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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Genericname2365 on February 12, 2012, 06:54:53 pm

Title: Question about motion
Post by: Genericname2365 on February 12, 2012, 06:54:53 pm: There doesn't appear to be any one thread to post in like the in Methods section, so I thought I'd just make this.

I'm having trouble figuring out this question:

6. A 200 g snooker ball with initial velocity 9.0 ms to the right collides with a stationary snooker ball of mass 100g. After the collision, both balls are moving to the right. The 200g ball has a speed of 3 m s while the 100g ball has a speed of 12 m s.
A) Determine the total kinetic energy of the system before the collision.

I tried using Ek=1/2 mv² (although admittedly the information provided only gives the initial velocity) but that didn't work. Can anyone tell me what I'm supposed to do? My theoretical grasp of this area is not particularly strong.
Title: Re: Question about motion
Post by: Phy124 on February 12, 2012, 07:00:14 pm: "A 200 g snooker ball with initial velocity 9.0 ms"

You have both mass and the velocity for your equation :)

$E = E_{k.b1} + E_{k.b2} =\frac{1}{2}(m)(v)^{2}+\frac{1}{2}(m)(v)^{2}= \frac{1}{2}(0.2)(9)^{2}+ \frac{1}{2}(0.1)(0)^{2}$ where .b1 = ball 1, .b2 = ball 2

$E= 8.1 J$

edit: put 7 in the calc, by accident...
Title: Re: Question about motion
Post by: Genericname2365 on February 12, 2012, 07:04:40 pm: Quote from: Phy124 on February 12, 2012, 07:00:14 pm
"A 200 g snooker ball with initial velocity 9.0 ms"

You have both mass and the velocity for your equation :)

E = E_k.b1 + E_k.b2 = $\frac{1}{2}(0.2)(9)^{2}+ \frac{1}{2}(0.1)(0)^{2}$

$E= 4.9 J$
I forgot to mention the answer is listed as 8.1 J.
Edit: Wait, I just got that answer...
Title: Re: Question about motion
Post by: b^3 on February 12, 2012, 07:06:57 pm: Phy124 your working is right but $\frac{1}{2}(0.2)(9)^{2}=8.1 \: J$

EDIT: nvm Genericname2365 worked it out.
Title: Re: Question about motion
Post by: Genericname2365 on February 12, 2012, 07:08:22 pm: Quote from: Phy124 on February 12, 2012, 07:00:14 pm
"A 200 g snooker ball with initial velocity 9.0 ms"

You have both mass and the velocity for your equation :)

$E = E_{k.b1} + E_{k.b2} =\frac{1}{2}(m)(v)^{2}+\frac{1}{2}(m)(v)^{2}= \frac{1}{2}(0.2)(9)^{2}+ \frac{1}{2}(0.1)(0)^{2}$ where .b1 = ball 1, .b2 = ball 2

$E= 4.9 J$
Quote from: b^3 on February 12, 2012, 07:06:57 pm
Phy124 your working is right but $\frac{1}{2}(0.2)(9)^{2}=8.1 \: J$
Ok thanks for the help, I forgot to square the velocity. *sigh*
Title: Re: Question about motion
Post by: rife168 on February 12, 2012, 07:09:43 pm: Because it asks for the total energy of the system, you have to analyse all of the different objects.
In this case, you only have to worry about the 200g ball, as the 100g ball is stationary. Were the 200g ball and 100g ball moving relative to the system, then you would have to consider the kinetic energy of both the balls.
Good to see you figured it out.
Title: Re: Question about motion
Post by: Phy124 on February 12, 2012, 07:15:32 pm: I accidently put 7 into the windows calculator instead of 9, my apologies...

*facepalm*
Title: Re: Question about motion
Post by: b^3 on February 12, 2012, 07:16:36 pm: Also note that if it is an isolated system, then the engery before and after should be the same. And doing the maths, it is.
Before
Quote from: Phy124 on February 12, 2012, 07:00:14 pm
"A 200 g snooker ball with initial velocity 9.0 ms"

You have both mass and the velocity for your equation :)

E = E_k.b1 + E_k.b2 = $\frac{1}{2}(0.2)(9)^{2}+ \frac{1}{2}(0.1)(0)^{2}$

$E= \mathbf{8.1} J$
After
E = E_k.b1 + E_k.b2 = $\frac{1}{2}(0.2)(3)^{2}+ \frac{1}{2}(0.1)(12)^{2}$
$E=0.9+7.2$
$E= 8.1 J$
Title: Re: Question about motion
Post by: Genericname2365 on February 25, 2012, 07:13:53 pm: Circular motion question. I'm going through Checkpoints and they have a number of questions about why people are lighter or heavier at the top/bottom of a vertical circular motion, and the answer is often in reference to $mg - R$ . How do you derive this formula?
I know the formulas $R= mv^2/ r - mg$ and $R = mv^2 / r + mg$ , but I don't understand how to get this.
Title: Re: Question about motion
Post by: Phy124 on February 25, 2012, 08:59:52 pm: I'm not exactly sure what "R" represents (is that the normal force?), but the formula is derived by forces upward against forces downwards.

(http://i43.tinypic.com/301ebd2.png)

*Note, the centripetal force is not a third force, but the net force.

For example, in figure 1, both the gravitational and normal forces are downward, so:

$F_{c}=F_{n}+F_{g}$

Where as, in figure 2, the normal force is up, whilst the gravitational force is down, so:

$F_{c}=F_{n}-F_{g}$
Title: Re: Question about motion
Post by: Genericname2365 on March 08, 2012, 06:14:46 pm: ls 0.054 an acceptable answer when the answer is 5.40 × 10^–2? (Gravity) If not, do you think I should get a newer calculator?
Title: Re: Question about motion
Post by: Phy124 on March 08, 2012, 06:20:29 pm: Quote from: Genericname2365 on March 08, 2012, 06:14:46 pm
ls 0.054 an acceptable answer when the answer is 5.40 × 10^–2? (Gravity) If not, do you think I should get a newer calculator?
If the question is looking for 3 significant figures and you wish to write it in the form you have above, you would need to write 0.0540 (as the 0's before the 5 do not count) which is the same as 5.40 x 10^-2
Title: Re: Question about motion
Post by: Genericname2365 on March 08, 2012, 06:25:19 pm: Quote from: ~My♥Little♥Pony~ on March 08, 2012, 06:20:29 pm
Quote from: Genericname2365 on March 08, 2012, 06:14:46 pm
ls 0.054 an acceptable answer when the answer is 5.40 × 10^–2? (Gravity) If not, do you think I should get a newer calculator?
If the question is looking for 3 significant figures and you wish to write it in the form you have above, you would need to write 0.0540 (as the 0's before the 5 do not count) which is the same as 5.40 x 10^-2
So the zero after the 54 because of the ^-2?
Title: Re: Question about motion
Post by: yawho on March 09, 2012, 08:08:47 pm: A 1000 kg carriage moves at constant speed of 15 m/s in a vertical circle of radius 25 m. What is the magnitude and direction of the net force on the carriage at (a) the top, (b) bottom and (c) mid way between the top and the bottom of the circle?
Title: Re: Question about motion
Post by: Genericname2365 on May 03, 2012, 08:35:51 pm: V40 kΩ = I total × 40 kΩ = 0.205 × 40 = 8.2 V

I'm probably missing an obvious concept here (as displayed in my previous posts in this thread :P) but I don't understand why 40kΩ is not being converted to 40,000Ω. Could anyone tell me why it's just left as 40kΩ?
Title: Re: Question about motion
Post by: Phy124 on May 03, 2012, 09:04:38 pm: Quote from: Genericname2365 on May 03, 2012, 08:35:51 pm
V40 kΩ = I total × 40 kΩ = 0.205 × 40 = 8.2 V

I'm probably missing an obvious concept here (as displayed in my previous posts in this thread :P) but I don't understand why 40kΩ is not being converted to 40,000Ω. Could anyone tell me why it's just left as 40kΩ?
Is the current in milliamps?

This would cause the 10³ and 10^-3 to cancel out.

i.e. 0.205 x 10^-3 x 40 x 10³ is the same as 0.205 x 40, if they were converted to Amps and Ohms respectively.
Title: Re: Question about motion
Post by: Lasercookie on May 03, 2012, 09:08:22 pm: I'm assuming you meant (bit confused with what I assumed to be subscripts):

$V_{40k \Omega} = I_{total} \cdot 40 k \Omega = 0.205 \text{ ?? } \cdot 40 \Omega = 8.2 V$
Assuming that the working out here is correct, my suspicion would be with the units of current being used here. It's possible that it's $10^{-3}$ for I, which would give the answer of 8.2 V:

$0.205 * 10^{-3} \cdot 40 * 10^{-3} = 8.2V$

But yeah, always work in SI Units, if what I guessed above is correct, then poor move by whoever wrote those solutions.

edit: damn, posted too slow, got beaten to it by a pony. :P
Title: Re: Question about motion
Post by: Genericname2365 on May 03, 2012, 10:46:42 pm: Thanks, that clears it up.