration of the rocket with reference to the law of conservation of momentum.   (5 marks)

Show Answers Only

→ The combustion of fuel and expulsion of gases causes propulsion of the rocket. The rocket exerts a force on these gases in order to expel them backwards.

→ An equal and opposite force is exerted back on the rocket by the gases (Newton’s Third Law).

→ As the rocket and gases form a closed system, the law of conservation of momentum applies, meaning the momentum of the rocket is equal in magnitude and opposite in direction to the momentum of the fuel.

→ The momentum of the rocket is given by  `p=mv`. Since the force (and hence momentum) of the fuel is constant, the velocity of the rocket must increase as its mass decreases due to the expulsion of its fuel.

→ The rocket will accelerate. Also, using  `F=ma`, as the force acting on the rocket due to the gases is constant, the acceleration of the rocket must increase as its mass decreases.

Show Worked Solution

→ The combustion of fuel and expulsion of gases causes propulsion of the rocket. The rocket exerts a force on these gases in order to expel them backwards.

→ An equal and opposite force is exerted back on the rocket by the gases (Newton’s Third Law).

→ As the rocket and gases form a closed system, the law of conservation of momentum applies, meaning the momentum of the rocket is equal in magnitude and opposite in direction to the momentum of the fuel.

→ The momentum of the rocket is given by  `p=mv`. Since the force (and hence momentum) of the fuel is constant, the velocity of the rocket must increase as its mass decreases due to the expulsion of its fuel.

→ The rocket will accelerate. Also, using  `F=ma`, as the force acting on the rocket due to the gases is constant, the acceleration of the rocket must increase as its mass decreases.


♦ Mean mark 52%.