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PHYSICS, M2 2023 VCE 4 MC

The diagram below shows the force versus time graph of the force on a tennis ball when it is hit by a tennis racquet. The tennis ball is stationary when the tennis racquet first comes into contact with the ball.
 

Which one of the following is closest to the impulse experienced by the tennis ball as it is hit by the tennis racquet?

  1. \(0.50 \text{ N s}\)
  2. \(5.0 \text{ N s}\)
  3. \(10 \text{ N s}\)
  4. \(50 \text{ N s}\)
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\(B\)

Show Worked Solution

→ The impulse is equal to the area under the curve \((I=Ft)\).

→ By splitting the areas into whole number of squares, there are about 10 squares.

→ Each square is \(50\ \text{N} \times 0.01\ \text{s}=0.5\ \text{N s}\)

→ Hence the impulse experienced by the tennis ball \(\approx 5.0\ \text{N s}\).

\(\Rightarrow B\)

♦ Mean mark 49%.

PHYSICS, M2 2017 VCE 8 MC

A model car of mass 2.0 kg is propelled from rest by a rocket motor that applies a constant horizontal force of 4.0 N, as shown below. Assume that friction is negligible.
 

Which one of the following best gives the magnitude of the impulse given to the car by the rocket motor in the first 5.0 seconds?

  1. \(4.0 \text{ N s}\)
  2. \(8.0 \text{ N s}\)
  3. \(20 \text{ N s}\)
  4. \(40 \text{ N s}\)
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\(C\)

Show Worked Solution
\(I\) \(=Ft\)  
  \(=4 \times 5\)  
  \(=20 \text{ N s}\)  

 

\(\Rightarrow C\)

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.