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

AI Thread Summary
In an inelastic collision where a bullet embeds itself in a wood block, the bullet's mass is 0.049 kg and its initial speed is 139 m/s, while the block, weighing 0.54 kg, is at rest. The combined velocity of the block and bullet immediately after the collision is calculated to be approximately 11.56 m/s. The discussion highlights the importance of considering the bullet and block as part of the system while neglecting external forces like gravity and friction due to the short duration of the collision. Participants express confusion over which forces can be neglected and the validity of their assumptions regarding the collision dynamics. Clarifying these assumptions is essential for understanding the momentum conservation principle in this scenario.
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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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