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PHYSICS, M5 EQ-Bank 2 MC

The horizontal and vertical components of the velocity of a projectile are respectively `v_x` and `v_y`.

Which pair of graphs best represents the velocity of the projectile?
 

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`B`

Show Worked Solution

→ There are no horizontal forces on a projectile so its horizontal velocity will remain constant.

→ The projectile experiences a constant downwards vertical acceleration due to gravity. So, its vertical velocity will decrease at a constant rate.

`=>B`

PHYSICS, M5 2018 HSC 27

  1. The diagram shows a camera and a ruler set up to obtain data about a projectile's motion along the trajectory shown. The entire trajectory is visible through the camera.
     
     
       
     
    Identify ONE of the errors in this set-up and describe the effect of this error on the results.   (3 marks)
  1. An experiment was set up based on the method described in part (a), but conducted so that the data obtained were valid.
     
    The image shows the trajectory of the ball.
     
       
     
    The graphs show data from this experiment.
     
       

    Using the graphs, describe the velocity and acceleration of the ball quantitatively and qualitatively.   (3 marks)

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a.   Experimental error:

→ The distance between the ruler and the camera is less than the distance between the trajectory and the camera.

→ The effect of this is that the distance measured on the ruler will be greater than the true distance the projectile travels. So, the calculated velocity will be inaccurate as it will be greater than the true velocity.

Other errors include:

→ The camera and the ruler are off centre.

→ The effect of this is also that the distance measured on the ruler will be inaccurate.
 

b.    → Ball is moving horizontally at a constant velocity (1st graph).

→ Quantitatively: `v=(-1.1-(-0.3))/(1-0.5)=-1.6  text{m s}^(-1).` So, the ball’s speed is 1.6 `text{m s}^(-1)`

→ The second graph shows that the ball is accelerating at a constant rate vertically downwards.

→ Quantitatively: `a=(-2.4-2.4)/(0.5)=-9.6\ text{m s}^(-2)`

Show Worked Solution

a.   Experimental error:

→ The distance between the ruler and the camera is less than the distance between the trajectory and the camera.

→ The effect of this is that the distance measured on the ruler will be greater than the true distance the projectile travels. So, the calculated velocity will be inaccurate as it will be greater than the true velocity.

Other errors include:

→ The camera and the ruler are off centre.

→ The effect of this is also that the distance measured on the ruler will be inaccurate.
 


♦ Mean mark (a) 45%.

b.    → Ball is moving horizontally at a constant velocity (1st graph).

→ Quantitatively: `v=(-1.1-(-0.3))/(1-0.5)=-1.6  text{m s}^(-1).` So, the ball’s speed is 1.6 `text{m s}^(-1)`

→ The second graph shows that the ball is accelerating at a constant rate vertically downwards.

→ Quantitatively: `a=(-2.4-2.4)/(0.5)=-9.6\ text{m s}^(-2)`


♦ Mean mark (b) 42%.

PHYSICS, M5 2022 HSC 11 MC

A projectile is launched vertically upwards. The displacement of the projectile as a function of time is shown.
 


 

Which velocity-time graph corresponds to this motion?
 

 

 

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`B`

Show Worked Solution

By elimination:

Initially, the projectile is moving upwards (positive velocity)

→ Eliminate`A`

At  `t=2` s, the projectile reaches its maximum height and has a velocity of zero.

→ Eliminate `D`

Projectile then moves downwards (negative velocity)

→ Eliminate `C`

`=>B`

PHYSICS, M5 2020 HSC 24

The graph shows the vertical displacement of a projectile throughout its trajectory. The range of the projectile is 130 m.
 

Calculate the initial velocity of the projectile.   (4 marks)

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37  m `text{s}^(-1)`, at 54° above the horizontal.

Show Worked Solution

From the graph, at `t=3`,  the projectile reaches a maximum height of 44 m:

`s_(y)` `=u_(y)t+(1)/(2)a_(y)t^(2)`  
`44` `=u_(y)(3)-(1)/(2)(9.8)(3^(2))`  
`u_(y)` `=29.4` m `text {s}^(-1)`  

  
Find `u_x` given time of flight = 6 s:

`u_(x)` `=(s_(x))/(t)`  
  `=(130)/(6)`  
  `=21.7` m `text{s}^(-1)`  

   
Using Pythagoras:

`u^(2)` `=u_(x)^(2)+u_(y)^(2)`  
  `=21.7^(2)+29.4^(2)`  
`u` `=37` m `text{s}^(-1)`  

  
Find launch angle (`theta)`:

`tan theta` `=(u_y)/(u_x)`  
`theta` `=54^(@)`  

  
So,  `u=37` m `text{s}^(-1)`, at 54° above the horizontal.