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PHYSICS, M2 EQ-Bank 3

The diagram below shows a top view of two objects moving across a frictionless surface. Object \(X\) has a mass of 3.5 kg, and object \(Y\) has a mass of 2.5 kg. Their initial velocities are indicated in the diagram.
 

  

The two objects collide and stick together after the collision.

  1. Calculate the velocity of the combined mass after the collision.   (4 marks)
  1. Determine the amount of kinetic energy lost due to the collision.   (2 marks)
Show Answers Only

a.    \(2.91\ \text{ms}^{-1}, \ \text{S}57.8^{\circ}\text{E}\).

b.    \(16.21\ \text{J}\).

Show Worked Solution

a.    The momentum of each object can be seen in the diagrams below:
 

  • By letting east and south be the positive directions we can sum the vectors by their horizontal and vertical components:

\(p_h = 14\cos 60 + 8.25\cos 20 = 14.75\ \text{Ns}\)

\(p_v = 14\sin 60-8.25\sin 20 = 9.30\ \text{Ns}\)
 

  • The total momentum of the system after the collision can be represented in the diagram below:
     

\(p_T^2=9.30^2 + 14.75^2=\sqrt{304.0525} = 17.44\ \text{Ns}\)
 

  • Total mass of the system after the collision = 6 kg
  • The speed after the collision \(=\dfrac{p_T}{m} = \dfrac{17.44}{6} = 2.91\ \text{ms}^{-1}\).
\(\tan \theta \) \(=\dfrac{9.30}{14.75}\)  
\(\theta\) \(=\tan^{-1}\left(\dfrac{9.30}{14.75}\right)=32.2^{\circ}\)  

  
\(\therefore\) The velocity of the combined mass after the collision is \(2.91\ \text{ms}^{-1}\), \(\text{S}57.8^{\circ}\text{E}\).

 
b.    The kinetic energy before the collision:

\(KE_i = \dfrac{1}{2} \times 3.5 \times 4^2 + \dfrac{1}{2} \times 2.5 \times 3.3^2 = 41.6125\ \text{J}\)

The kinetic energy after the collision:

\(KE_f = \dfrac{1}{2} \times 6 \times 2.91^2 = 25.4043\ \text{J}\)

\(\Delta KE = 25.4043-41.6125 = -16.21\ \text{J}\)

\(\therefore\) The kinetic energy lost in the collision was \(16.21\ \text{J}\).

PHYSICS, M2 EQ-Bank 2

A 2.00 kg block is moving right at 0.30 m/s. A second 0.40 kg block is moving in the opposite direction at 0.10 m/s, resulting in a collision.

After the collision, both blocks stick together and move to the right.
 

  1. Calculate the speed of the two blocks after the collision.   (2 marks)
  1. Determine the impulse that the smaller block applies to the larger block.   (2 marks)
Show Answers Only

a.    \(0.233\ \text{ms}^{-1}\)

b.    \(0.134\ \text{kg m s}^{-1}\) to the left.

Show Worked Solution

a.    Using the conservation of energy:

\(m_1u_1 + m_2u_2\) \(=v(m_1 + m_2)\)  
\(2 \times 0.3 + 0.4 \times -0.1\) \(=v \times (2 + 0.4)\)  
\(2.4v\) \(=0.56\)  
\(v\) \(=0.233\ \text{ms}^{-1}\)  

 

b.   Impulse can be calculated using: \(I = \Delta p = m_1v-m_1u_1\):

  •    \(I =2 \times 0.233- \times 0.3 = -0.134\ \text{kgms}^{-1}\)
  • The smaller block applies an impulse of \(0.134\ \text{kg m s}^{-1}\) on the larger block to the left.

PHYSICS, M2 EQ-Bank 4 MC

A white ice hockey puck of mass \(m\) with an initial speed \(u\) collides with a stationary black ice hockey puck also of mass \(m\). After the collision, the black puck moves off with speed \(v\).
 

The collision is elastic.

What is the speed of the white puck after the collision?

  1. \(0\)
  2. \(u\)
  3. \(v\)
  4. \(\dfrac{v}{2}\)
Show Answers Only

\(A\)

Show Worked Solution
  • Let the speed of the white puck after the collision be \(x\).
  • By the law of conservation of momentum: 
\(mu\) \(=mv + mx\)  
\(mu\) \(=m(v+x)\)  
\(u\) \(=v + x\ …\ (1)\)  
     
  •  As the collision is elastic, the kinetic energy of the system will be conserved during the collision:
\(\dfrac{1}{2}mu^2\) \(=\dfrac{1}{2}mv^2 + \dfrac{1}{2}mx^2\)  
\(\dfrac{1}{2}mu^2\) \(=\dfrac{1}{2}m(v^2 + u^2)\)  
\(u^2\) \(=v^2 + x^2\ …\ (2)\)  

  
Substitute equation (1) into equation (2):

\((v+x)^2\) \(=v^2+x^2\)  
\(v^2 +2vx +x^2\) \(=v^2 + x^2\)  
\(2vx\) \(=0\)  
     
  • The speed of the white puck after the collision \((x)\) must be \(0\) as \(v \neq 0\).

\(\Rightarrow A\)

PHYSICS, M2 EQ-Bank 1

Jordan is trying to push a heavy filing cabinet across the office floor, but it’s not budging. The cabinet has been used by physics students, and someone has stuck a post-it note on it with a “reminder” that reads:

“According to Newton’s third law, the cabinet pushes back with the same force you apply — so you’ll never move it!”

Jordan sighs and tries harder. Another note on the cabinet reads:

“Don’t bother! The law of conservation of momentum says that if the cabinet is at rest, its momentum must stay at zero forever.”

Identify and explain the two misunderstandings about Newton’s third law and the law of conservation of momentum. Use correct physics principles to explain how Jordan can, in fact, move the filing cabinet.    (5 marks)

Show Answers Only

Newton’s third law misconception:

  • Newton’s third law states that every action has an equal and opposite reaction, but these force pairs act on different objects.
  • Jordan pushes the cabinet, and the cabinet pushes back on him with equal force. These forces do not cancel because they act on separate objects.
  • The motion of the cabinet depends on the net force acting on it, according to Newton’s second law:  \(F_{\text{net}} = ma\).

 Conservation of momentum misconception:

  • Momentum is conserved only in a closed system with no external net forces.
  • Since Jordan exerts an external force on the cabinet and the floor provides friction, the cabinet is not a closed system.
  • Therefore, momentum conservation does not prevent Jordan from moving the cabinet.

 How Jordan can move the cabinet:

  • By bracing his feet, Jordan increases backward friction against the floor, allowing him to exert a greater forward force.
  • If this force exceeds static friction, the cabinet accelerates and its momentum changes:  \(\Delta p = F \Delta t\).
Show Worked Solution

Newton’s third law misconception:

  • Newton’s third law states that every action has an equal and opposite reaction, but these force pairs act on different objects.
  • Jordan pushes the cabinet, and the cabinet pushes back on him with equal force. These forces do not cancel because they act on separate objects.
  • The motion of the cabinet depends on the net force acting on it, according to Newton’s second law:  \(F_{\text{net}} = ma\).

 Conservation of momentum misconception:

  • Momentum is conserved only in a closed system with no external net forces.
  • Since Jordan exerts an external force on the cabinet and the floor provides friction, the cabinet is not a closed system.
  • Therefore, momentum conservation does not prevent Jordan from moving the cabinet.

 How Jordan can move the cabinet:

  • By bracing his feet, Jordan increases backward friction against the floor, allowing him to exert a greater forward force.
  • If this force exceeds static friction, the cabinet accelerates and its momentum changes:  \(\Delta p = F \Delta t\).

PHYSICS, M2 EQ-Bank 2-3 MC

Two identical blocks, each with a mass of 200 grams, are moving toward each other on a frictionless track at equal speeds.

Part 1

Which two identical physical quantities do the blocks share as they approach each other?

  1. Mass and velocity
  2. Mass and impulse
  3. Mass and momentum
  4. Mass and kinetic energy

 
Part 2

After the collision, the two blocks stick together and move as one object. What is the total final momentum of the combined gliders system?

  1. \(0.25\ \text{kgm/s}\)
  2. \(0\ \text{kgm/s}\)
  3. \(0.02\ \text{kgm/s}\)
  4. \(1.60\ \text{kgm/s}\)
Show Answers Only

\(\text{Part 1:}\ D\)

\(\text{Part 2:}\ B\)

Show Worked Solution

Part 1

  • Velocity, impulse and momentum are all vector quantities hence they will not be equal for the blocks as they will be occurring in the opposite direction.
  • Mass and kinetic energy are the only two scalar values.

\(\Rightarrow D\)
 

Part 2

  • The momentum’s of the two blocks will be equal and opposite and therefore the total momentum of the system sums together to be zero.
  • By the law of conservation of momentum, the final momentum of the system is zero.

\(\Rightarrow B\)

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 the figure below. The slope is 5° to the horizontal.

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

  1. The combined system of Maia and the skateboard is modelled as a small box with point \(\text{M}\) at the centre of mass, as shown below.
  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.
 

   

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. \( 6.0 \text{ m s} ^{-1}\)
     

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

Question 8

  • By the law of conservation of momentum:
\(m_Xu_X+m_Yu_Y\) \(=m_Xv_X+m_Yv_Y\)  
  \(=v(m_X + m_Y)\ \ \ (v_X=v_Y) \)  
\(10\ 000 \times 6 +0\) \(= v(10\ 000 + 5000)\)  
\(60\ 000\) \(=15\ 000v\)  
\(v\) \(=4\ \text{ms}^{-1}\)   

 
\( \Rightarrow C\)
 

Question 9

  • By the law of conservation of momentum, momentum in the collision is conserved.
  • Kinetic energy conservation:
       \(KE_{\text{init}}\) \(=\dfrac{1}{2}m_Xu_X^2+\dfrac{1}{2}m_Yu_Y^2\)
   

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

 

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

  
\(\therefore\) 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 the diagrams below.
 

  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 Δmomentum is 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 the diagram below. 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 \(3.2\ \text{m s}^{-2}\) 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 \(4.0\ \text{m s}^{-1}\) 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.

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

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

\(\therefore\) Since 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. \(F_{net}=ma=2.0 \times 3.2=6.4\ \text{N}\)

\(6.4\) \(=F-F_f\)  
\(F_f\) \(=8.3-6.4\)  
  \(=1.9\ \text{N}\)  
♦ 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.

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

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

\(\therefore\) Since 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  \(v= 2.0\ \text{m s}^{-1}\).
 


 

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

 
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

Question 6

  • By the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision.
\(m_Xu_X+m_Yu_Y\) \(=m_Xv_X+m_Yv_Y\)  
  \(=v(m_X + m_Y)\ \ \ (v_X=v_Y) \)  
\(10000 \times 3 +0\) \(= 2(10000 + m_Y)\)  
\(15000\) \(=10000 +m_Y\)  
\(m_Y\) \(=5000\ \text{kg}\)   
  \(=5\ \text{tonnes}\)   

\( \Rightarrow B\)

 
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, M5 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%.