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

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In a collision where a bullet of mass m embeds itself in a block of mass M at rest, the principle of conservation of momentum applies. The initial momentum of the system is solely from the bullet, calculated as mv. After the collision, the total momentum of the block and bullet combined must equal the initial momentum, leading to the equation mv = (M + m)v_f, where v_f is the final speed of the block and bullet together. Rearranging gives v_f = mv / (M + m), providing the speed of the block after the bullet embeds itself. This demonstrates the conservation of momentum in a closed system.
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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?
 
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Momentum is conserved. What would be the momentum of the "block+bullet" after the collision? (Write it mathematically.)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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