Recent content by digalumps

Thanks that makes sense - basically I have added 0.5 + 0.05 =0.55 joules from my muscles. Thank you.

Ok my first working out. An infinitely light person is sitting on top of a 'stationary' 10kg rocket in space KE = 0 (v=0) Then person throws a 1kg bowling ball of the back at 1 m/sec Conservation of momentum. m1v1=m2v2 Thus the rocket moves off in opposite direction at 0.1 m/sec KE=1/2 mv...

Thank you very much I will work on this now.

Or it could just be that I have not reached the place in my studies where I am ABLE to calculate what you are asking. If I could then I would. Everyone has to start somewhere. This is a basic homework thread, not a masterclass in differential calculus. I did do calculations to the limit of my...

I will have one last try then shut up as I don't want to waste people's time. I come home one morning after shopping. "Hey son, I got a 2 for 1 deal on rockets this morning.." Great Dad let's go down to the disused airfield and try them on my bike. (At airfield) OK son, attach both rockets to...

Why doesn't it? The same force is applied and therefore the same acceleration? a=f/m.

First of all a big thank you for everyone who posts on here it is incredibly kind of you to give up your time this way. I mean that. Also, although there have been many 'complicated' answers including calculus, advice to run the equations etc this is not pure mathematics or string theory where...

Hi - I did not say the work was the same as work=force x distance and the distance is more in the second case. I just could not understand intuitively how if you attach the same rocket to the same mass and burn it for eg 1 second in the two cases and get an acceleration the same (f=ma - nothing...

Hi everyone thank you for taking the time and trouble to reply. I still do not really understand but perhaps it is because I am just starting out in Physics maybe one day I will. My rocket produces F and thus accelerates my mass by a=F/m (F=ma) From 10->11 m/sec or 1000->1001 m/sec If I were...

Thank you for your help and I understand the equations above. I think what I don't understand in the intuitive bit - the thought experiment if you will. I accelerate the same mass by the same amount for the same time (1 sec) and get different KE increase depending on the initial velocity. The...

I think so - the work done by the 1 Newton rocket is used to increase the kinetic energy. The work done is the same in both scenarios (the rocket burns for 1 second and accelerates the mass) but the KE increases are not the same.

I have a problem regarding Kinetic Energy which as we know is 1/2 m v squared. Say I have a 1kg mass moving at 10 meters/second. I have a 1 Newton rocket which I attach to the back and it burns for 1 second accelerating the mass by 1 m/sec/sec to 11 m/sec. The KE originally was 50 joules and it...