Maximum Velocity
A shell flying with velocity 500 m/s bursts into three identical fragments so that the kinetic energy of the system increases 1.5 times. What maximum velocity can one of the fragments obtain?
Labels: funphysics
posted by Rajesh Lal at
1:33 PM
15 Comments:
Is it me or the questions are becoming very difficult now a days :(
Kinetic Energy: T=mv^2 -> Kin.Energy after: T'=3/2*mv^2
=3/2(m1v1^2+m2v2^2+m3v3^2)
-- v2 and v3 eq. 0 -->
v1^2=3v^2 --> v1 =Sqrt(3*v)=866m/s
The maximum velocity is 866m/s
last line should be : v1=Sqrt(3*v^2)=866m/s
This post has been removed by the author.
This post has been removed by the author.
This post has been removed by the author.
I don't understand the "burst". a bullet might break into three pieces, and the velocity could not increase. But if you mean a secondary energy impulse, or explosion, like and exploding bullet (like a rocket's second stage), then you would need to know how much energy was added to figure the resulting velocity.
The trick is to recognise that two of the fragments must be ejected relatively backwards and the third relatively forwards. Also, conservation of momentum must be honoured.
Let v = 500 m/s, f be the forward gain of speed of one fragment and b the the backward loss of speed of the other two fragments. Also let each fragment have mass m (so initial mass is 3m).
momentum conservation => 3mv = m(v+f) + 2m(v-b)
twiddling about => b = f/2
KE after explosion = (m/2)(v+f)^2 + 2(m/2)(v-b)^2
using b=f/2 and twiddling about =>
KE after = (3m/4)(2v^2 + f^2)
But KE after = (5/2) * KE before
and KE before = (3/2)mv^2
Equating and twiddling about => 3v^2 = f^2 => f = v sqrt(3)
Fastest frag has speed v + f = v(1 + sqrt(3)) = 1366 m/s approx.
Just for interest, the speed of the two slowed down particles
= 67 m/s approx.
there are 2 ways of solving this:
1.if the conesrvation of momentum must be honoured(which i believe is true)
then my calcucaltions give the biggest velocity of 1000 while the other 2 being 250.chris i believe your calculations are not well conducted because the KE of the part with the velocity of 1366 would be higher than 3/2*KE in the begining.
2.if it must not be honoured(because of wind etc)then the result is V1=sqert(V^2*9/2)=1060.66
Hi milos.
The problem stated that the KE increased by 1.5 times. I took this to mean add 1.5 times the original KE, so I used 1+1.5 = 5/2
That was my (only) bad. I can no longer see why that seemed sensible at the time.
The only change you need to my calculations, is to change the line "But KE after = (5/2) * KE before" to "But KE after = (3/2) * KE before".
Then following through, gives v = f = 500, so fastest fragment has speed 1000 m/s. And the other two fragments have speed 250 m/s, just as you say.
I don't see a way of doing the calculations without explicitly invoking the law of conservation of momentum. So I would be interested in how you did it.
I certainly didn't allow for wind drag as I took it that the assumed explosion was over and done with so quickly that the energy and momentum changes due to drag were negligible, and not simply overlooked.
Thanks for spotting a problem with my values.
Ragknot - r u having an off day or is there a fake Ragknot out there?
i quite agree with u in that calculation as u can see in my post i used the conservation of momentum.in the other one u can get the other result but as i stated i believe that the first one is the right one.
Sorry milos, I misinterpreted your gist in your 6:50 am post. You seemed to be suggesting that conservation of momentum was optional. I was having a senior moment.
...again.
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