Bullet hitting a block of wood

  • Thread starter Ry122
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  • #1
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Homework Statement


A bullet leaves the muzzle of a rifle at an unknown velocity and strikes a wooden block that is suspended by a piece of string.
If the block swings backwards to make an angle of 40 degrees, at what velocity must have the bullet been travelling when it struck the block?
mass of the bullet and the wood are, respectively, m=0.05 kg and M=2 kg, and the length of the string is L=1 m. Theta = 40 degrees
[PLAIN]http://img810.imageshack.us/img810/5295/bullet.jpg [Broken]


Homework Equations





The Attempt at a Solution


The block is given kinetic energy equal to the kinetic energy the bullet had.
Since the block experiences no friction, all of this kinetic energy is used up by moving against gravity.
s and f can be determined and so the energy can be calculated
with f x s = energy
Then use (1/2)mv^2 = energy to calculate the speed of the bullet

The problem is that I'm unsure how to calculate s. How can this be done? (s = vertical displacement of the block)
 
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Answers and Replies

  • #2
kuruman
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The block is given kinetic energy equal to the kinetic energy the bullet had.
Incorrect. See below.
Since the block experiences no friction, all of this kinetic energy is used up by moving against gravity.
The block experiences no friction only after the bullet comes to rest inside it. While the bullet is slowing down inside the block there is a lot of friction. So some of the initial kinetic energy of the bullet goes into heat generated by friction.
The problem is that I'm unsure how to calculate s. How can this be done? (s = vertical displacement of the block)
Can you find how fast the bullet+block system is moving after the bullet becomes fully embedded?
 
  • #3
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@ kuruman, i think at a high school/ introductory physics level we're allowed to assume that energy is conserved within the bullet-block system.

The solution is simple. It is a mere calculation using the conservation of energy.

0.5*m*v^2 = m*g*h

Theta is given to be 40 degrees. So in order to calculate the rise in the height of the block,
Lcos40 = h
g is a constant. The rest are all given.
working that out you should get
V = 24.5 m/s ( 3sf )
 
  • #4
Doc Al
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@ kuruman, i think at a high school/ introductory physics level we're allowed to assume that energy is conserved within the bullet-block system.
Absolutely not! You've missed a key point of the exercise.

The solution is simple. It is a mere calculation using the conservation of energy.
Simple, but incorrect. Listen to kuruman!

0.5*m*v^2 = m*g*h

Theta is given to be 40 degrees. So in order to calculate the rise in the height of the block,
Lcos40 = h
g is a constant. The rest are all given.
working that out you should get
V = 24.5 m/s ( 3sf )
Let the OP do their own work please.
 
  • #5
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Ry122, do you know what the answer to the question is?
 

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