Inelastic collision kinetic energy lost?

AI Thread Summary
In an inelastic collision involving a bullet and a block of wood, the bullet's initial speed is 896 m/s, and after passing through, it slows to 435 m/s. The final velocity of the wood was calculated to be approximately 11.2 m/s. The kinetic energy lost during the collision is determined using the formula for kinetic energy, which requires squaring the velocities. A common mistake noted in the discussion is the incorrect handling of signs when calculating energy loss, emphasizing the importance of using the correct formula for kinetic energy loss. Proper application of conservation of momentum and careful calculation will yield accurate results for both final velocity and energy lost.
girl52
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inelastic collision...kinetic energy lost?

Homework Statement


a bullet with a mass of 6 g is fired through a 1.25kg block of wood on a frictionless surface. initial speed of the bullet is 896 m/s and speed of bullet after it passes through it is 435 m/s. what's the final velocity? how much KE is lost?

Homework Equations


ke lost=ke initial-ke final
ke lost=1/2mvf-1/2mvi

i got 11.2 m/s for the final velocity of the wood

The Attempt at a Solution


KEi=1/2(.006)(896^2) + 0
KEf=1/2(.006)(435^2) + 1/2(1.25)(11.225^2)

i got 2408.228 - 488.925

thats not negative..where did i go wrong??
 
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Don't forget to square the velocity when calculating KE. But first try using conservation of momentum, and show your work, please.
 


edited
 


girl52 said:

Homework Statement


a bullet with a mass of 6 g is fired through a 1.25kg block of wood on a frictionless surface. initial speed of the bullet is 896 m/s and speed of bullet after it passes through it is 435 m/s. what's the final velocity? how much KE is lost?


Homework Equations


ke lost=ke initial-ke final
ke lost=1/2mvf-1/2mvi

i got 11.2 m/s for the final velocity of the wood
how did you arrive at this number?

The Attempt at a Solution


KEi=1/2(.006)(896^2) + 0
KEf=1/2(.006)(435^2) + 1/2(1.25)(11.225^2)

i got 2408.228 - 488.925

thats not negative..where did i go wrong??
The change in energy (final minus initial) is negative...if you use (initial minus final), you get your signs reversed. Don't do it that way, the minus signs are bad enough without compounding them.
 
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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