Solve Challenging Physics Question: Block + Bullet UP Motion

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A university-level physics problem involves a 3.00 kg block of wood and a 32.0 g bullet fired vertically upward at 415 m/s. The discussion focuses on determining the speed of the block and bullet after the collision, calculating their total kinetic energy, and finding the maximum height they reach together. Participants emphasize the importance of showing attempted solutions before seeking help, adhering to forum guidelines. One user provides a worked solution, calculating the post-collision speed as approximately 4.38 m/s and the total kinetic energy as 29.1 J, along with the maximum height of 0.98 m. The conversation highlights the collaborative nature of problem-solving in academic forums.
Willtd
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This question is one i got from uni. i just can't seem to get the right answers. Please post your worked solutions, i will tell you if you get the answer found on my sheet. Best of luck Guys, you might need it.

A 3.00 kg block of wood is suspended at rest by a thin thread attached to a supporting beam. A bullet of mass 32.0 g is fired vertica1ly upwards with velocity 415 m/s into the wood directly above it. The block and the bullet inside it move upwards together.


1 Determine the speed of the upward moving block and bullet immediately after the collision.
.
2 Calculate the total kinetic energy of the block and bullet immediately after the collision.

3 Calculate the maximum height the block and bullet reach.
 
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Sounds like a straightforward homework problem to me! :wink:

Why don't you show what you've done so far?
 
Welcome to the Forums Willtd,

According to our forum guidelines, one is expected to show one's own efforts before asking for help. So if you post your attempted solution, we might be able to show you where've you've gone wrong.

In addition, we have dedicated Homework & Coursework forums for such questions.

Edit: The Doc Strikes again.
 
Do you think you are some kind of a professor? hehe

Please obey the guidelines of these forums.
 
My Attempt

I got it! Here is a solution for anybody who wants it. :)

1. m1v1 = m2v2
.032x415 = 3.032v
v = 4.379

2. Total Ke = 0.5 x m x v^2
= 0.5x 3.032x4.38^2
= 29.1
3. v^2 = u^2 +2as
v (final speed) = 0
s = sqrt [(4.38^2) / (2x9.8)]
= 0.98m
 
hehe.. are you joking with us? :)
 
Yeh man ur cool

malawi_glenn said:
hehe.. are you joking with us? :)

Maybe you should help next time m8 :D
 
Willtd said:
Maybe you should help next time m8 :D

1. you did not show us your work so far

2. You wished us good luck, and said that we would need it.

That is not an attitude that I and others don't like. There is a guideline here that states that you shall show us work done so far, so we can HELP, not just solve the problem for you. That is probably why no one helped you this time, and you will meet much much more difficult problems than this one in the future...
 
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