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# tt1sol - (R-gx 2/2v i 2 2 x 2> R 2 → R 2 gRx 2/v i 2 g...

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PHY131F – Term Test 1 – October 6, 2009 Multiple-Choice Questions Version 1 1(A) 2(A) 3(E) 4(A) 5(E) Version 2 1(B) 2(B) 3(D) 4(D) 5(D) Long Answer Problem Part A Model – Employ the particle model. Air resistance is zero. Visualize – The ball’s initial position is (x i ,y i )=(0,R) and its trajectory is given by y=y i +v iy t+½a y t 2 =R+0-½gt 2 (1) x=x i +v ix t=0+v i t=v i t (2), where v ix =v i , v iy =0, and t=t-t i =t-0=t. Substitute t=x/v i from (2) into (1) to find the parabolic path of the ball y=R-gx 2 /2v i 2 (3). Solve – Points on the surface of the rock are defined by Y 2 +x 2 =R 2 . At t=0, the ball is at the surface of the rock. For t>0, the ball must remain above the hemisphere. This is ensured by setting y 2 +x 2 >R 2 (4), where y refers to the position of the ball. Substitute (3) into (4) to find (R-gx
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Unformatted text preview: (R-gx 2 /2v i 2 ) 2 + x 2 > R 2 → R 2- gRx 2 /v i 2 + g 2 x 4 /4v i 4 + x 2 > R 2 → x 2 (g 2 x 2 /4v i 4 + 1 - gR/v i 2 ) > 0. For x approaching 0, g 2 x 2 /4v i 4 + 1 - gR/v i 2 > 0 holds if 1 - gR/v i 2 > 0 → 1 > gR/v i 2 → v i 2 > gR → v i > (gR) ½ . This implies that the minimum speed is v i = (gR) ½ . Assess – The above expression for the minimum speed has the appropriate dimensions. Part B The ball will reach the ground at position (x,y)=(x,0), where x is given by Eq. (3) 0 = R-gx 2 /2v i 2 → gx 2 /2v i 2 = R → x 2 = 2v i 2 R/g → x 2 = 2(gR)R/g → x 2 =2R 2 → x= √ 2R. rock’s base distance ball lands from rock’s base (-R,0) x-(-R)= √ 2R+R=( √ 2+1)R (0,0) x-0= √ 2R (0,R) x-R= √ 2R-R=( √ 2-1)R...
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