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Conservation of energy, determining the speed from the change in position

  • Thread starter zakare
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  • #1
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Homework Statement



A bullet with a masse (m) is shot into a block of wood with a masse (M) at a velocity of (v). The bullet is lodged into the block of wood which is attached to a string of (l) length. After the impact where the velocity of the wood-bullet mass is equivalent to zero, the string forms an angle (a) with the vertical axis.

m = 0.02kg
M = 15kg
angle a = 20 degrees
l = 1 meter
v = ?m/s

Homework Equations



E(mec initial) = E(mec final) = mgh + 1/2*mv²

The Attempt at a Solution



E(mec final) = E(mec initial)
1/2*mv² = (m + M)gh

v = sqrt((2(M + m)gh)/m)

h = 2sin²(a/2)

and I get the answer: 42m/s
The correct answer is 825m/s.

I'd appreciate any help. Thanks
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

E(mec final) = E(mec initial)
1/2*mv² = (m + M)gh
Hi zakare! Welcome to PF! :smile:

This is not an elastic collision.

The bullet embeds in the block, and energy is lost.

You can only use conservation of momentum for the collision (but you can use conservation of energy after the collision).

Use momentum to find the velocity immediately after the collision, then use energy! :smile:
 
  • #3
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To reiterate what Tiny Tim just said, you need to see up two equations. One for conservation of mechanical energy after the collision and one for conservation of momentum during.

By doing so, you can solve for both the initial velocity of the bullet and final velocity.

Also, "1/2*mv² = (m + M)gh" should be [tex]\frac{1}{2}(m+M)v^{2}_{f}=(m+M)gh[/tex] since the bullet is lodged in the wood block so the mass is the total mass of the bullet and the wood block.

Also, how did you get h? You are given [tex]\theta[/tex] with the vertical, l=length of the string. Now find h (the height reached at the top of the swing) using a simple trig identity.
 
Last edited:
  • #4
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I don't see why I should use momentum because at the end of the sequence the bullet-block combination is motionless (vfinal = 0) and the angle is the angle max.

And to konthelion:
Why wouldn't it be what I wrote if the initial velocity of the bullet is what contains all the energy of the system (in kinetic form), and afterwards all the energy is in potential gravitational form...
 
  • #5
tiny-tim
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Homework Helper
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Hi zakare! :smile:
I don't see why I should use momentum because at the end of the sequence the bullet-block combination is motionless (vfinal = 0) and the angle is the angle max.
You always have conservation of momentum in collisions.
… afterwards all the energy is in potential gravitational form...
No it isn't!

Some of the energy has gone into the bullet pushing into the block, also heat and sound.

Energy is not conserved.
 

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