Rigel Posted April 8, 2011 Report Share Posted April 8, 2011 Thanks Gene!!! I thought you had to make a triangle . Don't know where that came from...I got another question.A rocket leaves the earth with a constant acceleration of 12m/s2, in the first 1000m. Suddenly, the engines shut off, and the rocket falls down by free fall. Ignore air resistance.¿What would be the maximum height of the rocket, how much time would it take the rocket to reach it's maximum height?The book says that the height is 2220m and the time would be 28.7s.. Thanks! Reply Link to post Share on other sites More sharing options...
genepeer Posted April 8, 2011 Report Share Posted April 8, 2011 Assuming g = 9.81ms-2Let the time take to cover those 1000m be t1.s = at12/21000 = 12*t12/2t12 = 166.667t1 = 12.91Speed at 1000m = at = 12*12.91 = 154.92ms-1From here to the maximum height, we have the:- initial velocity (u) = 154.92ms-1 - final velocity (v) = 0ms-1- acceleration (a) = -9.81ms-2, free-fall and negative because it's downwards while the rockets still moving up - the unknowns, distance travelled (s) and time taken (t2)a)we use the third law of motion, which has an expression for the given variables and sv2 = u2 + 2as0 = 24000 + 2*(-9.81)*ss = 1223.2mMaximum height = 1000 + 1223.2 = 2220m to 3sf.b)we use one of the laws of motion, which has an expression for the given variables and t2Try this yourself 1 Reply Link to post Share on other sites More sharing options...
Rigel Posted April 11, 2011 Report Share Posted April 11, 2011 Gotcha Gene, that helped a lot!I'm sorry.. but i have to ask another question, hehe. (I'm sorry if i'm starting to get annoying, i want to be sure if the answer i put was correct).An object is thrown in the air with an inicial velocity of 30ms-1. Then, 0.5s after that, another object is thrown 60 meters ahead of the ball. Find: At what height will the objects collide? In how much time will that happen?THANK YOU IB SURVIVAL for all your help! Reply Link to post Share on other sites More sharing options...
flamicecream Posted April 29, 2011 Report Share Posted April 29, 2011 (edited) This seems like it should be really simple but I can't get it...A ball is suspended from a ceiling by a string of length 7.5m. The ball is kicked horizontally and rises to a maximum height of 6.0m. a) Assuming that the air resistance is negligible, show that the initial speed of the ball is 11 m/sb) The mass of the ball is 0.55kg and the impact of the kicker's foot with the ball is 150ms. Estimate the average force exerted on the ball by the kick.c) i) Explain why the tension in the string increases immediately after the ball is kicked. ii) Calculate the tension in the string immediately after the ball is kicked. Assume that the string is vertical. for b) I'm guessing you use F=Δp/Δt...I worked it out to be 40.3 newtons. Thanks for any help.Btw this is from the 2010 SL exam, paper 2 Edited April 29, 2011 by flamicecream Reply Link to post Share on other sites More sharing options...
CocoPop Posted April 29, 2011 Report Share Posted April 29, 2011 (edited) This seems like it should be really simple but I can't get it...A ball is suspended from a ceiling by a string of length 7.5m. The ball is kicked horizontally and rises to a maximum height of 6.0m. a) Assuming that the air resistance is negligible, show that the initla speed of the ball is 11 m/sb) The mass of the ball is 0.55kg and the impact of the kicker's foot with the ball is 150ms. Estimate the average force exerted on the ball by the kick.c) i) Explain why the tension in the string increases immediately after the ball is kicked. ii) Calculate the tension in the string immediately after the ball is kicked. Assume that the string is vertical. for b) I'm guessing you use F=Δp/Δt...I worked it out to be 40.3 newtons. Thanks for any help.Btw this is from the 2010 SL exam, paper 2I don't know why I'm taking a break from studying by doing more physics, but here's a quick go.a) Equate maximum kinetic energy (1/2*m*v^2) and maximum potential energy (mgh). Maximum velocity occurs when h=0 (where h is measured from the rest position) and your answer should round up to 11.b) Force equals rate of change of momentum. F = m(v-u)/t. v = final velocity = 11. u = initial velocity = 0. t = 150ms.c i) Centripetal force.c ii) Resolve forces due to weight (mg) and centripetal acceleration (mv^2/r).EDIT - Since the string is vertical this is quite simple. Just add the forces. Edited April 29, 2011 by CocoPop Reply Link to post Share on other sites More sharing options...
Rigel Posted April 29, 2011 Report Share Posted April 29, 2011 This seems like it should be really simple but I can't get it...A ball is suspended from a ceiling by a string of length 7.5m. The ball is kicked horizontally and rises to a maximum height of 6.0m. a) Assuming that the air resistance is negligible, show that the initial speed of the ball is 11 m/sb) The mass of the ball is 0.55kg and the impact of the kicker's foot with the ball is 150ms. Estimate the average force exerted on the ball by the kick.c) i) Explain why the tension in the string increases immediately after the ball is kicked. ii) Calculate the tension in the string immediately after the ball is kicked. Assume that the string is vertical. for b) I'm guessing you use F=Δp/Δt...I worked it out to be 40.3 newtons. Thanks for any help.Btw this is from the 2010 SL exam, paper 2I can only work out a), because i'm doing Dynamics right now. 1 Reply Link to post Share on other sites More sharing options...
CocoPop Posted April 29, 2011 Report Share Posted April 29, 2011 This seems like it should be really simple but I can't get it...A ball is suspended from a ceiling by a string of length 7.5m. The ball is kicked horizontally and rises to a maximum height of 6.0m. a) Assuming that the air resistance is negligible, show that the initial speed of the ball is 11 m/sb) The mass of the ball is 0.55kg and the impact of the kicker's foot with the ball is 150ms. Estimate the average force exerted on the ball by the kick.c) i) Explain why the tension in the string increases immediately after the ball is kicked. ii) Calculate the tension in the string immediately after the ball is kicked. Assume that the string is vertical. for b) I'm guessing you use F=Δp/Δt...I worked it out to be 40.3 newtons. Thanks for any help.Btw this is from the 2010 SL exam, paper 2I can only work out a), because i'm doing Dynamics right now.b) F = m(v-u)/r = 0.55(11-0)/(150*10^-3) = 40.3Nc) i) The ball's movement is restricted to a circular motion due to the string. This causes a centripetal acceleration with force F=mv^2/r which causes an increase in tension in the string.c) ii) T = mg + mv^2/r = 0.55(9.81 + 11^2/7.5) = 14.3N. 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted May 1, 2011 Report Share Posted May 1, 2011 taken from SL past paper Nov 07the answer is C. 8 but idk why.what formula should I use? I tried W=Fs cosθ but I don't think I'm doing the right thing Reply Link to post Share on other sites More sharing options...
genepeer Posted May 1, 2011 Report Share Posted May 1, 2011 (edited) You were in the right direction. Initial velocity is 0, sov2 = 2ass = v2/2aand F = ma.Work Done by F1 = (m a1)(v2/2a1) = mv2/2Work Done by F2 = (2m a2)((2v)2/2a2) = 8mv2/2Work Done by F2/Work Done by F1 = 8Oh my, I just reinvented the wheel. Work Done by a force to move a body from rest is equal to the body's kinetic energy. Edited May 1, 2011 by Gene-Peer 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted May 1, 2011 Report Share Posted May 1, 2011 so basicly I just need to use W=½mv²?and this v²=2as formula, is it in the data booklet? I totally forgot.. and used v=s/t instead thank you! Reply Link to post Share on other sites More sharing options...
genepeer Posted May 1, 2011 Report Share Posted May 1, 2011 so basicly I just need to use W=½mv²?Yup.about the formula, don't know about the data booklet but you may want to know these three by heart:v = u + ats = ut + at2/2v2 = u2 + 2as Reply Link to post Share on other sites More sharing options...
Rigel Posted May 1, 2011 Report Share Posted May 1, 2011 (edited) This one also (The least used):s = ((v+u)/2)*t Edited May 1, 2011 by Ipos Manger Reply Link to post Share on other sites More sharing options...
ElLos Posted May 3, 2011 Report Share Posted May 3, 2011 I need help on this problem please (Exam is in 4 hours!!!):Thank you IBS! Reply Link to post Share on other sites More sharing options...
Drake Glau Posted May 4, 2011 Author Report Share Posted May 4, 2011 This drops increase in distance between them.I feel like you can do B...If it drops every 5.6s you can take 5.6/0.8=7 drops then count to the 7th drop and see how far away that is.For Bi you can take the distance between the last two drops and that took 0.8sec and your speed is d/tBii, use one of the suvat equations. You have 3 of the variables now so you can always use the them Reply Link to post Share on other sites More sharing options...
ElLos Posted May 4, 2011 Report Share Posted May 4, 2011 This drops increase in distance between them.I feel like you can do B...If it drops every 5.6s you can take 5.6/0.8=7 drops then count to the 7th drop and see how far away that is.For Bi you can take the distance between the last two drops and that took 0.8sec and your speed is d/tBii, use one of the suvat equations. You have 3 of the variables now so you can always use the them Well i got a. and b.But, when i count the dots and work out the distance, i get something aroudn 50-60, but, the markscheme states that the distance is 37.6m.. Reply Link to post Share on other sites More sharing options...
genepeer Posted May 4, 2011 Report Share Posted May 4, 2011 I'm getting the distance as 57.6m (14.4cm on grid). I think that was a typo meant to be a '5' not a '3'. Reply Link to post Share on other sites More sharing options...
Chronofox Posted May 8, 2011 Report Share Posted May 8, 2011 (edited) A converging mirror has a focal length of 15 cm. Where would you place an object in order to produce an erect virtual image twice as tall as the object?So using the magnification formula for concave mirrors, hi / ho = -di / do (where i = image and o = object), I get that hi / ho = 2, so that means -di / do must equal 2 as well. What do I do next? Edited May 8, 2011 by Chronofox Reply Link to post Share on other sites More sharing options...
Drake Glau Posted May 8, 2011 Author Report Share Posted May 8, 2011 A converging mirror has a focal length of 15 cm. Where would you place an object in order to produce an erect virtual image twice as tall as the object?So using the magnification formula for concave mirrors, hi / ho = -di / do (where i = image and o = object), I get that hi / ho = 2, so that means -di / do must equal 2 as well. What do I do next?Converging lens, convex right? The only way to get a virtual image with one is to have the object inside the focal length and for it to be twice as big it needs to be at the 0.5f mark I think. But then you said concave in your next part? I'm so confused Reply Link to post Share on other sites More sharing options...
Chronofox Posted May 8, 2011 Report Share Posted May 8, 2011 A converging mirror has a focal length of 15 cm. Where would you place an object in order to produce an erect virtual image twice as tall as the object?So using the magnification formula for concave mirrors, hi / ho = -di / do (where i = image and o = object), I get that hi / ho = 2, so that means -di / do must equal 2 as well. What do I do next?Converging lens, convex right? The only way to get a virtual image with one is to have the object inside the focal length and for it to be twice as big it needs to be at the 0.5f mark I think. But then you said concave in your next part? I'm so confused Converging is actually concave. Reply Link to post Share on other sites More sharing options...
Drake Glau Posted May 8, 2011 Author Report Share Posted May 8, 2011 No? Reply Link to post Share on other sites More sharing options...
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