Question about motion

[Genericname2365]

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.

[Phy124]

"A 200 g snooker ball with initial velocity 9.0 ms"

You have both mass and the velocity for your equation

where .b1 = ball 1, .b2 = ball 2



edit: put 7 in the calc, by accident...

[Genericname2365]

"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 =

I forgot to mention the answer is listed as 8.1 J.
Edit: Wait, I just got that answer...

[b^3]

Phy124 your working is right but

EDIT: nvm Genericname2365 worked it out.

[Genericname2365]

"A 200 g snooker ball with initial velocity 9.0 ms"

You have both mass and the velocity for your equation

where .b1 = ball 1, .b2 = ball 2

Phy124 your working is right but

Ok thanks for the help, I forgot to square the velocity. *sigh*

[rife168]

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.

[Phy124]

I accidently put 7 into the windows calculator instead of 9, my apologies...

*facepalm*

[b^3]

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

"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 =

After
E = E_k.b1 + E_k.b2 =

[Genericname2365]

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 . How do you derive this formula?
I know the formulas and , but I don't understand how to get this.

[Phy124]

I'm not exactly sure what "R" represents (is that the normal force?), but the formula is derived by forces upward against forces downwards.



*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:



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

[Genericname2365]

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?

[Phy124]

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

[Genericname2365]

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?

[yawho]

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?

[Genericname2365]

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 ) 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?