How Do You Calculate the Velocity After a Bullet Embeds in a Block?

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SUMMARY

The discussion focuses on calculating the velocity of a bullet embedded in a wood block after an inelastic collision. The bullet, with a mass of 0.049 kg and an initial speed of 139 m/s, collides with a stationary block of mass 0.54 kg. The resulting velocity of the combined bullet and block system immediately after the collision is determined to be approximately 11.5637 m/s. Participants also debated the assumptions made regarding the neglect of gravitational and frictional forces during the brief collision period.

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Westin
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Poster informed about mandatory use of homework template
A bullet is fired directly into a wood block (along the x-axis), which rests on a rough table. The bullet embeds itself in the block. (This is an inelastic collision: the block deforms and the block and bullet heat up as a result of the collision).

a) If the bullet has a mass 0.049kg and is initially moving with a speed of 139m[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngs, while the block has a mass 0.54kg and is initially at rest, determine the velocity of the block and bullet just after the collision.

v[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char7E.png=[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmex10/alpha/144/char1C.png 11.5637 m/s , 0 m/s , 0 m/s https://s3.lite.msu.edu/adm/jsMath/fonts/cmex10/alpha/144/char1D.png

You are correct.
Your receipt no. is 147-6323

b) To determine the velocity of the bullet and block, what did you include in your system?
True: The bullet
True: The block
False: The table

You are correct.
Your receipt no. is 147-1738

c) Which of the following is true about your assumptions? (Be careful -- some are incorrect because the physics is incorrect!)

Select all that are True.
1)I neglected the gravitational force because the collision was short.
2)I neglected the gravitational force because it point in the negative y-direction and therefore doesn't affect the momentum in the x-direction.
3)I neglected the frictional force due to the table because the collision was short.
4)I neglected the upward force of the table force because it points in the y-direction and therefore doesn't affect the momentum in the x-direction.
5)I neglected the upward force of the table force because the collision was short.

Incorrect. Tries 2/3

I am confused with last question. To neglect a force since the collision was only .037s seems very iffy to me. Previous attempts were I said all were true, and #2, #4 were true. Both were wrong. My 3rd attempt is telling me #1 #3 #5 are true, what do you think?
 
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I would think that you are only guessing at this point. For each alternative, try to argue why it is true or false and we might be able to help you clear up misunderstandings rather than just giving you the answers.
 

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