Space Time & Motion A.2 Forces & Momentum (id: 1f61c05e5)

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 cannonball is fired horizontally from a cannon positioned at the edge of a cliff.
1.In the context of Newton's Laws, explain why the cannon recoils to the left when the cannonball is fired.
2.The initial recoil speed of the cannon is 1.0$ms^{-1}$ and its mass is 1200 kg.
(1)The total of opposing forces acting on the cannon during the motion of the cannon is 8.0 kN. Determine the distance covered by the cannon after the firing.  m
(2)The mass of the cannonball is 9.0 kg. Calculate the magnitude of the firing velocity of the ball is approximately   $ms^{-1}$.
(3)Calculate the magnitude of the displacement of the cannonball after 5.0s. Air resistance can be ignored.  m

参考答案:
空格1: 0.074  空格2: 113  空格3: 677 


本题详细解析:

1. By Newton's 3rd law, there is a force to the right acting on the cannonball $\mathrm{✓}$

   There is an equal and opposite force (to the left) on the cannon $\mathrm{✓}$

   $\underset{‾}{\text{Further Explanation}}$: 
   There is a force acting on the cannonball during the firing. According to Newton's third law, there must be an equal and opposite force on the cannon.

   

   

   Any reference to the conservation of momentum

   $v_2=\frac{(-m_1)(v_1)}{m_2}=\frac{-(1200)(1.0)}{9.0}=\ll -133m{s}^{-1}\gg$ $\mathrm{✓}$

   
   

   $\underline{\text{Further Explanation}}$: 
   Due to the conservation of momentum, the total momentum of the objects before the firing should be equal to the total momentum of the objects after the firing.

   The total momentum of the objects before the firing is zero due to the absence of motion. Therefore there total momentum of the objects after firing is also zero

   

   The negative sign represents the direction of the velocity. Therefore it is opposite to the direction of the motion of the cannon.

   

   $\underline{\text{Further Explanation}}$: 
   Due to the absence of air resistance, the horizontal velocity of the ball is constant and vertical velocity increases with the acceleration of free fall. The horizontal displacement x of the cannonball

   • $x=vt$ where $v=133m{s}^{-1}$ and $t=3.0s$

   • $x=\left(133\right)\left(5.0\right)=666m$

   The vertical displacement y of the cannonball

   • $y=ut+\frac{1}{2}a{t}^2$

   where

   • $u=0$,
     $a=9.81m{s}^{-1}$
     $t=5.0s$

   • $y={\displaystyle \frac{1}{2}}\left(9.81\right)\left(5^2\right)=122m$

   The magnitude of the displacement s can be found by using Pythagoras's theorem;

   $s=\sqrt{x^2+y^2}$

   $s=\sqrt{{666^2}+{122^2}}$=677m

   
   

   $s=\sqrt{x^2+y^2}$

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