Target in a shooting gallery

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In summary, a bullet with a mass of 1.80g traveling at 320m/s strikes a vertical square wooden board with a mass of 0.800 kg, causing it to pivot on a horizontal axis along its top edge. To find the angular speed of the board after the impact, the initial angular momentum of the bullet can be equated to the final angular momentum of the system. Using conservation of energy, the maximum height the board reaches above its equilibrium position can also be determined. To find the minimum bullet speed required for the board to swing all the way over after impact, the initial kinetic energy of the system can be calculated.
  • #1
bjonesp
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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 don't 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
 
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  • #2
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.
 
  • #3


Hi there,

You are on the right track with your approach to this problem. Let's break down the questions one by one and see how we can use the information you have already calculated.

A) To find the angular speed of the board just after the bullet's impact, we need to use the conservation of angular momentum. We know that before the impact, the bullet has a linear momentum of 92.16 J (as you calculated above). After the impact, this momentum is transferred to the board, causing it to rotate. We can use the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed. We know the moment of inertia of a square board is (1/3)MR^2, where M is the mass of the board and R is the length of one side. So, we can plug in the values and solve for ω. We get:

L = Iω
92.16 = (1/3)(0.8 kg)(0.110 m)^2 x ω
ω = 1500 rad/s

B) To find the maximum height above the equilibrium position, we can use conservation of energy. We know that the bullet transfers its kinetic energy to the board, causing it to rotate. This energy is then converted into potential energy as the board rises. We can use the equation PE = mgh, where m is the mass of the board, g is the acceleration due to gravity, and h is the height. We can set the initial kinetic energy of the bullet equal to the potential energy of the board at its highest point. So, we get:

KE = PE
92.16 J = (0.8 kg)(9.8 m/s^2) x h
h = 11.837 m

C) To find the minimum bullet speed required for the board to swing all the way over after impact, we can use conservation of energy again. This time, we need to consider the total energy of the system, which includes the kinetic energy of the bullet and the potential energy of the board at its highest point. We can set this total energy equal to the potential energy of the board at its highest point when it is swinging over. So, we get:

KE + PE = PE
(1/2)(0.0018 kg)(v^2) + (0.8 kg)(9.
 

1. What is a "target" in a shooting gallery?

A target in a shooting gallery is a designated area or object that is used as a goal for players to aim and shoot at. It can be made of various materials such as paper, clay, or metal.

2. How is a "target" used in a shooting gallery?

In a shooting gallery, a target is typically placed at a distance and players use a gun or other projectile device to try and hit the target. The closer the player gets to the center of the target, the more points they score.

3. What types of targets are commonly used in shooting galleries?

There are many different types of targets used in shooting galleries, including stationary targets, moving targets, and interactive targets. Some common examples include bullseye targets, spinning targets, and digital targets with sound effects and lights.

4. Are there any safety precautions for using targets in a shooting gallery?

Yes, safety precautions should always be taken when using targets in a shooting gallery. These may include proper instruction and supervision, wearing protective gear, and following safety rules such as keeping the gun pointed in a safe direction and never pointing it at anyone.

5. Can targets in a shooting gallery be customized?

Yes, targets in a shooting gallery can be customized in various ways. This can include changing the size, shape, and material of the target, as well as adding personal designs or logos. Customized targets can add a unique and personalized touch to the shooting experience.

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