What is the speed of the block after a bullet embeds itself in it?

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SUMMARY

The discussion centers on the conservation of momentum in a collision scenario where a bullet of mass m embeds itself into a block of mass M at rest on a nearly frictionless surface. The initial momentum of the bullet is calculated as mv, where v is the bullet's velocity. After the collision, the combined system of the block and bullet conserves momentum, leading to the equation (m + M)v_f = mv, where v_f is the final velocity of the block and bullet together. Thus, the speed of the block after the bullet embeds itself can be determined using the formula v_f = mv / (m + M).

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Homework Statement


A bullet of mass m traveling horizontally at a very high speed v embeds itself in a block of mass M that is sitting at rest on a nearly frictionless surface. What is the speed of the block after the bullet embeds itself in the block?


Homework Equations


momentum=(mass)(velocity)


The Attempt at a Solution


momentum (p) of the bullet = mv.
But I don't know how to figure out the speed of the block..since momentum is conserved, would it just be v=p/m?
 
Physics news on Phys.org
Momentum is conserved. What would be the momentum of the "block+bullet" after the collision? (Write it mathematically.)
 

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