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Mechanics, EXT2 M1 2015 HSC 15a

A particle `A` of unit mass travels horizontally through a viscous medium. When  `t = 0`, the particle is at point `O` with initial speed `u`. The resistance on particle `A` due to the medium is `kv^2`, where `v` is the velocity of the particle at time `t` and `k` is a positive constant.

When  `t = 0`, a second particle `B` of equal mass is projected vertically upwards from `O` with the same initial speed `u` through the same medium. It experiences both a gravitational force and a resistance due to the medium. The resistance on particle `B` is `kw^2`, where `w` is the velocity of the particle `B` at time `t`. The acceleration due to gravity is `g`.

  1. Show that the velocity `v` of particle `A` is given by  
  2.     `1/v = kt + 1/u.`   (2 marks)

  3. By considering the velocity  `w`  of particle  `B`, show that
  4.     `t = 1/sqrt(gk) (tan^-1(u sqrt(k/g))-tan^-1 (w sqrt(k/g))).`   (3 marks)

  5. Show that the velocity  `V`  of particle  `A`  when particle  `B`  is at rest is given by
  6.     `1/V = 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g)).`   (1 mark)

  7. Hence, if  `u`  is very large, explain why  
  8.     `V ~~ 2/pi sqrt(g/k).`   (1 mark)

Show Answers Only

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

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

d.    `text(See Worked Solutions)`

Show Worked Solution

a.    `text(Particle)\ A:`

`ddot x` `= -kv^2`
`(dv)/(dt)` `= -kv^2`
`(dt)/(dv)` `=- 1/(kv^2)`
`t` `= -1/k int 1/v^2\ dv`
`-kt` `= -1/v + c`

 
`text(When)\ \ t=0,\ \ v=u\ \ => \ \ c=1/u`

`-kt` `= -1/v + 1/u`
`:.1/v` `= kt + 1/u`

 
b.    `text(Particle)\ B:`

`ddot x` `= -g-kw^2`
`(dw)/(dt)` `= -g-kw^2`
`(dt)/(dw)`   `=-1/(g + kw^2)`
`t`   `=-int (dw)/(g + kw^2)`
   `= -1/k int (dw)/(g/k + w^2)`
   `= -1/k xx 1/sqrt (g/k) tan^-1(w/sqrt (g/k)) + c`
  `= -1/sqrt (gk)\ tan^-1 ((sqrt k w)/sqrt g) + c`

 

`text(When)\ \ t = 0,\ w = u:`

`=>c= 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`

`:. t` `= -1/sqrt (gk) tan^-1 ((sqrt k w)/sqrt g) + 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`
  `=1/sqrt (gk) (tan^-1 (u sqrt(k/g))-1/sqrt (gk) tan^-1 (w sqrt (k/g)))`

 

c.    `B\ \ text(at rest when)\ \ w = 0`

`t = 1/sqrt (gk) (tan^-1 (u sqrt (k/g)))`

`:.1/V` `= k xx 1/sqrt(gk) tan^-1 (u sqrt (k/g)) + 1/u,\ \ \ \ text{(part (a))}`
  `= 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g))`

 

d.    `1/V = 1/u + sqrt (k/g) tan^-1 (u sqrt (k/g))`

`text(As)\ \ u -> oo,\ \ tan^-1 (u sqrt (k/g)) -> pi/2`

`:.\ text(If)\ u\ text(is very large,)`

`1/V` `~~ 0 + sqrt (k/g) xx pi/2`
`:.V` `~~ 2/pi sqrt (g/k)`