- #1
devanlevin
a cannon with a mass of M is let free from the top of a slope with an incline of [tex]\vartheta[/tex] the friction between the cannon and the slope is given as [tex]\mu[/tex]N.
after the cannon has passed a dstance of L it shoots a shell, with a mass of m horizontally, what is the minimal velocity of the shell so that the cannon will stop,?
im not sure if this means that M is the mass of the cannon and the shell, or of the cannon alone, but form the answer i think that they mean that it is just the mass of the cannon, and somehow they don't take the mass of the shell into consideration until its shot.
the answer is
Umin=(M+m)/m(cos[tex]\vartheta[/tex]-[tex]\mu[/tex]sin[tex]\vartheta[/tex])*[tex]\sqrt{2Lg(sin[tex]\vartheta[/tex]-[tex]\mu[/tex]cos[tex]\vartheta[/tex])}[/tex]
what i did was,
I)drew a diagram,-- http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5278474258602595442
II)found the cannons acceleration
III)found its velocity after it had passed L
IV)found the minimal Velocity of th shell
II)to find the acceleration
using[tex]\sum[/tex]F=ma=mgsin[tex]\vartheta[/tex]-[tex]\mu[/tex]N
acceleration downhill a=g(sin[tex]\vartheta[/tex]-[tex]\mu[/tex]cos[tex]\vartheta[/tex])
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III)to find the cannons velocity
V^2=2a[tex]\Delta[/tex]x
velocity downhill V=[tex]\sqrt{2Lg(sin[tex]\vartheta[/tex]-[tex]\mu[/tex]cos[tex]\vartheta[/tex])}[/tex]
----------------------------------------------------------------------
IV) to find the minimum shooting velocity
(M+m)V=mu*cos[tex]\vartheta[/tex]+MU (when MU=0)
u=V(M+m)/mcos[tex]\vartheta[/tex]
which would be my final answer except that the answer according to the book is
u=V(M+m)/m(cos[tex]\vartheta[/tex]-[tex]\mu[/tex]sin[tex]\vartheta[/tex])
where does this extra bit come from? where have i gone wrong?
after the cannon has passed a dstance of L it shoots a shell, with a mass of m horizontally, what is the minimal velocity of the shell so that the cannon will stop,?
im not sure if this means that M is the mass of the cannon and the shell, or of the cannon alone, but form the answer i think that they mean that it is just the mass of the cannon, and somehow they don't take the mass of the shell into consideration until its shot.
the answer is
Umin=(M+m)/m(cos[tex]\vartheta[/tex]-[tex]\mu[/tex]sin[tex]\vartheta[/tex])*[tex]\sqrt{2Lg(sin[tex]\vartheta[/tex]-[tex]\mu[/tex]cos[tex]\vartheta[/tex])}[/tex]
what i did was,
I)drew a diagram,-- http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5278474258602595442
II)found the cannons acceleration
III)found its velocity after it had passed L
IV)found the minimal Velocity of th shell
II)to find the acceleration
using[tex]\sum[/tex]F=ma=mgsin[tex]\vartheta[/tex]-[tex]\mu[/tex]N
acceleration downhill a=g(sin[tex]\vartheta[/tex]-[tex]\mu[/tex]cos[tex]\vartheta[/tex])
-----------------------------------------------------------------------
III)to find the cannons velocity
V^2=2a[tex]\Delta[/tex]x
velocity downhill V=[tex]\sqrt{2Lg(sin[tex]\vartheta[/tex]-[tex]\mu[/tex]cos[tex]\vartheta[/tex])}[/tex]
----------------------------------------------------------------------
IV) to find the minimum shooting velocity
(M+m)V=mu*cos[tex]\vartheta[/tex]+MU (when MU=0)
u=V(M+m)/mcos[tex]\vartheta[/tex]
which would be my final answer except that the answer according to the book is
u=V(M+m)/m(cos[tex]\vartheta[/tex]-[tex]\mu[/tex]sin[tex]\vartheta[/tex])
where does this extra bit come from? where have i gone wrong?
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