**ANSWER**

BALLS COLLIDE

Where to make them collide? (Same Place, Same Time)

You
are given two ramps at different slope facing one another. ( A short ramp at a
steep angle, and a long ramp at a shallower angle) You are given a ball. You
will roll the ball down each ramp and make measurements. Then the balls will be
taken away from you. Using only your information and formulas, predict where to
put TWO balls, one on each ramp, so they will collide at the bottom. Have
teacher verify results

_{(Vi = 0)}

RAMP1 LENGTH TIME A_{1}
=*2D _{1}/T_{1}^{2}*

D=1/2AT^{2}

RAMP2 LENGTH TIME A_{2}
=*2D _{2}/T_{2}^{2}*

D=1/2AT^{2}

^{}^{}

^{ }

*Ramp1 Ramp2*

*Distance1 up ramp__D _{1}______ *Acceleration2__A_{2}_______
(from measurements)*

**Acceleration1___A _{1}______(from measurements) *Time1
down ramp ___T_{2}=T_{1}_ (same as ramp1) *

**Time1 down ramp__T _{1}_____ *Distance
up ramp2___^{ }D_{2}=1/2A_{2}T^{2}*

* *

*sample numbers: first ramp is 2 meters long, has a time of
.6 seconds, so its acceleration is*

* A _{1}= 2(2)/.6^{2}=11.11 m/s^{2}*

* Second
ramp is also 2 meters long to start, has a time of 1.2 seconds so its
acceleration is A _{2}= 2(2)/(1.2)^{2}= 2.78 m/s^{2}….*

* To
place the ball on ramp 2 so it hits at the same time as ramp 1….*

* D _{2}=1/2A_{2}T^{2}
or D_{2 }= ˝* 2.78 * (.6)^{2} = 0.5 meters*

*SAMPLE ANSWERS*

^{Where
will they collide?}

Where will they collide???? (Same Place Same time)

You
are given one ball and a ramp. You will roll the ball down the ramp and make
measurements. THEN, you will be given a second ball. BEFORE you are given the
second ball, you will be asked to predict where a ball from the top and the
middle of the ramp will collide along the floor, if at all. Show your work and
calculations. (HINT: you can use algebra, calculus, or the good old graphing
method. Assume that the ball after the ramp goes a constant speed). Have
teacher verify results

D=1/2AT^{2}

Vi=0 RAMP
(fixed angle) LENGTH TIME

D=1/2AT^{2} Vf^{2}=2AD D^{after}=Vf*T_{after}

^{ }

^{}

^{ t=
0 t=t2 t
= t1 t=tafter
+ t3}

^{* =
calculations}

*BALL1: BALL2:
(first down ramp)*

*MEASURE :Length
of ramp:__ ___ 2m Length
down ramp (half of ball one)__ _ D _{2} = ˝*D_{1}=1m*

*MEASURE: Time1
down ramp__ .6 sec_ *Acceleration
of ramp (same as ball1)_A_=11.11_m/s ^{2}_*

**Acceleration
of ramp_A=2D _{1}/T1^{2} =2*2/(.6^{2})=11.11_m/s^{2}_ *Time_{2}
down ramp__T_{2}=SQRT(2D_{2}/A)= SQRT(2*1/11.11)=.424 sec *

**Velocity1
at bottom of ramp__ _ Vf _{1}=SQRT (2AD_{1}) *Velocity_{2}
at bottom of ramp Vf_{2}=SQRT(2AD_{2})=2Vavg=2(1/.424)*

*Vf _{1}
=2 Vavg= 2 D_{1}/T_{1} = 2(2)/.6=6.667m/s Vf_{2}=4.714
m/s *

* *

**Time difference(Time along
floor before ball1)TD=(T _{1} - T_{2})=
(.6-.424) =.1754 sec*

*Distance
along floor=Velocity _{1} * Time after ramp*

*Distance along floor=Velocity _{2} *( Time
after ramp + Time difference)*

* *

* D? =
V1 * T?
D? =
V2 ( T? + TD)*

* two
equations, two unknowns, solve for distance after ramp. Only known numbers are
D _{1},T_{1}*

* *

*Sample numbers : Distance =
2 meters, Time = .6 seconds……Ball 1 is at 2 meters, its acceleration is A _{1}=
2(2)/.6^{2}=11.11 m/s^{2} . Its velocity at the end of the ramp
is V_{1}= 2 Vavg = 2*2/.6 = 6.67 m/s . Its distance along the floor is
D= 6.67 * T (T is time along the
floor).*

* *

*Ball 2 is at 1 meter, its
acceleration is also 11.11 . Its time on the ramp is T ^{2}= 2 D/A =
2/11.11 = .42426 seconds = T_{2} . Its velocity at the end of the ramp
is V_{2} =2 Vavg = 2 * 1/.42426 = 4.714 m/s . It spends a longer amount
of time on the floor by the time difference T_{1}-T_{2} or
.6-.42426 seconds = .17574 seconds. So its distance along the floor is D= V_{2}*
(T+TD) = 4.714 (T+.1754)*

* *

*Since they collide D=D or 6.67
T = 4.714 (T + .1754) T=
.82844/ ( 1.956) = .423537 sec*

* D
= 2.82499 meters*

-------------------------------

*ALGEBRA:*

*BALL1: BALL2:
(first down ramp)*

*MEASURE :Length
of ramp:__ ___ D _{1} Length
down ramp (half of ball one)__ _ D_{2} = ˝*D_{1}*

*MEASURE: Time1
down ramp__ T _{1}__ *Acceleration
of ramp (same as ball1)_A_ __*

**Acceleration
of ramp_A=2D _{1}/T1^{2} __ *Time_{2}
down ramp__T_{2}=SQRT(2D_{2}/A)= T_{1}/(SQRT(2)) *

**Velocity1
at bottom of ramp__ _ Vf _{1}=SQRT (2AD_{1})*

* *Velocity _{2}
at bottom of ramp Vf_{2}=SQRT(2AD_{2})=SQRT
(AD_{1})=SQRT(2)*D_{1}/T_{1}*

*Vf _{1}
= 2 D_{1}/T_{1} *

* *

**Time difference(Time along
floor before ball1)TD=(T _{1} - T_{2})*

*Distance
along floor=Velocity _{1} * Time after ramp*

*Distance along floor=Velocity _{2} *( Time
after ramp + Time difference)*

* *

* D? =
V1 * T?
D? =
V2 ( T? + TD)*

* two
equations, two unknowns, solve for distance after ramp. Only known numbers are
D _{1},T_{1}*

* *

* SQRT
(2AD _{1}) * T =
SQRT (2A ˝*D_{1}) * (T+(T_{1}-T_{2}))*

* SQRT
(2 (2D _{1}/T_{1}^{2})D_{1}) * T =
SQRT ( A D_{1}) * (T+(T_{1}-T_{2}))*

* SQRT(
4 D _{1}^{2}/T_{1}^{2} ) * T =
SQRT ((2D_{1}/T_{1}^{2})D_{1}) * (T+(T_{1}-T_{2}))*

* 2
D _{1}/T_{1} * T = SQRT(2) * D_{1}/T_{1} *
( T + T1 – SQRT (2 *˝*D_{1}/A) )*

* =
SQRT (2) * D _{1}/T_{1} * (T + T1 – SQRT ( D_{1}/A)
)*

* =
SQRT (2) * D _{1}/T_{1} * (T + T1 – SQRT ( D_{1}/(2D_{1}/T_{1}^{2}))
)*

* =
SQRT (2) * D _{1}/T_{1} * (T + T1 –* T_{1}/
SQRT (2))*

^{ }*= SQRT (2) * D _{1}/T_{1}
* (T) + SQRT (2) * D_{1}/T_{1} * T1 – SQRT (2) * D_{1}/T_{1}
*T_{1}/ SQRT (2)*

^{ }*2 D _{1}/T_{1}
* T =
SQRT (2) * D_{1}/T_{1} * (T) + SQRT (2) * D_{1}–D_{1}*

^{ }*2 D _{1}/T_{1}
* T - SQRT (2) * D_{1}/T_{1} * (T) = + SQRT (2) * D_{1}–D_{1}*

^{ }(*2 - SQRT (2)) * D _{1}/T_{1}
)* (T) =
+ (SQRT (2)–1) * D_{1}*

* T = T _{1}*((
SQRT(2)-1))) / (2-SQRT(2))*

* T
= T _{1}/(
SQRT(2))*

* D = (2
D _{1}/T_{1}) * ( T_{1}/( SQRT(2)))*

* D = SQRT(2)
*D _{1}*

* *