monkeyseemonkeydo Posted April 4, 2010 Report Share Posted April 4, 2010 I concur with root 2 and 3.25root2 4.5m/s? That's a really shit gun... I could throw the bullet faster than that! Quote Link to comment Share on other sites More sharing options...
manuel Posted April 4, 2010 Report Share Posted April 4, 2010 4.5m/s? That's a really shit gun... I could throw the bullet faster than that! To be shuuuurrre that's just what I get - anyway, you are the engineer I'm just a postman... Quote Link to comment Share on other sites More sharing options...
Ash-Kennard Posted April 4, 2010 Author Report Share Posted April 4, 2010 4.5m/s? That's a really shit gun... I could throw the bullet faster than that! thats why i think this is wrong. have a go dave? im sure your good with stuff similar to this Quote Link to comment Share on other sites More sharing options...
DannyBazz (: Posted April 4, 2010 Report Share Posted April 4, 2010 4.5m/s? That's a really shit gun... I could throw the bullet faster than that! Yeah but it's not really to do with the gun, its to do with the velocity of the bullet. The answers never actually relate to real life if you get me... I think they do it to throw you off. Quote Link to comment Share on other sites More sharing options...
Otacon Posted April 4, 2010 Report Share Posted April 4, 2010 Speak to Ed Emus (forteh) on here. He's a engineer, i reckon he'll sort you out. To me it means absolutely nothing lmao. Quote Link to comment Share on other sites More sharing options...
PeanuckleJive Posted April 4, 2010 Report Share Posted April 4, 2010 thats why i think this is wrong. have a go dave? im sure your good with stuff similar to this try adjusting the number i suggested! your equation is wrong remember. Quote Link to comment Share on other sites More sharing options...
Ash-Kennard Posted April 4, 2010 Author Report Share Posted April 4, 2010 try adjusting the number i suggested! your equation is wrong remember. is the equation actually wrong? and what do you mean adjust the numbers? cos if i do then ill defo have the wrong answer? Quote Link to comment Share on other sites More sharing options...
PeanuckleJive Posted April 4, 2010 Report Share Posted April 4, 2010 is the equation actually wrong? and what do you mean adjust the numbers? cos if i do then ill defo have the wrong answer? remember the post I made? The question says it ends 0.1m above starting point and you put it in as 0.01, read your initial post again... Quote Link to comment Share on other sites More sharing options...
Ash-Kennard Posted April 4, 2010 Author Report Share Posted April 4, 2010 (edited) remember the post I made? The question says it ends 0.1m above starting point and you put it in as 0.01, read your initial post again... ahh yeah. i did change that. i had already done the working then messed it up slightly typing it up on here RIGHT, i believe this one is solved. thanks everyone for your help Edited April 4, 2010 by Ash-Kennard Quote Link to comment Share on other sites More sharing options...
forteh Posted April 6, 2010 Report Share Posted April 6, 2010 Nah Im crap at theoretical bollocks like this Besides which its been so long since Ive done any calcs like this and have forgotten most of it Quote Link to comment Share on other sites More sharing options...
Tom Booth Posted April 6, 2010 Report Share Posted April 6, 2010 ??? Quote Link to comment Share on other sites More sharing options...
AdamR28 Posted April 6, 2010 Report Share Posted April 6, 2010 Pretty sure you can do it very simply using Kinetic and Potential energy: 1) KE of bullet = KE of bullet + block of wood. The combined items have a higher mass therefore lower speed. (Probably get this from working back from the solution to question 2?) 2) Amount of increase in PE of bullet+block (by rising 0.1m) is the amount of energy the bullet had when it hit the block. Edit: Been 10 years since I did any of this stuff... haha. So that's probably wrong Quote Link to comment Share on other sites More sharing options...
monkeyseemonkeydo Posted April 8, 2010 Report Share Posted April 8, 2010 Pretty sure you can do it very simply using Kinetic and Potential energy: 2) Amount of increase in PE of bullet+block (by rising 0.1m) is the amount of energy the bullet had when it hit the block. Edit: Been 10 years since I did any of this stuff... haha. So that's probably wrong I tried that but got an even lower velocity for the bullet so didn't bother putting it up... I'm sure it should work though! Quote Link to comment Share on other sites More sharing options...
AdamR28 Posted April 8, 2010 Report Share Posted April 8, 2010 Hmm, yeah. I got ~12.4m/s... weird...! Quote Link to comment Share on other sites More sharing options...
ManxTrialSpaz Posted April 8, 2010 Report Share Posted April 8, 2010 (edited) I worked through it last night as something to do and finished up with 4.55ms-1 Edited April 8, 2010 by ManxTrialSpaz Quote Link to comment Share on other sites More sharing options...
ManxTrialSpaz Posted April 8, 2010 Report Share Posted April 8, 2010 I was originally writing a reply but in hindsight it needn't be aimed towards anyone in particular - instead, here is my working with a bit of explanation of the steps for clarity. Assuming 100% efficiency, we can say that all of the kinetic energy after impact (so mass and velocity of block and bullet) is transferred into potential energy at its highest point on the string as where as v=0, Ek obviously also becomes 0. So we can say ⌂Ek=⌂Eg ½mv2 = mg⌂h => v=sqrt*(2g⌂h). As ⌂h=0.1m and g=9.81ms-1, v=sqrt*(2x9.81ms-2x0.1m), so the answer to A; the velocity of the block and bullet just after impact is 1.4ms-1 The calculations for initial velocity require use of conservation of momentum, the way I write it is as m1u1+m2u2=m1v1+m2v2 m1=0.04kg u1=? (Initial velocity of the bullet) m2=0.09kg u2=0 But since we know the masses coalesce after impact and that the wooden block has no velocity, therefore no momentum before the impact, we can write m1u1=m3v, where m3 = m1+m2 So from there, we can rearrange for u=(m3v)/m1 = (0.13kg x 1.4ms-1)/0.04kg = 4.55ms-1 Quote Link to comment Share on other sites More sharing options...
AdamR28 Posted April 8, 2010 Report Share Posted April 8, 2010 Haha, I forgot to include g in PE = mgh, lol, fail. The solution above sounds 'right', but the numbers sound all wrong. Probably to throw you off... Quote Link to comment Share on other sites More sharing options...
RobinJI Posted April 8, 2010 Report Share Posted April 8, 2010 (edited) From what I can see, manxtrialspaz is correct. Also that question's pretty poor. If it was my piece of work I'd make a point of noting in the answer that you feel this result appears to abnormally low for a realistic speed for a bullet to travel, but that you feel the answer had been found in the correct manner. That you have assumed 'just after' to be referring to the instant after impact at which the 2 body's mass's are first combined*, and that you have assumed zero influence from external factors. * Just after is hardly a accurate term, If I said 'just after lunch', you'd think in terms of a few minutes, but if I said 'just after the ice age', you probably wouldn't think I meant a couple of minutes after. Edited April 8, 2010 by RobinJI Quote Link to comment Share on other sites More sharing options...
Shaun H Posted April 8, 2010 Report Share Posted April 8, 2010 I was originally writing a reply but in hindsight it needn't be aimed towards anyone in particular - instead, here is my working with a bit of explanation of the steps for clarity. Assuming 100% efficiency, we can say that all of the kinetic energy after impact (so mass and velocity of block and bullet) is transferred into potential energy at its highest point on the string as where as v=0, Ek obviously also becomes 0. So we can say ⌂Ek=⌂Eg ½mv2 = mg⌂h => v=sqrt*(2g⌂h). As ⌂h=0.1m and g=9.81ms-1, v=sqrt*(2x9.81ms-2x0.1m), so the answer to A; the velocity of the block and bullet just after impact is 1.4ms-1 The calculations for initial velocity require use of conservation of momentum, the way I write it is as m1u1+m2u2=m1v1+m2v2 m1=0.04kg u1=? (Initial velocity of the bullet) m2=0.09kg u2=0 But since we know the masses coalesce after impact and that the wooden block has no velocity, therefore no momentum before the impact, we can write m1u1=m3v, where m3 = m1+m2 So from there, we can rearrange for u=(m3v)/m1 = (0.13kg x 1.4ms-1)/0.04kg = 4.55ms-1 Always make sure you stick this sorta shit in your final answers. Lecturers love it when you show you understand the details. I haven't bothered working through it but I can't see any problems with your initial answer and ManxTrialSpaz's either. Give us some more Quote Link to comment Share on other sites More sharing options...
monkeyseemonkeydo Posted April 8, 2010 Report Share Posted April 8, 2010 Always make sure you stick this sorta shit in your final answers. Lecturers love it when you show you understand the details. I'd have gone for 'assuming no losses' rather than 100% efficiency but there you go . Quote Link to comment Share on other sites More sharing options...
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