SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

PHYSICS, M2 EQ-Bank 5

Two identical 1200 kg cars travelling at 72 kmh\(^{-1}\) in opposite directions collide head-on and then rebound with exactly half of their original speeds.

  1. Is this collision elastic or inelastic? Give your reasons.   (2 marks)
  1. Given that the cars were in contact with each other for a total of 0.5 seconds, calculate the average force that acts on each car whilst they are in contact.   (2 marks)
Show Answers Only

a.    A collision will be elastic if kinetic energy is conserved in the reaction.

  \(KE_i = \dfrac{1}{2} \times 1200 \times 20^2 + \dfrac{1}{2} \times 1200 \times 20^2 = 480\ \text{kJ}\)

  \(KE_f = \dfrac{1}{2} \times 1200 \times 10^2 + \dfrac{1}{2} \times 1200 \times 10^2 = 120\ \text{kJ}\)

  • As kinetic energy is not conserved in the collision, the collision is inelastic.
     

b.    There is 72 kN acting on each car in the direction of their change in motion.

Show Worked Solution

a.    A collision will be elastic if kinetic energy is conserved in the reaction.

  \(KE_i = \dfrac{1}{2} \times 1200 \times 20^2 + \dfrac{1}{2} \times 1200 \times 20^2 = 480\ \text{kJ}\)

  \(KE_f = \dfrac{1}{2} \times 1200 \times 10^2 + \dfrac{1}{2} \times 1200 \times 10^2 = 120\ \text{kJ}\)

  • As kinetic energy is not conserved in the collision, the collision is inelastic.
     

b.    Using the impulse equation, \(I= \Delta p = F \Delta t\):

\(F\) \(=\dfrac{\Delta p}{\Delta t}\)  
  \(=\dfrac{m \Delta v}{\Delta t}\)  
  \(=\dfrac{1200 \times (10-(-20))}{0.5}\)  
  \(=-72\,000\ \text{N}\)  
     
  • There is 72 kN acting on each car in the direction of their change in motion.

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 2017 VCE 12

Students are using two trolleys, Trolley \(\text{A}\) of mass 4.0 kg and Trolley \(\text{B}\) of mass 2.0 kg, to investigate kinetic energy and momentum in collisions.

Before the collision, Trolley \(\text{A}\) is moving to the right at 5.0 m s\(^{-1}\) and Trolley \(\text{B}\) is moving to the right at 2.0 m s\(^{-1}\), as shown in Diagram A. The trolleys collide and lock together, as shown in Diagram B.
 

Determine, using calculations, whether the collision is elastic or inelastic. Show your working and justify your answer.   (3 marks)

Show Answers Only

By the conservation of momentum:

\(m_Au_A+m_Bu_B\) \(=v(m_A+m_B)\)  
\(4 \times 5 + 2 \times 2\) \(=v(4 +2)\)  
\(24\) \(=6v\)  
\(v\) \(=4\ \text{ms}^{-1}\)  
 

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

\(KE_{\text{init}}=\dfrac{1}{2} m_Au_A^2+ \dfrac{1}{2} m_Bu_B^2 =\dfrac{1}{2} \times 4 \times 5^2+\times \dfrac{1}{2} \times 2 \times 2^2=54\ \text{J}\)

\(KE_{\text{final}}=\dfrac{1}{2} (m_Am_B)v^2=\dfrac{1}{2} \times (4+2) \times 4^2=48\ \text{J}\)

\(\therefore\) Since the kinetic energy of the system decreases after the collision, the collision is inelastic.

Show Worked Solution

By the conservation of momentum:

\(m_Au_A+m_Bu_B\) \(=v(m_A+m_B)\)  
\(4 \times 5 + 2 \times 2\) \(=v(4 +2)\)  
\(24\) \(=6v\)  
\(v\) \(=4\ \text{ms}^{-1}\)  
 

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

\(KE_{\text{init}}=\dfrac{1}{2} m_Au_A^2+ \dfrac{1}{2} m_Bu_B^2 =\dfrac{1}{2} \times 4 \times 5^2+\times \dfrac{1}{2} \times 2 \times 2^2=54\ \text{J}\)

\(KE_{\text{final}}=\dfrac{1}{2} (m_A + m_B)v^2=\dfrac{1}{2} \times (4+2) \times 4^2=48\ \text{J}\)

\(\therefore\) Since the kinetic energy of the system decreases after the collision, the collision is inelastic.

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.