Niha afzal Posted March 28, 2016 Share Posted March 28, 2016 I used the kinematic equations To first find mass from the k.e given And in the secound part I used the mass to calculate the new k.e .... I attached my working Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2016 Share Posted March 28, 2016 (edited) Calculations refer to the position of the centre (of gravity) of the ball. They give you the ball diameter because this is not negligable. For instance the ball falls from where the top is at +80 cm to where the bottom is at 0 cm. (reckoning up as positive). So how far does the ball fall, ie what is its loss of PE? Edited March 28, 2016 by studiot 1 Link to comment Share on other sites More sharing options...
Niha afzal Posted March 28, 2016 Author Share Posted March 28, 2016 Oh so il take the height as .8-0.04 meters .... Then the p.e will be mass× -9.81 × .8-0.04 ? Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2016 Share Posted March 28, 2016 Oh so il take the height as .8-0.04 meters .... Then the p.e will be mass× -9.81 × .8-0.04 ? Er Hem How big is the ball? 1 Link to comment Share on other sites More sharing options...
Niha afzal Posted March 28, 2016 Author Share Posted March 28, 2016 Diameter 8 cm .. 0.08 m Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2016 Share Posted March 28, 2016 (edited) So how far does the ball fall, ie what is its loss of PE? Let me rephrase this So how far does the centre of the ball fall, ie what is its loss of PE? Edited March 28, 2016 by studiot 1 Link to comment Share on other sites More sharing options...
Niha afzal Posted March 28, 2016 Author Share Posted March 28, 2016 Okay okay ojaay maybe we take the height as 80 -8 cm.... bcz we start the height frm center of gravity of baallll and end it at COG which makes a complete diameter Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2016 Share Posted March 28, 2016 (edited) OK so why do we need to calculate all these kinetic quantities? Walk me through the sequence of energies (KE and PE) at the important stages in the path of the ball. See if you can puzzle it out first, before I tell you. Edited March 28, 2016 by studiot 1 Link to comment Share on other sites More sharing options...
Niha afzal Posted March 28, 2016 Author Share Posted March 28, 2016 (edited) Hmn so the potential energy at thestart of motion convert to kinetic energy (ke ) the ke is max at the bottom when ball is about to hit the groud so the potential energy at the begning converted to k.e Now since the balll after rebound didnt go back to 80 cm some energy must have been lost as heat light .. But in the upwards motion to the k.e getconverted potential energy. Thats all I know about k.e and g.e Edited March 28, 2016 by Niha afzal Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2016 Share Posted March 28, 2016 What you know is correct. you know the height of the drop and the height of the bounce back, and the PEs must be in the same ratio as these heights. you know that all the drop PE is converted to KE and similarly all the bounce back PE starts off as KE, again in the same ratio Can you put this into figures now? 1 Link to comment Share on other sites More sharing options...
Niha afzal Posted March 28, 2016 Author Share Posted March 28, 2016 Yessssss! !! :D Link to comment Share on other sites More sharing options...
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