Bullet passing through block of wood. Find max height of the wood.

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SUMMARY

The problem involves a bullet of mass 0.005 kg shot at a suspended block of wood with mass 1 kg, initially traveling at 200 m/s. The bullet passes through the block, doing 50 J of work in the process. To find the maximum height the block will rise after the bullet passes through, the formula used is (v * mb / (mb + mw))^2 * 1/2g, resulting in a maximum height of 0.05 m. This calculation incorporates the conservation of momentum and the work-energy principle.

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demenius
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Homework Statement


A bullet of mass 0.005kg is shot at a suspended block of wood with mass 1kg. The bullet initially has a speed of 200m/s, and passes through the block of wood. The bullet passes through the wood, and 50J of work is done deforming the shape of the bullet and block. If the block is suspended from a very long string, what is the maximum height to which it will rise?

http://imageshack.us/photo/my-images/266/bulletblock.png/"

Homework Equations


None given.


The Attempt at a Solution


I tried solving it using momentum before and after. But there is no mass given for the block and I do not know what to do with the 50J of work.
 
Last edited by a moderator:
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demenius said:

Homework Statement


A bullet of mass 0.005kg is shot at a suspended block of wood with mass 1kg. The bullet initially has a speed of 200m/s, and passes through the block of wood. The bullet passes through the wood, and 50J of work is done deforming the shape of the bullet and block. If the block is suspended from a very long string, what is the maximum height to which it will rise?

http://imageshack.us/photo/my-images/266/bulletblock.png/"

Homework Equations


None given.


The Attempt at a Solution


I tried solving it using momentum before and after. But there is no mass given for the block and I do not know what to do with the 50J of work.

In red
 
Last edited by a moderator:
I cannot believe I did not see that. :S. Thank you.
 
So would the height be equal to (v*mb/(mb+mw))^2 * 1/2g?

So. (200*0.005/(0.005+1))^2 *1/2(9.81) = 0.05m?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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