Curling stone coming to rest - friction

[ghostbuster25]

This is just a revision question for an exam i have in a couple weeks.

I think i kow the answer but just want to double check.

a curling stone is released with an instial speed of 3.1 m/s
the coefficent of sliding friction between the stone and ice is 0.017
What is the distance the stone will come to rest?
A 14m B 19m C 29m D 58m E Impossible to tell without mass



I believe it to be E

IF i did have the mass what equation would i need?
$$\mu$$$$_{}slide$$ * mg ??


Many thanks

 

[proculation]

you sure ? If a ton of steel is on a ice and I push it to give it a momentum, will it stop at the same place than a curling stone of 10kg ?

Think about it.

 

[proculation]

btw it was a joke. You had it almost right :P

coefficient of dynamics friction * Force which equals Mass times Acceleration.

 

[ghostbuster25]

ok re-looked at it and have decided on D 58m as D=V²/the coefficent of sliding friction * gravity (9.8 m/s)

That look correct?

 

[proculation]

How can you tell the acceleration ? (which is negative) you conclude it is the gravity ?

 

[ghostbuster25]

well the equation i have is D=V²/$$\mu$$slide * g

I just put in my values of 3.1 m/s²/ 0.017*9.8 m/s

It gave me an answer of 57.6 which is the same as answer D 58m (2sf)

 

[proculation]

I already told you you were right.

It's just my way to teach. To make sure you understood and looked at your response again to be sure.

 

[ghostbuster25]

ha ha oh right, ok thanks
Do you know much about collisions in three dimensions between 2 objects? i.e (vbx,vby)?
:)

 

[proculation]

It's the same thing then between two objects in 2 dimensions. You just add another dimension.

You separate the forces in x, y, z instead of only x and y. That's it.

 

[ghostbuster25]

sorry I've been an idiot, i meant in 2 dimensions.
ill tell you the question...

2 objects with masses A 0.60kg and B 0.45kg are traveling towards one another parallel and anti parallel at uAx 1.5m/s and uBx -2.0m/s
the velocity of A after collision is (0,12)m/s what is (vBx,vBy) of B after collision?

I just can't find in my text how to do it for (x,y)

 

[proculation]

Hint: 3rd law of Newton.

 

[ghostbuster25]

so if object A has an opposite but equal in magnitude force on object B when they collide then the force acting on B would be f=ma which would be 0.9N, if i rearrnged to find for v i get v=f/m which gives me 2m/s but as B is traveling at -2m/s it would reusult in 0m/s so B would come to a stand still and have (vBx,vBy) (0,0)m/s

is that sort of the way to go?

 

[proculation]

You "can" do it that way. But the best way is to go with the 3rd law of Newton: the conservation of energy.

Or I may say, the momentum. m1 x v1 = m2 x v2

 

[ghostbuster25]

yer i was going to ask if there was a better way of doing it. That makes sense :) Thanks