smc-1179-25-Momentum

Vectors, SPEC1 2021 VCAA 1

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

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

Show Answers Only

`1/2 underset~i + 6/5 underset~j`
`1/2 t underset~i + (6/5 t – 3) underset~j`
`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

0.04
0.06
0.10
0.24
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

`3 underset ~i - underset ~j`
`2 underset ~i - 12 underset ~j`
`4 underset ~i - 7 underset ~j`
`2 underset ~i + 5 underset ~j`
`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`

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)`