Solving Collision Problem: 10.0g Bullet & 5.00lg Block

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A 10.0-g bullet embeds into a stationary 5.00 kg block, resulting in a combined speed of 0.600 m/s after the collision. The original speed of the bullet needs to be determined using the conservation of momentum. The provided equation is incorrect for this scenario, as it calculates the speed of the combined mass rather than the bullet's initial speed. Instead, the conservation of momentum should be applied, and the equation should be rearranged to isolate the bullet's speed. The discussion emphasizes the importance of correctly applying physics principles to solve the problem.
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Homework Statement


a 10.0-g bullet is fired into a stationary block of wood (m=5.00 lg). the bullet imbeds into the block. the speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. what was the original speed of the bullet


Homework Equations



v= (m1)(v1a)/((m1)+(m2)
I=delat p
i am not 100% if you would even use that, or if that is what you need. i tryed it out and i will post it in about 10min. but the answer was given to me, and when i tested it out, i was wrong

The Attempt at a Solution



to be added soon.
 
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Dejey said:
v= (m1)(v1a)/((m1)+(m2)

If you tried to work out the answer using this then you would be wrong. You're trying to find the speed of the bullet. the equation you've given is arranged for the speed of the bullet and block which you already know. Start with the conservation of momentum equation and try rearranging it again.
 

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