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Target in a shooting gallery

  • Thread starter bjonesp
  • Start date
2
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A target in a shooting gallery consists of a vertical square wooden board, 0.220m on a side and with mass 0.800 kg, that pivots on a horizontal axis along its top edge. The board is struck face-on at its center by a bullet with mass 1.80g that is traveling at 320m/s and that remains embedded in the board.
A)What is the angular speed of the board just after the bullet's impact?

B)What maximum height above the equilibrium position does the center of the board reach before starting to swing down again?

C)What minimum bullet speed would be required for the board to swing all the way over after impact?

I'm just a little confused about how to approach this problem. I wanted to know if I am on the right track and maybe where I should go with my information.

I started off finding the KE of the bullet before it strikes the board. Using KE = (1/2)mv^2 I got (1/2)(.0018kg)(320 m/s)^2 to get KE = 92.16 and I took this to be the force on the board that caused it to rotate

Then I wanted to know the Torque about the square so I used the equationg Radius X Force. I got (.110m) X (92.16 J) = 10.1376

I'm not exactly sure where to take this information. I thought about using Conservation of Energy but I dont know the PE of the system afterwards because I don't know how high it rises. If someone could just point in the right direction to find (A) I would greatly appreciate it. thank you
 

Answers and Replies

Shooting Star
Homework Helper
1,975
4
It is not the KE that is conserved at the moment of impact but the angular momentum.

Equate the initial angular momentum of the bullet to the final ang mom of the system just after impact, and do the calc based on this. This will give you the initial KE of the system. From that, you can find out how much the board rises using consvn of energy.
 

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