Solving for the Speed of an 8kg Bullet

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The problem involves an 8 kg bullet embedded in a stationary 9 kg block of wood, which moves at 40 cm/s post-impact. The collision is classified as inelastic since the bullet becomes lodged in the wood. The conservation of momentum principle is applied to determine the bullet's initial speed. Key equations involve momentum before and after the collision, emphasizing that kinetic energy is not conserved in inelastic collisions. The discussion seeks clarification on the collision type and its implications for momentum conservation.
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Homework Statement



An 8 kg bullet is fired into a stationary 9 kg block of wood which is free to move. The bullet is stopped in the wood which has a velocity of 40 cm/s after the impact. What was the speed of the bullet?

Homework Equations





The Attempt at a Solution

 
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use the conservation of momentum
 
Well we know from the question that the bullet gets lodged into the wood upon impact. What kind of collision would that be? Elastic or Inelastic? What kind of properties, specifically with conservation of momentum/kinetic energy does that collision contain?

Let me know if you need more help.
 
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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