Find the speed of the bullet as it emerges from the block

In summary, the problem involves a 10.0g bullet being shot through a 1.0 kg wood block suspended on a string 2.0m long. The center of mass of the block rises a distance of 1.0cm. The task is to find the speed of the bullet as it emerges from the block if its initial speed is 450m/s. The equations that can be used are linear momentum and conservation of energy. After some confusion, the correct equation is determined to be mgh + 1/2mv^2 = mgy + 1/2mv^2. This is because the bullet has to spend its initial kinetic energy on raising the wood block and bullet, resulting in a lower
  • #1
innightmare
35
0

Homework Statement


1. Homework Statement [/b]
A 10.0g bullet is shot through a 1.0 kg wood block suspended on a string 2.0m long. The center of mass of the block rises a distance of 1.0cm. find the speed of the bullet as it emerges from the block if its initial speed is 450m/s.

Homework Equations


I think this has to do with linear momentum, p(i)=p(f)(mv=mv). Either that or conservation of energy: mgh +1/2mv^2=mgy +1/2mv^2

The Attempt at a Solution


Im confused as to how to go about solving this problem. for the initial mv, I plugged in the values for the bullets mass and used 450 m/s as the initial velocity. for the final mv, i plugged in the the mass value for the block and bullet, and used that to divide my initial mv, which was good for linear momentum but still not good enough, i don't think its right. so when i use the conservation of energy,i know my h=2.0m and y=.01m

Am I on the right track. Which equation should i use? Thanks[
 
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  • #2
Okay think about you started out with this much energy then you had to spend that energy on other things then how much energy do you have left?
 
  • #3
I think its this equation: mgh +1/2mv^2=mgy +1/2mv^2, using my kinetic energy as my velocity
should be:(I THINK) v=radical of 2*g*h

Is this correct?
 
  • #4
okay so, you start out with kinetic energy of the bullet then you hit the wood of certain mass, that slows down the bullet, but it slows down by raising the wood and bullet up some hight. then after that the bullet comes out of the wood at lower kinetic energy thus comes out slower then ititiallly started. Can you form an equation from that info?
 
  • #5
Is it mv(initial)=mv(final)?
 
  • #6
Antineutron said:
slows down the bullet, but it slows down by raising the wood and bullet up some hight.

No. Read again, this part. It must have done some work? What did the bullet do work on? What kind of collision is it when this occured?
 
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1. How is the speed of the bullet calculated?

The speed of the bullet can be calculated using the equation v = √(2gh), where v is the velocity, g is the acceleration due to gravity, and h is the height of the block.

2. What factors can affect the speed of the bullet as it emerges from the block?

The speed of the bullet can be affected by factors such as the mass and shape of the bullet, the type of gun used, the angle at which the bullet is fired, and air resistance.

3. Can the speed of the bullet be accurately determined?

Yes, the speed of the bullet can be accurately determined by conducting multiple trials and using precise measurement tools such as a chronograph.

4. How can the speed of the bullet be increased?

The speed of the bullet can be increased by using a more powerful gun, reducing air resistance by using a streamlined bullet, or increasing the angle at which the bullet is fired.

5. What are the real-world applications of determining the speed of a bullet?

Determining the speed of a bullet can be useful in forensic investigations, improving shooting accuracy and performance, and designing bullet-resistant materials for protection against firearms.

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