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Vectors, SPEC1 2021 VCAA 1

The net force acting on a body of mass 10 kg is  `underset~F = 5underset~i + 12underset~j`  newtons.

  1. Find the acceleration of the body in `text(ms)^(-2)`.   (1 mark)

  2. The initial velocity of the body is  `-3underset~j\ \ text(ms)^(-1)`.
    Find the velocity of the body, in `text(ms)^(-1)`, at any time  `t`  seconds.   (2 marks)

  3. Find the momentum of the body, in kg `text(ms)^(-1)`, when  `t = 2`  seconds.   (1 mark)

Show Answers Only
  1. `1/2 underset~i + 6/5 underset~j`
  2. `1/2 t underset~i + (6/5 t-3) underset~j`
  3. `10 underset~i-6underset~j`
Show Worked Solution

a.   `text(Using)\ underset~F = m underset~a:`

`10underset~a` `= 5underset~i + 12underset~j`
`underset~a` `= 1/10 (5underset~i + 12underset~j)`
  `= 1/2 underset~i + 6/5 underset~j`

 

b.    `underset~v(t)` `= int underset~a\ dt`
    `= int 1/2 underset~i + 6/5 underset~j\ dt`
    `= 1/2 t underset~i + 6/5 t underset~j + c`

 
`text(When)\ t = 0, v(t) = -3underset~j`

`=> c = -3underset~j`

`underset~v(t)` `= 1/2 t underset~i + 6/5 t underset~j-3underset~j`
  `= 1/2 t underset~i + (6/5 t-3) underset~j`

 

c.    `underset~v(2) = underset~i-3/5 underset~j`

`underset~p(2)` `= m underset~v(2)`
  `= 10(underset~i-3/5 underset~j)`
  `= 10 underset~i-6underset~j`

Vectors, SPEC2 2020 VCAA 19 MC

A cricket ball of mass 0.02 kg, moving with velocity  `2underset~i - 10 underset~j\ \ text(ms)^(−1)`,  is hit and after impact travels with velocity  `2underset~i - 7underset~j\ \ text(ms) ^(−1)`.

The magnitude of the change in momentum of the cricket ball, in  `text(kg ms)^(−1)`, is closest to

  1. 0.04
  2. 0.06
  3. 0.10
  4. 0.24
  5. 0.34
Show Answers Only

`B`

Show Worked Solution
`Delta underset~p` `= m Delta underset~v`
  `= 0.02((2underset~i – 7underset~j) – (2underset~i – 10underset~j))`
  `= 0.02(3underset~j)`
  `= 0.06underset~j`

 
`|Delta underset~p| = 0.06\ text(kg ms)^(−1)`

`=>B`

Vectors, SPEC2-NHT 2017 VCAA 15 MC

A particle of mass 2 kg has an initial velocity of   `underset ~i - 6 underset ~j\ \ text(ms)^(-1)`.

After a change of momentum of  `6 underset ~i - 2 underset ~j\ \ text(kg ms)^(-1)`,  the particle's velocity, in `text(ms)^(-1)`,  is

  1. `3 underset ~i - underset ~j`
  2. `2 underset ~i - 12 underset ~j`
  3. `4 underset ~i - 7 underset ~j`
  4. `2 underset ~i + 5 underset ~j`
  5. `11 underset ~i + 2 underset ~j`
Show Answers Only

`C`

Show Worked Solution
`m underset ~(v_i)` `= 2(underset ~i – 6 underset ~j)`
`m Delta underset ~v` `= m underset ~(v_f) – m underset ~(v_i)`
  `= m(underset ~(v_f) – underset ~(v_i))`
`6 underset ~i – 2 underset ~j` `= 2(underset ~ (v_f )- (underset ~i – 6 underset ~j))`
`2 underset ~ (v_f )` `= 6 underset ~i – 2 underset ~j + 2 underset ~i + – 12 underset ~j`
  `= 8 underset ~i -14 underset ~j`
`:. underset ~ (v_f )` `= 4 underset ~i -7 underset ~j`

 

`=>   C`

Vectors, SPEC1 2018 VCAA 6

A particle of mass 2 kg moves under a force  `underset ~ F`  so that its position vector  `underset ~ r`  at anytime `t` is given by  `underset ~r = sin(t) underset ~i + cos(t) underset ~j + t^2 underset ~k`.  Distances are measured in metres and time is measured in seconds.

Find the change in momentum, in kg ms¯², from  `t = pi/2`  to  `t = pi`.   (3 marks)

Show Answers Only

`2 (- underset ~i + underset ~j +  pi underset ~k)`

Show Worked Solution
`underset ~ dot r (t)` `= cos(t) underset ~i-sin(t) underset ~j + 2t underset~k`
`underset ~ dot r (pi)` `= cos(pi) underset ~i-sin(pi) underset ~j + 2 pi underset ~k`
  `=-underset ~i + 2 pi underset ~k`
   
`underset ~ dot r (pi/2)` `= cos(pi/2) underset ~i-sin (pi/2) underset ~j + pi underset ~k`
  `=-underset ~j + pi underset ~k`
   
`:. m Delta r` `= 2(underset ~ dot r (pi)-underset ~ dot r (pi/2))`
  `= 2(- underset ~i + 2 pi underset ~k-(- underset ~j + pi underset ~k))`
  `= 2 (- underset ~i + underset ~j +  pi underset ~k)`