/0's physics phread

[/0]

I don't know if I've already made a physics thread but in any case either I can't find it or I never made one in the first place  :crazy2:

Are the following statements true:

1. A collision is elastic iff ; i.e. the departing speed equals the negative of the approaching speeds.

2. A collision is inelastic iff

Also, is 'inelastic' reserved only for collisions where the objects stick together completely, or is it simply defined as not an elastic collision?

Thanks

[appianway]

Elastic and inelastic collisions refer to the conservation of kinetic energy, methinks.

[methodsboy]

- An elastic collision is where the kinetic energy is conserved throughout the collision
- An inelastic collision is one where some kinetic energy is transformed into heat and sound.

[/0]

I know that about energy, but the expressions above are derived from both conservation of momentum and conservation of energy. It would be much more efficient if I could use those expressions to determine the elasticity of a collision, rather than calculating the KE before and after, and I want to know if it is acceptable.

[appianway]

Just keep in mind that momentum's a vector and that energy's a scalar quantity (thus the negative in your first equation shouldn't exist).

[/0]

Another:

1. A golf ball rolls at 4m/s at the top of a small hill of radius 2m. Will the ball lose contact with the surface? Show your calculations and reasoning. A force diagram is advised.

I have proved that at the top of the hill the ball is still in contact, but could it lose contact on the sides of the hill? The centripetal force there would be smaller, wouldn't it?


2. In vertical circular motion, is the speed still constant? Or is it greater at the bottom than at the top? If so, then does change with time?
What information do you need to find the speeds at different times?


3. What is the action-reaction pair to the centripetal force?

4. What is the advantage of lowering the centre of mass when cornering?

[/0]

Another two:

5. Calculate the gravitational force attracting two masses of 1 kg, separated by 0.40m. Take


I used the formula but I'm wondering, does that formula only give the force on one body by the other ?

So would the total gravitational force attracting be the force on both bodies = ? Thanks

6. A rocket is disabled at a height of 9 Earth radii above the surface of the earth. It is moving very slowly. By how much will its velocity change in the next 200s? (Take g at the surface of the earth as 10m/s, and assume that the local value of g does not change in the 200s)

[TrueTears]

Another two:

5. Calculate the gravitational force attracting two masses of 1 kg, separated by 0.40m. Take


I used the formula but I'm wondering, does that formula only give the force on one body by the other ?

So would the total gravitational force attracting be the force on both bodies = ? Thanks

6. A rocket is disabled at a height of 9 Earth radii above the surface of the earth. It is moving very slowly. By how much will its velocity change in the next 200s? (Take g at the surface of the earth as 10m/s, and assume that the local value of g does not change in the 200s)

i'll have a go at 6.

Looking at the formula , let GM = k so



so we have g = 10 and R = 1R when the rocket is at the surface this gives k = 10 x
later when the rocket is at 9 Earth radii we have (notice its 10 because you have to add the earth radius as well)

equate the 2 k's and we have therefore g = 0.1.

So we know the a = 0.1 t = 200 and u = 0. By subbing into the equation v = u + at we get v = 20

[TrueTears]

For question 5, the formula only gives the force on one body by the other, but by Newton's third Law, the other body would exert an equal force on the other object. So you don't need to do

[TrueTears]

Another:
2. In vertical circular motion, is the speed still constant? Or is it greater at the bottom than at the top? If so, then does change with time?
What information do you need to find the speeds at different times?

For question 2, in vertical circular motion speed is not constant. Effects by gravity causes the object to go slower at the top than at the bottom. does change when the object is at the top compared to when its at the bottom. When the object is at the top, you have its Normal force acting down towards the centre, the force due to gravity acting down and acting down towards the centre. . Say you were in a roller coaster, and you were at the top of a circular track, you would feel slightly lighter.

But if the object was on the bottom, Its normal force is acting up towards the centre, the force due to gravity is acting down and the is acting up towards the centre, so you would get . Again if you were in a roller coaster, and you were at the bottom of the circular track, you would feel slightly heavier.

[TrueTears]

For question 3:

All forces have action-reaction  pairs. But the NET force on an object is not the same as a force on the object. Centripetal force is the force pulling the object towards the centre of the curve, and this force is the net force on that object, so it doesn't have a opposite force.

lets say 2 people are pulling a box in opposite directions. One with 2 newtons of force and the other with 6 newtons of force.  The net force is 4 newtons in the direction of the 6 newton force.  There is no "equal and opposite"  force to this 4 newton net force.

[/0]

thanks truetears

[TrueTears]

Also i think lowering the centre of mass has something to do with gravity...? Since gravity acts from the centre of mass, and lowering the centre of mass while turning a corner would 'pin' you more to the ground? lol i dono someone clear this up for me.

[/0]

hmmm

If a car is moving at 40m/s and a truck collides with it, stopping the car, would you say the work the truck does on the car is negative or positive? The answers say it's positive but how could that be, seeing as the car's change in kinetic energy is negative?

When you say work is done ON an object what do you mean?
e.g. If a truck hits a car, stopping it, is it the work exerted by the truck, or the work done [/i]to the system which is the car?

[pHysiX]

hmmm...here's my theory:

the way i'm thinking, if you take the reference from the car then yes, the work done by the truck is negative. however, work is an energy, hence, getting negative energy is "invalid". the direction really does not matter in this case because the work done is how far the truck has moved and the force it has applied. if you think about it, it's actually a negative distance and a negative net force by the truck, hence positive.

well that's my random thoughts...not sure how right or how wrong i am. open to more suggestions =]