VCE Physics Question Thread!

[Capristo]

I have a uniform circular motion question.
Just a mass attached to a string moving in a horizontal circular motion. (Diagram attached). At the bottom of the vertical line of string there is some weights of mass M. At the end of the diagonal section of string that moves in the circular path, there is a rubber stopper of mass m.

The tension, F, is constant. Show the relationship between L and T (period) by completing the following.
Write a proportionality expression for L
Show an equation with constant, K. (L=K.....)
Write an expression for K in terms of pi, m, M and g (gravity).

Thanks!

[clıppy]

A car of mass 600kg is being towed upward along an inclined plane at 5 degrees to the horizontal. Assume the frictional force on the car is constant throughout it's motion with a magnitude of 400N. The car is accelerating from rest at a rate of 0.2ms^-2
a) Find the magnitude of the tension in the towing cable
b) After traveling a distance of 100m the cable breaks. Determine the distance required for the car to come to a complete stop from the moment the cable breaks

I've found part a as 1043N which is correct but I have no idea how to find the answer to part b (15m)

[RKTR]

a student is twirling a weight on the end of a string 50cm long in a vertical circle
speed of weight is 4.0ms^-1 at the top

to find speed at the bottom ,i let change in ke = change in gpe . is this right?

[rhinwarr]

Yes, KE final = KE initial + change in GPE, then solve for speed.

[RKTR]

Yes, KE final = KE initial + change in GPE, then solve for speed.

erm if the question says the person does not add any further energy into the system as he twirls it, is the method still correct? tension at top is found . will tension at bottom in this case be the same ?

[rhinwarr]

That only works if the person does not add further energy. If it is in the vertical plane, tension at the bottom will be greater than the tension at the top because of the weight force acting against the centripetal force. Drawing force diagrams with the directions of the force will help you understand it.
At the top: centripetal force = tension + weight
At the bottom: centripetal force = tension - weight

[RKTR]

That only works if the person does not add further energy. If it is in the vertical plane, tension at the bottom will be greater than the tension at the top because of the weight force acting against the centripetal force. Drawing force diagrams with the directions of the force will help you understand it.
At the top: centripetal force = tension + weight
At the bottom: centripetal force = tension - weight

thanks . i was confused because my friend says tension will be the same since the person doesnt add any energy.

[RKTR]

http://a.imageshack.us/img831/1715/2shipprojectile.jpg

 had this as one of my sac questions today. 1. which ship will get hit 1st? 2.explain by referring to relevant physics principle. can anyone explain? i keep thinking there's not enough info.

[rhinwarr]

I think they would both get hit at the same time.
Vertical acceleration is the same for both so if the horizontal distances were the same (which they are not), ship B would get hit first as it reaches half the height of ship A's projectile.
However, since we need to take into account horizontal velocity (which remains constant for both), then the projectile for ship B actually has to travel twice as far so should land at the same time as ship A's projectile.
I'm not really sure so better check with someone else.

[RKTR]

found the explanation online. vertical displacement =0 
using x=ut+1/2 at^2
 0= vsin(theta)t -5t^2
t=vsin(theta)/5

v for both are the same . when theta is larger, sin theta is also larger , time is also increased. so B will get hit first.  ohno already lost 3 marks out of 40

[lzxnl]

There's not enough information. You haven't been told about the speeds and angles of each missile.

[RKTR]

There's not enough information. You haven't been told about the speeds and angles of each missile.

really? that's what i thought. but today the question on the sac was which ship will get hit 1st? not multiple choice. and how about the explanation i heard from other people and found online in my previous post?

[lzxnl]

I'm going to have a go at this and see what happens
Let the speed of the first projectile be v1 and the angle be s1
Then, the vertical displacement is 0 = v1*sin s1 t - 1/2 gt^2 => t = 2v1 sin s1/g
Hmm. The only way the two times are the same is if v1 sin s1 = v2 sin s2 and that's not always true. They haven't given you any numbers, so you really can't make a judgement.

[09Ti08]

*for the speed: the key information in the problem is the shells are fired "simultaneously", so I think it's reasonable to conclude that they fired at the same speed.

*for the angle: it is obvious that ship A is closer to the battleship, so the angle must be bigger for the shell to hit it.

[lzxnl]

Were the shells fired from the same cannon though? I know I'm being pedantic but I don't think I noticed that. Besides if they were, and the back shell was fired too hard, it could easily transfer some of its momentum to the shell in front. Imagine two bouncing basketballs on top of each other.