How Fast Was the Bullet Going if the Block Rises 75cm Above the Table?

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Homework Help Overview

The problem involves a 2kg block that is shot with a 15g bullet, resulting in the block rising 75cm after the bullet lodges in it. The discussion centers around the application of conservation of momentum and conservation of energy principles to determine the bullet's initial velocity.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to apply conservation of momentum for the bullet-block collision and conservation of energy for the block's rise. There are attempts to set up equations for both parts of the problem, with some questioning the relationship between the variables involved.

Discussion Status

Participants are actively engaging with the problem, exploring different equations and concepts. Some have provided guidance on how to approach the problem, suggesting the separation of the momentum and energy equations. There is a sense of progress as one participant expresses understanding after rearranging the equations.

Contextual Notes

There are two unknowns in the equations being discussed, which adds complexity to the problem-solving process. Participants are navigating the challenge of substituting variables to find a solution.

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Homework Statement


question: a 2kg block rests over a small hole in a table. janelle is breathe the table and shoots a 15g bullet through the hole and into the block, where it lodges. How fast was the bullet going if the block rises 75cm above the table?




Homework Equations


m1v1i +m1v2i = m1v1f + m2v2f


The Attempt at a Solution


i tired working this out with the straight line motion equation of s= vt -at^{2}
but this did not workout

i tired solving for t but got .153s
from there i am lost
 
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There are 2 parts to the problem...

1) bullet being fired into block... this requires conservation of momentum.

2) bullet+block rising up to 75cm... this requires conservation of energy.

try to write the equations for part 1) and part 2)... then you should be able to solve for the initial velocity of the bullet.
 
okay so m1v1i = m1v1f + m2v2f
but (because they are combined and therefor have the same combined mass and velocity), couldn't it be written as: m1v1i = (m1 + m2) vf


the conservation of energy? as in Ke=Ep or 1/2 mv(squared) = mgh?
or as in initial Ke= final Ke?

there are two unknows here. i guess what your saying is work them (the the separate equations) out and substitute it for one unknown. but could you please give me a bit of guidance? one i see it once i'll be good from there on.
 
Last edited:
turnip said:
okay so m1v1i = m1v1f + m2v2f
but (because they are combined and therefor have the same combined mass and velocity), couldn't it be written as: m1v1i = (m1 + m2) vf


the conservation of energy? as in Ke=Ep or 1/2 mv(squared) = mgh?
or as in initial Ke= final Ke?

there are two unknows here. i guess what your saying is work them (the the separate equations) out and substitute it for one unknown. but could you please give me a bit of guidance? one i see it once i'll be good from there on.

you're almost there:

equation 1:

m1v1i = (m1 + m2)vf

equation 2:

(1/2)(m1 + m2)vf^2 = (m1 + m2)gh (as Astronuc mentioned, all the energy is converted to gravitational potential energy)

solve these two equations... you can cancel the masses in the second equation. your two unknowns are v1i and vf...
 
right, that makes a lot of sense to me now
just rearranged for vf and subs into v1i and i get the right answer
thanks to both of you
 

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