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Mechanics, EXT2 M1 2022 HSC 10 MC

A particle is moving vertically in a resistive medium under the influence of gravity. The resistive force is proportional to the velocity of the particle.

The initial speed of the particle is NOT zero.

Which of the following statements about the motion of the particle is always true?

  1. If the particle is initially moving downwards, then its speed will increase.
  2. If the particle is initially moving downwards, then its speed will decrease.
  3. If the particle is initially moving upwards, then its speed will eventually approach a terminal speed.
  4. If the particle is initially moving upwards, then its speed will not eventually approach a terminal speed.
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`C`

Show Worked Solution

`text{Case 1: particle moving downwards}`

`ddotx=g-kv\ \ (k>0)`

`text{Terminal velocity occurs when}\ \ ddotx=0\ \ =>\ \ v=g/k`

`text{Whether the particle’s speed increases, decreases or stays}`

`text{constant depends on whether}\ \ v_o<=g/k.`

`→\ text{Eliminate A and B.}`
 


♦ Mean mark 42%.

`text{Case 2: particle moving upwards}`

`ddotx=-g-kv\ \ (k>0)`

`text{→ Acceleration of gravity and resistance against motion}`

`text{→ Particle will eventually hit a peak and then move downwards}`

`text{→ Once moving downwards}\ \ ddotx=g-kv\ \ (k>0)`

`text{→ Particle will hit terminal velocity (see Case 1)}`

`=>C`

Mechanics, EXT2 M1 2019 HSC 14b

A parachutist jumps from a plane, falls freely for a short time and then opens the parachute. Let t be the time in seconds after the parachute opens, `x(t)`  be the distance in metres travelled after the parachute opens, and  `v(t)`  be the velocity of the parachutist in `text(ms)^(-1)`.

The acceleration of the parachutist after the parachute opens is given by

`ddot x = g-kv,`

where `g\ text(ms)^(-2)` is the acceleration due to gravity and `k` is a positive constant.

  1. With an open parachute the parachutist has a terminal velocity of  `w\ text(ms)^(-1)`.
  2. Show that  `w = g/k`.   (1 mark)

  3. At the time the parachute opens, the speed of descent is `1.6 w\ text(ms)^(-1)`.
  4. Show that it takes `1/k log_e 6` seconds to slow down to a speed of `1.1w\ text(ms)^(-1)`.   (4 marks)

  5. Let  `D`  be the distance the parachutist travels between opening the parachute and reaching the speed `1.1w\ text(ms)^(-1)`.
  6. Show that  `D = g/k^2 (1/2 + log_e 6)`.   (3 marks)

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

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

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

Show Worked Solution

i.    `v_T=w\ \  text(when)\ \ ddot x = 0`

`0` `= g-kw`
`w` `= g/k`

 

ii.    `text(Show)\ \ t = 1/k log_e 6\ \ text(when)\ \ v = 1.1w`

`(dv)/(dt)` `= g-kv`
`(dt)/(dv)` `= 1/(g-kv)`
`t` `= int 1/(g-kv)\ dv`
  `= -1/k ln(g-kv) + C`

 
`text(When)\ \ t = 0,\ \ v = 1.6w`

`0` `= -1/k ln(g-1.6 kw) + C`
`C` `= 1/k ln(g-1.6 kw)`
`t` `= 1/k ln (g-1.6kw)-1/k ln(g-kv)`
  `= 1/k ln((g-1.6 kw)/(g-kv))`

 
`text(Find)\ t\ text(when)\ \ v = 1.1w`

`t` `= 1/k ln((g-1.6 k xx g/k)/(g-1.1k xx g/k))`
  `=1/k ln((g-1.6 g)/(g-1.1g))`
  `=1/k((-0.6g)/(-0.1g))`
  `= 1/k ln 6`

 

iii.    `v ⋅ (dv)/(dx)` `= g-kv`
  `(dv)/(dx)` `= (g-kv)/v`
  `(dx)/(dv)` `= v/(g-kv)`
  `x` `= int v/(g-kv)\ dv`
    `= 1/k int (kv)/(g-kv)\ dv`
    `= -1/k int 1-g/(g-kv)\ dv`

 

`:. D` `= -1/k int_(1.6w)^(1.1w) 1-g/(g-kv)\ dv`
  `= 1/k int_(1.1w)^(1.6w) 1-g/(g-kv)\ dv`
  `= 1/k[v + g/k ln (g-kv)]_(1.1w)^(1.6w)`
  `= g/k^2[(kv)/g + ln (g-kv)]_(1.1w)^(1.6w)`
  `= g/k^2[((1.6kw)/g + ln (g-1.6kw))-((1.1 kw)/g + ln (g-1.1kw))]`
  `= g/k^2[1.6 + ln ((g-1.6kw)/(g-1.1kw))-1.1]`
  `= g/k^2(0.5 + ln 6)`