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                                          A1 =2D1/T12

D=1/2AT2

RAMP2                                     LENGTH                                 TIME                                          A2 =2D2/T22

D=1/2AT2

Ramp1                                                                       Ramp2

Distance1 up ramp__D1______                                          *Acceleration2__A2_______ (from measurements)

*Acceleration1___A1______(from measurements)          *Time1 down ramp ___T2=T1_ (same as ramp1)

*Time1 down ramp__T1_____                                            *Distance up ramp2___          D2=1/2A2T2

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

A1= 2(2)/.62=11.11 m/s2

Second ramp is also 2 meters long to start, has a time of 1.2 seconds so its acceleration is A2= 2(2)/(1.2)2= 2.78 m/s2….

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

D2=1/2A2T2 or D2 = ˝* 2.78 * (.6)2 = 0.5 meters

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=1/2AT2

Vi=0                         RAMP (fixed angle)         LENGTH                                 TIME

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)__ _ D2 = ˝*D1=1m

MEASURE:              Time1 down ramp__ .6 sec_                  *Acceleration of ramp (same as ball1)_A_=11.11_m/s2_

*Acceleration of ramp_A=2D1/T12 =2*2/(.62)=11.11_m/s2_            *Time2 down ramp__T2=SQRT(2D2/A)= SQRT(2*1/11.11)=.424 sec

*Velocity1 at bottom of ramp__ _ Vf1=SQRT (2AD1)     *Velocity2 at bottom of ramp  Vf2=SQRT(2AD2)=2Vavg=2(1/.424)

Vf1 =2 Vavg= 2 D1/T1 = 2(2)/.6=6.667m/s                                                                                                     Vf2=4.714 m/s

*Time difference(Time along floor before ball1)TD=(T1 - T2)= (.6-.424) =.1754 sec

Distance along floor=Velocity1 * Time after ramp

Distance along floor=Velocity2 *( 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 D1,T1

Sample numbers : Distance = 2 meters, Time = .6 seconds……Ball 1 is at 2 meters, its acceleration is A1= 2(2)/.62=11.11 m/s2 . Its velocity at the end of the ramp is V1= 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 T2= 2 D/A = 2/11.11 = .42426 seconds = T2 . Its velocity at the end of the ramp is V2 =2 Vavg = 2 * 1/.42426 = 4.714 m/s . It spends a longer amount of time on the floor by the time difference T1-T2 or .6-.42426 seconds = .17574 seconds. So its distance along the floor is D= V2* (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

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ALGEBRA:

BALL1:                                                                                              BALL2: (first down ramp)

MEASURE               :Length of ramp:__ ___ D1                                      Length down ramp (half of ball one)__ _ D2 = ˝*D1

MEASURE:              Time1 down ramp__ T1__                                        *Acceleration of ramp (same as ball1)_A_ __

*Acceleration of ramp_A=2D1/T12 __                                                     *Time2 down ramp__T2=SQRT(2D2/A)= T1/(SQRT(2))

*Velocity1 at bottom of ramp__ _ Vf1=SQRT (2AD1)

Vf1 = 2 D1/T1

*Time difference(Time along floor before ball1)TD=(T1 - T2)

Distance along floor=Velocity1 * Time after ramp

Distance along floor=Velocity2 *( 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 D1,T1

SQRT (2AD1) * T                                        = SQRT (2A ˝*D1)  *              (T+(T1-T2))

SQRT (2 (2D1/T12)D1) * T                        = SQRT ( A D1)       *                (T+(T1-T2))

SQRT( 4 D12/T12 ) * T                                = SQRT ((2D1/T12)D1) *          (T+(T1-T2))

2 D1/T1 * T                                                    =  SQRT(2) * D1/T1 * ( T + T1 – SQRT (2 *˝*D1/A) )

= SQRT (2) * D1/T1 * (T + T1 – SQRT ( D1/A) )

= SQRT (2) * D1/T1 * (T + T1 – SQRT ( D1/(2D1/T12)) )

= SQRT (2) * D1/T1 * (T + T1 –* T1/ SQRT (2))

= SQRT (2) * D1/T1 * (T) + SQRT (2) * D1/T1 * T1 – SQRT (2) * D1/T1 *T1/ SQRT (2)

2 D1/T1 * T                = SQRT (2) * D1/T1 * (T) + SQRT (2) * D1–D1

2 D1/T1 * T - SQRT (2) * D1/T1 * (T)      = + SQRT (2) * D1–D1

(2 - SQRT (2)) * D1/T1 )* (T)                   = + (SQRT (2)–1) * D1

T                =                T1*(( SQRT(2)-1))) / (2-SQRT(2))

T                =                T1/( SQRT(2))

D               =                (2 D1/T1) * ( T1/( SQRT(2)))

D               =                SQRT(2) *D1