Space Time & Motion A.2 Forces & Momentum (id: 864e30ed8)

本题目来源于试卷: Space Time & Motion A.2 Forces & Momentum，类别为 IB物理学

[填空题]
A ball of mass 1.6kg is dropped from a very high building. The only forces acting on the ball are the force of gravity and air resistance. The graph shows the variation of the vertical velocity v of the ball with time t.

• 
  
  • Define terminal speed.
    1
  • State the terminal speed of the ball.   $ms^{-1}$
    1
    
• Estimate the vertical displacement of the ball during the t=2.0s  $ms^{-2$}
  2
• Determine the magnitude of the vertical component of air resistance at t=2.0s  N
  3
  

参考答案:
空格1: 18  空格2: 2.8±17%  空格3: 11 


本题详细解析:

1. 1. The maximum speed/velocity obtained by an object during motion in a fluid/gas/air/liquid $\mathrm{✓}$

   2. 18
      $\ll m{s}^{-1}\gg$ $\mathrm{✓}$

      $\underset{‾}{\text{Further Explanation}}$: 
      According to the graph, the velocity of the ball increases and then stays constant at $18m{s}^{-1}$.

2. Any reference to the area under the graph being equal to the displacement. $\mathrm{✓}$

   $125\pm 10$
   $\ll m,downwards\gg$ $\mathrm{✓}$

   $\underset{‾}{\text{Further Explanation}}$: 
   The area under the velocity-time graph is equal to the ball's displacement. The area can be calculated by using the small square approximation. The area $A_s$ of each small square:

   • $A_s=(0.2)(1)=0.2m$

   There are approximately 625 small squares. Therefore, the total area A or displacement:

   • $A=(625)(0.2)=125m$

   Note that displacement is a vector quantity and so a direction should be stated. In this case, the ball is falling so the direction of the displacement is downwards.

3. 1. Reference to use of the gradient of the graph to find instantaneous acceleration. $\mathrm{✓}$

      $2.8\pm 0.5$
      $\ll m{s}^{-2}\gg$ $\mathrm{✓}$

      $\underset{‾}{\text{Further Explanation}}$: 
      The gradient of the graph at a given time gives the instantaneous acceleration of the car. To find the gradient, the tangent line should be drawn.

      

      
      

      
      

      

      
      

      • $F_{net}$

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