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AP Physics Help

  • Thread starter rukawa1107
  • Start date
  • #1
A 1250 kg cannon, which fires a 55 kg shell with a speed of 566 m/s relative to the muzzle, is set at an elevation angle of 39° above the horizontal. The cannon is mounted on frictionless rails, so that it recoils freely.

What is the speed of the shell with respect to the Earth?

Vse=Vsm+Vme, Vme=Vse-Vsm, Vsm= 566m/s

MVi= 0 = 55kg(Vse)+1250kg(Vse-566m/s)

so, Vse=542m/s and Vme = 542m/s-566m/s = -24m/s

I have solved the first part by using the conservation of momentum.

However, I have some diffculty on the second part.


At what angle with the ground is the shell projected?

This angle of the shell relative to the ground can be given by the ratio

calculation however im not so sure.

Can anyone please solve and show me how to solve the second part?
 

Answers and Replies

  • #2
Doc Al
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Careful: Momentum is conserved in the horizontal direction only.

Once you correctly find the recoil speed of the cannon, you can find the velocity of the ball with respect to the earth via:
[tex]\vec{V}_{b/e} = \vec{V}_{b/c} + \vec{V}_{c/e} [/tex]

(It seems that Latex is not displaying. To find the velocity of the ball with respect to the earth find the vector sum of the velocity of the cannon wrt earth plus the velocity of the ball wrt the cannon.)
 
Last edited:

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