Need help with a problem(Energy and Momentum)

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The discussion revolves around a physics problem involving a bullet passing through a block and the resulting effects on energy and momentum. Key questions include determining the bullet's speed after exiting the block, the height the block rises, the work done by the bullet, and the outcome if the bullet lodged in the block instead. Participants suggest starting with a diagram to visualize the scenario and emphasize using conservation of momentum to solve for the velocities. The problem requires applying principles of impulse, kinetic energy, and work-energy relationships. Understanding these concepts is crucial for solving the homework effectively.
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Homework Statement


I don't even know where to start, completely lost. Can someone at least get me started on this thing??


A 3g bullet with a speed of 300m/s passes right through a 400g block suspended on a long cord. The bullet imparts an impulse to the block and a speed of 1.5m/s


1. draw a picture

2. what is the speed of the bullet after it leaves the block?

3. how high does the block rise after the bullet passes through?

4. what is the work done by the bullet by passing through the block?

5. if the bullet did not pas through the block but instead lodged in the block, what would be the velocity of the block and the bullet after the collision?
 
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Start with the conservation of momentum, m1v1=m2v2 (the v's are vectors but since in this case its a linear problem it doesn't matter)
 
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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