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Mechanics, EXT2 M1 2022 HSC 12c

A particle with mass 1 kg is moving along the `x`-axis. Initially, the particle is at the origin and has speed `u` m s-1 to the right. The particle experiences a resistive force of magnitude  `v+3 v^2`  newtons, where `v` m s-1 is the speed of the particle after `t` seconds. The particle is never at rest.

  1. Show that  `(dv)/(dx)=-(1+3v)`   (1 mark)

  2. Hence, or otherwise, find `x` as a function of `v`.   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `x=1/3 ln((1+3u)/(1+3v))`
Show Worked Solution

i.    `F=m ddotx=ddotx\ \ (m=1)`

`ddotx` `=-(v+3v^2)`  
`v*(dv)/(dx)` `=-(v+3v^2)`  
`:. (dv)/(dx)` `=-(1+3v)\ \ text{… as required}`  

  

ii.     `(dx)/(dv)` `=- 1/(1+3v)`
  `x` `=-int1/(1+3v)\ dv`
    `=-1/3 ln(1+3v)+c`

 
`text{When}\ \ t=0, \ v=u\ \ =>\ \ c=1/3 ln(1+3u)`

`:.x` `=1/3 ln(1+3u)-1/3 ln(1+3v)`  
  `=1/3 ln((1+3u)/(1+3v))`  

Mechanics, EXT2 M1 2014 HSC 14c

A high speed train of mass `m` starts from rest and moves along a straight track. At time `t` hours, the distance travelled by the train from its starting point is `x` km, and its velocity is `v` km/h.

The train is driven by a constant force `F` in the forward direction. The resistive force in the opposite direction is `Kv^2`, where `K` is a positive constant. The terminal velocity of the train is 300 km/h.

  1. Show that the equation of motion for the train is
  2.     `m ddot x = F[1-(v/300)^2]`.   (2 marks)

  3. Find, in terms of `F` and `m`, the time it takes the train to reach a velocity of 200 km/h.   (4 marks)

Show Answers Only

a.    `text{Proof (See Worked Solutions)}`

b.    `(150 m ln5)/F`

Show Worked Solution

a.    `m ddot x = F-Kv^2,\ \ v_T = 300`

`text(At)\ \ v_T,\ \ ddotx=0:`

`m(0)=F-K(300^2)\ \ =>\ \ K=F/300^2`

`:. m ddot x= F-F/300^2 v^2=F[1-(v/300)^2]\ \ \ text(… as required)`
 

b.     `m*(dv)/(dt)` `= F[1-(v/(300))^2]`
  `(dv)/(1-(v/300)^2)` `=F/m\ dt`
  `(dv)/(300^2-v^2)` `=F/(300^2 m)\ dt`
  `dt` `=(300^2 m)/F xx (dv)/(300^2-v^2)`

 

`int_0^t dt` `=(300 m)/F  int_0^200  300/((300+v)(300-v))\ dv` 
 `:. t` `=(300 m)/F  int_0^200  (1/2)/(300+v) + (1/2)/(300-v)\ dv`
  `=(150 m)/F [ln(300+v)-ln(300-v)]_0^200`
  `=(150 m)/F [ln ((300+v)/(300-v))]_0^200`
  `=(150 m)/F  (ln5-ln1)`
  `=(150 m ln5)/F`