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Mechanics, EXT2 M1 2022 HSC 16b

A projectile of mass `M` kg is launched vertically upwards from a horizontal plane with initial speed `v_0\ text{m s}^(-1)` which is less than 100`\ text{m s}^(-1)`. The projectile experiences a resistive force which has magnitude `0.1 M v` newtons, where `v\ text{m s}^(-1)` is the speed of the projectile. The acceleration due to gravity is 10`\ text{m s}^(-2)`.

The projectile lands on the horizontal plane 7 seconds after launch.

Find the value of `v_0`, correct to 1 decimal place.   (4 marks)

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`39.1\ text{m s}^(-1)`

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`text{Motion is only vertical.}`

`Mddoty=M(-g-0.1v)`

`(dv)/(dt)` `=-(g+0.1v)`  
`(dt)/(dv)` `=- 1/(g+0.1v)`  
`t` `=-int1/(g+0.1v)\ dv=-10ln abs(g+0.1v)+c`  

 
`text{When}\ \ t=0, \ v=v_0\ \ =>\ \ c=10ln abs(g+0.1v_0)`


♦ Mean mark 45%.

`t10ln abs(g+0.1v_0)-10ln abs(g+0.1v)=10ln abs((g+0.1v_0)/(g+0.1v))`

`e^(t/10)=(g+0.1v_0)/(g+0.1v)`

`(g+0.1v)` `=(g+0.1v_0)*e^(- t/10)\ \ \ (text{note}\ (g+0.1v)>0\ text{as}\ v<100)`  
`0.1v` `=-g+(g+0.1v_0)*e^(- t/10)`  
`v` `=10[-g+(g+0.1v_0)*e^(- t/10)]`  

 
`y=intv\ dt=10[-g t-10(g+0.1v_0)*e^(- t/10)]+c`

`text{When}\ \ t=0, \ y=0\ \ =>\ \ c=100(g+0.1v_0)`

`:.y=10[-g t-10(g+0.1v_0)*e^(- t/10)]+100(g+0.1v_0)`
  

`text{Find}\ v_0,\ text{given}\ y=0\ text{at}\ t=7:`

`0` `=10[-70-10(10+0.1v_0)*e^(-0.7)]+100(10+0.1v_0)`  
`0` `=-700-100(10+0.1v_0)*e^(-0.7)+1000+10v_0`  
`0` `=-700-1000e^(-0.7)-10v_0e^(-0.7)+1000+10v_0`  
`0` `=300-1000e^(-0.7)+v_0(10-10e^(-0.7))`  
`:.v_0` `=(1000e^(-0.7)-300)/(10-10e^(-0.7))=39.0503…=39.1\ text{m s}^(-1)\ \ text{(1 d.p.)}`  

Mechanics, EXT2 M1 EQ-Bank 28

A canon ball of mass 9 kilograms is dropped from the top of a castle at a height of `h` metres above the ground.

The canon ball experiences a resistance force due to air resistance equivalent to  `(v^2)/500`, where `v` is the speed of the canon ball in metres per second. Let  `g=9.8\ text(ms)^-2`  and the displacement, `x` metres at time `t` seconds, be measured in a downward direction.

  1. Show the equation of motion is given by
  2. `ddotx = g-(v^2)/4500`   (1 mark)

  3. Show, by integrating using partial fractions, that
  4. `v = 210((e^(7/75 t)-1)/(e^(7/75 t) + 1))`   (5 marks)

  5. If the canon hits the ground after 4 seconds, calculate the height of the castle, to the nearest metre.   (3 marks)

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a.    `text(See Worked Solutions)`

b.    `text(See Worked Solutions)`

c.    `78\ text(metres)`

Show Worked Solution

a.    `text(Newton’s 2nd Law:)`

`text(Net Force)` `= mddotx`
`mddotx` `= mg-(v^2)/500`
`9ddotx` `= 9g-(v^2)/500`
`ddotx` `= g-(v^2)/4500`

 
b.    `(dv)/(dt)= g-(v^2)/4500= (44\ 100-v^2)/4500`

`(dt)/(dv) = 4500/(44\ 100-v^2)`
 

`text(Using Partial Fractions:)`

`1/(44\ 100-v^2) = A/(210- v) + B/(210-v)`

`A(210-v) + B(210 + v) = 1`
 

`text(If)\ \ v = 210,`

`420B = 1 \ => \ B = 1/420`
 

`text(If)\ \ v = −210,`

`420A = 1 \ => \ A = 1/420`
 

`t` `= int 4500/(210^2-v^2)\ dv`
  `= 4500/420 int 1/(210 + v) + 1/(210-v)\ dv`
  `= 75/7 [ln(210 + v)-ln(210-v)] + c`
  `= 75/7 ln((210 + v)/(210-v)) + c`

 

`text(When)\ \ t = 0, v = 0:`

`0= 75/7 ln(210/210) + c\ \ =>\ \ c=0`

` t` `= 75/7 ln((210 + v)/(210-v))`
`7/75 t` `= ln((210 + v)/(210-v))`
`e^(7/75 t)` `= (210 + v)/(210-v)`
`e^(7/75 t) (210-v)` `= 210 + v`
`210e^(7/75 t)-210` `= ve^(7/75 t) + v`
`210(e^(7/75 t)-1)` `= v(e^(7/75 t) + 1)`
`:. v` `= 210((e^(7/75 t)-1)/(e^(7/75 t) + 1))`

 

c.     `v · (dv)/(dx)` `= (44\ 100-v^2)/4500`
  `(dx)/(dv)` `= (4500v)/(44\ 100-v^2)`
  `int (dx)/(dv)\ dv` `= −4500/2 int (−2v)/(44\ 100-v^2)\ dv`
  `x` `= −2550 log_e(44\ 100-v^2) + c`

 
`text(When)\ \ x = 0, v = 0:`

`0` `= −2250 log_e(44\ 100) + c`
`c` `= 2250 log_e(44\ 100)`
`x` `= −2250 log_e(44\ 100-v^2) + 2250 log_e44\ 100`
  `= 2250 log_e((44\ 100)/(44\ 100-v^2))`

 
`text(When)\ \ t = 4:`

`v` `= 210((e^(28/75)-1)/(e^(28/75) + 1))`
  `= 38.7509…`

 

`:. h` `= 2250 log_e((44\ 100)/(44\ 100-38.751^2))`
  `= 77.94…`
  `= 78\ text(metres)`