Question

A curling stone weighing 19.96 kilograms slides down the ice with an initial velocity of 1.7 m/s for 30 seconds. A) how far will the curling stone go before coming to a stop? B) the curling stone hits a second curling stone weighing 18.7 kilograms initially at rest. If the velocity of curling stone A is still 1.7 m/s, what will the final speed and direction of the two curling stones be, assuming it is a perfectly elastic collision.

Answer 1

I'm actually not sure about part a.
For part b, since we're talking about elastic collisions, there is conservation of linear momentum.
\[\huge m_1{v_1}_i +m_2{v_2}_i = m_1{v_1}_f +m_2{v_2}_f \]

Answer 2

really? i thought the equation for elastic collisions was KE\[\frac{ 1 }{ 2 }mv _{Ai}^{2} + \frac{ 1 }{ 2} mv _{bi}^{2} = \frac{ 1 }{ 2 }mv _{Af}^{2} + \frac{ 1 }{ 2} mv _{\bf}^{2}\]

Answer 3

Oh, forgot to mention that. Ok, yes you are partially correct.
When you have a perfectly inelastic collision both energy and linear momentum must be conserved.
The equation I wrote was the conservation of linear momentum.
What you wrote was the conservation of kinetic energy.
You need BOTH to solve this problem.
Sorry bout that.

Answer 4

OOPS i mean elastic

Answer 5

I'm confused. how do i do this problem then. sorry its just my teacher is really fast and she's expecting us to make up our own problems but we need to solve it and i just don't know how to do this

Answer 6

ok so you have an ELASTIC collision. when that happens you need to conserve two things: kinetic energy and momentum

Momentum:
\[\huge m_1{v_1}_i +m_2{v_2}_i =m_1{v_1}_f +m_2{v_2}_f \]

Kinetic energy:
\[\huge {\frac 1 2} m_1{v_1}_i^2 +{\frac 1 2 } m_2{v_2}_i^2 ={\frac 1 2 } m_1{v_1}_f^2 +{\frac 1 2 } m_2{v_2}_f^2 \]

The basic idea is 2 equations 2 unknowns.
Make sense?

Answer 7

sort of. its just that i got weird answers for it. for the final velocity of A i got 0.915 m/s and that doesn't look right

Answer 8

ok, so are you familiar with the concept needing the same number of equations as you have unknown variables?

Answer 9

yes

Answer 10

This is essentially what you need here.
Both of the final velocities are unknown so you need two equations to solve them.

Conservation of momentum
\[\huge m_A{v_A}_i + m_B{v_B}_i =m_A{v_A}_f + m_B{v_B}_f\]
Conservation of kinetic energy
\[\huge {\frac 1 2}m_A{v_A}_i^2 + {\frac 1 2}m_B{v_B}_i^2 ={\frac 1 2}m_A{v_A}_f^2 + {\frac 1 2}m_B{v_B}_f^2\]

Answer 11

\[\large 19.96(1.7) + 0 =19.96{v_A}_f + 18.7{v_B}_f\]
\[\large {\frac 1 2}(19.96){(1.7)}^2 + 0 ={\frac 1 2}(19.96){v_A}_f^2 + {\frac 1 2}(18.7){v_B}_f^2\]

Answer 12

Were these your two equations?

Answer 13

YES

Answer 14

but then it simplified to \[19.96(1.7)^{2} = 19.96 V _{af}^{2}+18.7V_{\bf}^{2}\]

Answer 15

and then i got \[57.6844= 19.96 vaf ^{2} + 18.7 vbf ^{2}\]
which then made 0= -57.6844 +19.96 vaf2 + 18.7 vbf2
then i used the quadratic formula and got vaf= 1.31 m/s

Answer 16

you can't do that because you're trying to solve two unknowns using one equation. The kinetic energy one. You have to solve both simultaneously

Answer 17

okay how do u do that

Answer 18

are you allowed to use a calculator?

Answer 19

yep

Answer 20

graphing?

Answer 21

no

Answer 22

ok then we'll have to do this the old fashioned way

Answer 23

oh u mean a graphing calculator yes, sorry. tired

Answer 24

oh ok then
TI- which number do you have?

Answer 25

84

Answer 26

84 plus

Answer 27

aww; oh well;
84 can't solve multi-variable equations
only 89 and 92 I think
never mind, we can simplify it to a single variable
brb

Answer 28

okay

Answer 29

\[\large 19.96(1.7) =19.96{v_A}_f + 18.7{v_B}_f => 33.932 = 19.96v_Af+18.7v_Bf\]
\[\ {\frac 1 2}(19.96){(1.7)}^2 ={\frac 1 2}(19.96){v_A}_f^2 + {\frac 1 2}(18.7){v_B}_f^2=>28.8422=9.98v_Af^2+9.35v_Bf^2\]

Answer 30

Take both equations and simplify them as much as possible first.

Answer 31

Then using the momentum equation, solve for either v_A f or v_B f.
Take that and substitute it into the energy equation so that you now have only one variable. From there you can use the solve function on your calculator.

Answer 32

Make sense?

Answer 33

by my calculation you should get
Va_f = 0.0554m/s
Vb_f=1.755 m/s

Answer 34

i didn't get that

Answer 35

\[\huge vAf={33.932-18.7vBf\over 19.96}\]
did you get that at least?
or did you solve for vbf?

Answer 36

no i got that

Answer 37

\[\large 28.8422=9.98[{33.932-18.7vBf\over 19.96}]^2 + 9.35vBf^2\]

Answer 38

this?

Answer 39

yep

Answer 40

i just don't know where to go from there

Answer 41

umm, are you any good with algebra?....

Answer 42

ya

Answer 43

its just I'm stuck bc i haven't done this in a while

Answer 44

not sure what to say...it's all just math now

Answer 45

i got 0.223 bc i took everything in parentheses, squared it, multiplied it all by 9.98. then i made the 9.35 vbf have a common denominator and got 37176.0488vbf2/3976.048 now tell me what i did wrong

Answer 46

PLEASE. i need this for tomorrow

Answer 47

one sec; let me finish dinner

Answer 48

okay let's see

Answer 49

\[\huge 28.8422 = 9.98({32.932-18.7v_B\over 19.96})^2 +9.35v_B\]

Answer 50

umm the last term is 9.35v_B

Answer 51

omg i got it thank u solo much!

Answer 52

\[\large 28.8422=9.98(0.878v_B^2-3.09v_B+2.72)+9.35v_B\]
can you get here at least?

Answer 53

Part A seems like a dubious question. It might be interpreted as how far will it go in that 30s, otherwise, you'll need a coefficient of friction for stone on ice to find the resisting force.