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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: kdgamz on September 17, 2009, 09:10:57 pm

Title: dynamics/kinematics questions
Post by: kdgamz on September 17, 2009, 09:10:57 pm: i cant seem to get my head around these two questions.. anyone?

Question1

A particle of mass 20 kg at rest is acted on by a force that decreases uniformly with the
distance travelled. The force is initially 100 N and 30 N after 20 metres. The speed of the
particle at this time is:

A 3√230 m/s
B 130 m/s
C 230 m/s
D √130 m/s
E 5√193 m/s

Question 2

An object of mass 10 kg is subjected to a variable force described by the expression
(100 − 2x) newtons, where x is the distance up a ramp that makes an angle of 10° with the
horizontal. Assume the ramp is smooth and that the object starts from rest from the bottom
of the ramp.
(a) Find the speed, correct to two decimal places, of the object
(i) 3 m up the ramp

thanks
Title: Re: dynamics/kinematics questions
Post by: Flaming_Arrow on September 17, 2009, 09:32:27 pm: is the answer D for q1?
Title: Re: dynamics/kinematics questions
Post by: kdgamz on September 17, 2009, 09:33:59 pm: yeh but how?
Title: Re: dynamics/kinematics questions
Post by: Flaming_Arrow on September 17, 2009, 09:44:21 pm: a= 5 when x=0 a=3/2 x=20

find the equation of acceleration in terms of x

$\frac{5-1.5}{0-20} = \frac{-7}{40}$

$a - 0 = -\frac{7}{4}(x-0)$

$a= -\frac{7x}{40} + 5$

we know that

$\frac{d}{dx} (\frac{v^2}{2}) = a}$

so

$\frac{d}{dx} (\frac{v^2}{2} ) = -\frac{7x}{40} + 5$

$\frac{v^2}{2} = \int -\frac{7x}{40} + 5 dx$

$\frac{v^2}{2} = -\frac{7x^2}{80} + 5x +c$

$c=0$

$v^2 = -\frac{7x^2}{40} + 10x$

$v = \sqrt{-\frac{7x^2}{40} + 10x}$

$v(20) = \sqrt{130} ms^{-1}$
Title: Re: dynamics/kinematics questions
Post by: Flaming_Arrow on September 17, 2009, 09:55:50 pm: whats the answer to 2a?
Title: Re: dynamics/kinematics questions
Post by: Mao on September 18, 2009, 08:33:24 am: Quote from: kdgamz on September 17, 2009, 09:10:57 pm
Question 2

An object of mass 10 kg is subjected to a variable force described by the expression
(100 − 2x) newtons, where x is the distance up a ramp that makes an angle of 10° with the
horizontal. Assume the ramp is smooth and that the object starts from rest from the bottom
of the ramp.
(a) Find the speed, correct to two decimal places, of the object
(i) 3 m up the ramp

I'm assuming the variable force points up the ramp

Then, $ma = \sum F_{net} = (100 - 2x) - mg \sin 10^o$

$\begin{align*}\implies a & = 10 - 0.2 x - g\sin 10^o\\ a & = v \frac{dv}{dx} \\ \implies v \frac{dv}{dx} &= 10 - 0.2 x - g\sin 10^o \\ \int_{0}^{v_1} v\; dv & = \int_0^3 10 - 0.2 x - g\sin 10^o \; dx \\ \frac{(v_1)^2}{2} & = 3(10 - g\sin 10^o) - 0.9 \\ v_1 & \approx 6.23\\ \end{align*}$
Title: Re: dynamics/kinematics questions
Post by: kdgamz on September 18, 2009, 11:04:53 am: Quote from: Flaming_Arrow on September 17, 2009, 09:44:21 pm
a= 5 when x=0 a=3/2 x=20

find the equation of acceleration in terms of x

$\frac{5-1.5}{0-20} = \frac{-7}{40}$

$a - 0 = -\frac{7}{4}(x-0)$

$a= -\frac{7x}{40} + 5$

we know that

$\frac{d}{dx} (\frac{v^2}{2}) = a}$

so

$\frac{d}{dx} (\frac{v^2}{2} ) = -\frac{7x}{40} + 5$

$\frac{v^2}{2} = \int -\frac{7x}{40} + 5 dx$

$\frac{v^2}{2} = -\frac{7x^2}{80} + 5x +c$

$c=0$

$v^2 = -\frac{7x^2}{40} + 10x$

$v = \sqrt{-\frac{7x^2}{40} + 10x}$

$v(20) = \sqrt{130} ms^{-1}$

nice wrk! thanks, my mistake was putting acceleration as 1.5 without even using the other 5...
Title: Re: dynamics/kinematics questions
Post by: kdgamz on September 18, 2009, 11:15:18 am: Quote from: Mao on September 18, 2009, 08:33:24 am
Quote from: kdgamz on September 17, 2009, 09:10:57 pm
Question 2

An object of mass 10 kg is subjected to a variable force described by the expression
(100 − 2x) newtons, where x is the distance up a ramp that makes an angle of 10° with the
horizontal. Assume the ramp is smooth and that the object starts from rest from the bottom
of the ramp.
(a) Find the speed, correct to two decimal places, of the object
(i) 3 m up the ramp

I'm assuming the variable force points up the ramp

Then, $ma = \sum F_{net} = (100 - 2x) - mg \sin 10^o$

$\begin{align*}\implies a & = 10 - 0.2 x - g\sin 10^o\\ a & = v \frac{dv}{dx} \\ \implies v \frac{dv}{dx} &= 10 - 0.2 x - g\sin 10^o \\ \int_{0}^{v_1} v\; dv & = \int_0^3 10 - 0.2 x - g\sin 10^o \; dx \\ \frac{(v_1)^2}{2} & = 3(10 - g\sin 10^o) - 0.9 \\ v_1 & \approx 6.23\\ \end{align*}$

thanks man (the answers 6.93 right?)

one more thing...i used
$\frac{d}{dx} (\frac{v^2}{2} ) = a$
and got the same answer .. can i do this???
.. and also when im doin trig, when do i use DEGREE mode and when do i use RADIAN mode... in this case only DEGREE mode works

thanks
Title: Re: dynamics/kinematics questions
Post by: Flaming_Arrow on September 18, 2009, 03:06:04 pm: Quote from: kdgamz on September 18, 2009, 11:15:18 am
Quote from: Mao on September 18, 2009, 08:33:24 am
Quote from: kdgamz on September 17, 2009, 09:10:57 pm
Question 2

An object of mass 10 kg is subjected to a variable force described by the expression
(100 − 2x) newtons, where x is the distance up a ramp that makes an angle of 10° with the
horizontal. Assume the ramp is smooth and that the object starts from rest from the bottom
of the ramp.
(a) Find the speed, correct to two decimal places, of the object
(i) 3 m up the ramp

I'm assuming the variable force points up the ramp

Then, $ma = \sum F_{net} = (100 - 2x) - mg \sin 10^o$

$\begin{align*}\implies a & = 10 - 0.2 x - g\sin 10^o\\ a & = v \frac{dv}{dx} \\ \implies v \frac{dv}{dx} &= 10 - 0.2 x - g\sin 10^o \\ \int_{0}^{v_1} v\; dv & = \int_0^3 10 - 0.2 x - g\sin 10^o \; dx \\ \frac{(v_1)^2}{2} & = 3(10 - g\sin 10^o) - 0.9 \\ v_1 & \approx 6.23\\ \end{align*}$

thanks man (the answers 6.93 right?)

one more thing...i used
$\frac{d}{dx} (\frac{v^2}{2} ) = a$
and got the same answer .. can i do this???
.. and also when im doin trig, when do i use DEGREE mode and when do i use RADIAN mode... in this case only DEGREE mode works

thanks

look at what the angle is given in, in this case its in degrees so u have to use degrees, but you can convert it to radians and use radian mode if u want to
Title: Re: dynamics/kinematics questions
Post by: kdgamz on September 18, 2009, 03:19:05 pm: so the angle is 10 degrees therefore i use DEGREE mode... if the angle was π/6, then i would have to use RADIAN mode..right?
Title: Re: dynamics/kinematics questions
Post by: Flaming_Arrow on September 18, 2009, 03:20:57 pm: Quote from: kdgamz on September 18, 2009, 03:19:05 pm

so the angle is 10 degrees therefore i use DEGREE mode... if the angle was π/6, then i would have to use RADIAN mode..right?

by n/6 if u mean pi/6 then yes
Title: Re: dynamics/kinematics questions
Post by: kdgamz on September 18, 2009, 03:22:41 pm: Quote from: Flaming_Arrow on September 18, 2009, 03:20:57 pm
Quote from: kdgamz on September 18, 2009, 03:19:05 pm

so the angle is 10 degrees therefore i use DEGREE mode... if the angle was π/6, then i would have to use RADIAN mode..right?

by n/6 if u mean pi/6 then yes

lol..yeh thats it, thanks man
Title: Re: dynamics/kinematics questions
Post by: kdgamz on September 18, 2009, 03:29:35 pm: one more.. its continued from question 2

Question 2

(b) What is the maximum distance the object can travel up the ramp?

The velocity function I got was v = √((20x)-0.2x^2- 2gxsin10), I found its derivative and made it equal to zero and solved for x, i got 41.49 m, but the answer is double that: 82.98 m

thanks
Title: Re: dynamics/kinematics questions
Post by: kdgamz on September 18, 2009, 03:51:26 pm: Quote
maximum distance occurs when v=0 and not a=0, if they tell you to find the maximum velocity then ur way is right

yeh that gives the right answer..thanks 8-)
Title: Re: dynamics/kinematics questions
Post by: Flaming_Arrow on September 18, 2009, 03:55:43 pm: Quote from: kdgamz on September 18, 2009, 03:29:35 pm
one more.. its continued from question 2

Question 2

(b) What is the maximum distance the object can travel up the ramp?

The velocity function I got was v = √((20x)-0.2x^2- 2gxsin10), I found its derivative and made it equal to zero and solved for x, i got 41.49 m, but the answer is double that: 82.98 m

thanks

maximum distance occurs when v=0 and not a=0, if they tell you to find the maximum velocity then ur way is right

sorry accidently deleted my post
Title: Re: dynamics/kinematics questions
Post by: kdgamz on September 18, 2009, 03:57:51 pm: heres another..i have no clue about this question except for the fact that you need to use standard units

Question 3

The period T seconds of a pendulum is known to vary with some power of its length,
l metres, some power of its mass, m grams and some power of its acceleration due to
gravity, g m/s2. Use the equality of dimensions method to find a variance expression
for T.

thanks
Title: Re: dynamics/kinematics questions
Post by: biggzee on September 18, 2009, 07:26:31 pm: Quote from: kdgamz on September 18, 2009, 03:57:51 pm
heres another..i have no clue about this question except for the fact that you need to use standard units

Question 3

The period T seconds of a pendulum is known to vary with some power of its length,
l metres, some power of its mass, m grams and some power of its acceleration due to
gravity, g m/s2. Use the equality of dimensions method to find a variance expression
for T.

thanks

All of your units should be equal on each side of the variation/ equality sign.

T = k* L^(p) * M^(q) * g^(r) (T- period, L - length, M - mass, g -accel. due to grav., k p q and r are constants)
= [1]*[m^p][kg^q][m^r*s^-2r] k is dimensionless
-2r = 1 (r= -1/2), as it is the only expression on the RHS involving the unit of time
= [m^p][kg^q][m^-0.5 * s]
p - 0.5 = 0 (p=0.5), as m^p and m^0.5 are multiplied to give a unit free of length.
and q=0, as the LHS has no mass component.

therefore T is proportional to L^0.5 * g^-0.5