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lecture_2_07b

# lecture_2_07b - Falling ball seen from platform...

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05/11/09 Physics 13 - Fall 07 - G.R. Goldstein 1 Falling ball seen from platform distance along platform (meters) height (meters) Elapsed time is 1 sec V=100 Km/hr =27.8 m/sec 0.2 sec 0.6 sec 0.8 sec 0.4 sec

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05/11/09 Physics 13 - Fall 07 - G.R. Goldstein 2 acceleration Newton’s 1st Law: body remains in uniform motion unless acted upon by (net) force Non-uniform means v changes (magnitude or direction) Changing v over time is acceleration a = dv dt = d 2 x dt 2 a due to gravity = 9.8 m/s 2 = g always downward Vertical motion on earth
05/11/09 Physics 13 - Fall 07 - G.R. Goldstein 3 Galilean relativity Ball seen from platform follows trajectory for initial velocity in x- direction u x = 100 Km/hr = V of train (at time t=0) [and no initial velocity in y & z directions] No force in x direction so u x of ball unchanged and x increases x = u x t Gravity in -y direction, so downward u y increases and y falls from say h = 5.0 m to 0 via y = h - gt 2 /2 y = 0 for 1 2 gt 2 = h So t = 2 h g = 10. m 9.8 m /sec 2 =1.0 sec

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05/11/09 Physics 13 - Fall 07 - G.R. Goldstein 4 Galilean relativity (cont’d) Acceleration same in either Frame of Reference - none in x-direction & - g in y-direction So F =m a = m a Newton’s 2nd Law unchanged There is no way to determine which Frame of Reference is “at rest” by doing experiments! (Invariance of Laws of Physics)
05/11/09 Physics 13 - Fall 07 - G.R. Goldstein 5 Galilean relativity & 2nd Law Ball seen from platform follows trajectory for initial velocity u x = 100 Km/hr = V and u y =0,u z =0 x ( t ) = x ( t ) + Vt u x ( t ) = dx dt = d dt ( x ( t ) + Vt ) = u x ( t ) + V so u x = u x + V addition of velocities Also a x ( t ) = d 2 x dt 2 = d dt ( u x ( t ) + V ) = a x ( t ) Hence F =m a = m a - Newton’s 2nd Law unchanged

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05/11/09 Physics 13 - Fall 07 - G.R. Goldstein 6 Lateral bouncing ball in 2 inertial frames  S : frame on train seen from above L V x y Position of ball on upward path x =0, y =u y t until y =L So for round trip t R =2L/u y Let u y =c the speed of ball and note that u y ’= u y in S S: platform frame seen from above L x y d Vt R /2 L 2 + ( Vt R /2) 2 = d 2 = L 2 (1+V 2 /u y 2 ) Now d/(t R /2) = c ball speed seen in platform frame c =d u y /L = u y (1+V 2 /u y 2 ) = (u y 2 +V 2 ) or c= (c 2 +V 2 ) or c =c (1-V 2 /c 2 )
05/11/09 Physics 13 - Fall 07 - G.R. Goldstein 7 Swimmer in flowing river

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