How can I determine the speed of a bullet fired into a block of wood?

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Homework Help Overview

The discussion revolves around determining the speed of a bullet fired into a block of wood, involving concepts from momentum conservation in physics. The original poster presents a problem where a bullet embeds itself in a block, and the resulting velocity of the block is given.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the conservation of momentum principle but questions their calculations when the initial answer does not match the provided options. Some participants suggest checking the calculations and clarify the mass conversion from grams to kilograms.

Discussion Status

Participants are engaging in a back-and-forth to clarify the calculations involved in the momentum conservation approach. There is an acknowledgment of the original poster's understanding of the concept, but also a correction regarding the mass conversion, indicating a productive direction in the discussion.

Contextual Notes

The problem involves specific mass values and a scenario where the bullet is embedded in the block, which may lead to confusion regarding unit conversions and the application of momentum principles.

Crashdowngurl
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Bullet speed- Please Help!

I am not to great at physics and I was hoping someone could point me where I am going wrong in this problem! :) My answer I came up with is 2.5 m/s but that isn't one of my options so I know something has to be off.


5. To find the speed with which a gun fires bullets, a bullet of mass 5 gm is fired horizontally into a 10 Kg block of wood resting on a frictionless table. The bullet gets embedded in the block and the block is found to have a velocity of 0.5 m/s after collision. What is the speed of the bullet?

Bullet Speed= 2.5 m/s

Mv1i + 0 = (m +M)V
Mv1i + 0 = (5gm +10 kg) .5 m/s
m1 u1 + m2 u2 = (m1 + m2) V
(5gm u1) + (10kg *0m/s) = (5gm + 10kg) .5m/s
(5gm u1) + (0) =7.5
u1 = 2.5 m/s
 
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Think of the conservaton off momentum. Remember the bullet is embeded in the block.

added by edit: Never mind. I didn't look at your work well enough. You've got the idea. But (5gm + 10kg)=1005gm not 15gm.

regards
 
Last edited:
So... :)

So would this be more on track:


Mv1i + 0 = (m +M)V
Mv1i + 0 = (5gm +10 kg) .5 m/s
m1 u1 + m2 u2 = (m1 + m2) V
(5gm u1) + (10kg *0m/s) = (5gm + 10kg) .5m/s
(5gm u1) + (10kg *0m/s) = (1005gm) .5m/s
(5gm u1) + (0) = 502.5
u1 = 100.5 m/s


Bullet Speed= 100.5 m/s


Thank you so much for your help! :)

Emma
 
Any time. And welcome to the forums.
 

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