Ballistics Pendulum Homework: Solving for Velocity After Bullet Emerges

AI Thread Summary
A 23 g bullet traveling at 235 m/s penetrates a 3.8 kg block of wood and exits at 195 m/s. The problem requires calculating the block's velocity after the bullet emerges, using conservation of momentum. The relevant equation is mv = (m+M)v', which equates the momentum before and after the bullet's exit. The initial attempt yielded velocities of 1.14 m/s and 230 m/s, but further clarification suggests that only the first equation is necessary for solving the problem. The discussion emphasizes a step-by-step approach to understanding the momentum conservation principle.
12boone
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Homework Statement



A 23 g bullet traveling 235 m/s penetrates a 3.8 kg block of wood and emerges cleanly at 195 m/s.

If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?


Homework Equations



(1) mv= (m+M)v'

momentum before= momentum after

(2) 1/2(m+M)v'^2 = (m+M)gh

The Attempt at a Solution



I attempted to solvethis using these two equations and I received an answer for 1.14 m/s and 230 m/s. I am having a lot of trouble even determining how to start this problem. I don't want a quick easy answer but I would like help working through it! Thank You!
 
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12boone said:
A 23 g bullet traveling 235 m/s penetrates a 3.8 kg block of wood and emerges cleanly at 195 m/s.

If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

(1) mv= (m+M)v'

momentum before= momentum after

(2) 1/2(m+M)v'^2 = (m+M)gh

I attempted to solvethis using these two equations …

Hi 12boone! :smile:

However did you use equation (2)? :confused:

Equation (1) should be enough.

Try again! :smile:
 
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