Page 2 Introductory Exercise 7.3 1.   v 1 ® = u and   v 2 ® = v (say) Dv v v ® ® ® = +  2 1 ( ) =  ® ® v v 2 1 D D v u u v v u ® ® ® ® ® ® × =  ×  ( ) ( ) 2 1 2 1 = × + × ® ® ® ® u v u v 2 2 1 1   ® ® 2 2 1 v v     v v 2 2 1 2 2 2 ® ® + = + v u =  + u gL u 2 2 2   ( ) Dv ® =  2 2 2 u gL   ( ) Dv ® =  2 2 u gL 2. Ball motion from A to B : 0 2 2 2 2 = +  u g R min ( )( ) Þ u gR min 2 4 = Ball motion from C to A : v u g h 2 2 2 = +  min ( ) =  4 2 gR gh Þ v g R h =  2 2 ( ) 3. Decrease in KE of bob = Increase in PE of bob L v 2 – v 1 Di g v 1 ® ® ® ® h v C A 2R B PE of bob = K (say) KE of bob 2 = 1/2 mv 0 v 0 Page 3 Introductory Exercise 7.3 1.   v 1 ® = u and   v 2 ® = v (say) Dv v v ® ® ® = +  2 1 ( ) =  ® ® v v 2 1 D D v u u v v u ® ® ® ® ® ® × =  ×  ( ) ( ) 2 1 2 1 = × + × ® ® ® ® u v u v 2 2 1 1   ® ® 2 2 1 v v     v v 2 2 1 2 2 2 ® ® + = + v u =  + u gL u 2 2 2   ( ) Dv ® =  2 2 2 u gL   ( ) Dv ® =  2 2 u gL 2. Ball motion from A to B : 0 2 2 2 2 = +  u g R min ( )( ) Þ u gR min 2 4 = Ball motion from C to A : v u g h 2 2 2 = +  min ( ) =  4 2 gR gh Þ v g R h =  2 2 ( ) 3. Decrease in KE of bob = Increase in PE of bob L v 2 – v 1 Di g v 1 ® ® ® ® h v C A 2R B PE of bob = K (say) KE of bob 2 = 1/2 mv 0 v 0 1 2 0 2 mv mgh = v gh 0 2 2 = v gl 0 2 1 =  ( cos ) q = ´ ´ ´  ° 2 5 1 60 9.8 ( cos ) =7 ms 1 h PE of bob = K + mghRead More
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