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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 Figure 10a. The trolleys collide and lock together, as shown in Figure 10b.
 

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

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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.

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

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

→ As 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.

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

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

→ As 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 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)

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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)
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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.