What is the kinetic energy now?

AI Thread Summary
The discussion revolves around calculating the kinetic energy of a body that initially moves in one direction with kinetic energy K and then reverses direction at three times its initial speed. The kinetic energy formula, K = 1/2mv², is applied to find the new kinetic energy, resulting in K' = 9K. Participants clarify that the final kinetic energy can be expressed as nine times the initial kinetic energy, emphasizing the relationship between the two states. The conversation highlights the importance of recognizing how changes in speed affect kinetic energy. Ultimately, the conclusion is that the final kinetic energy is nine times the initial kinetic energy.
genu
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Homework Statement



Initially a body moves in one direction and has kinetic energy K. Then it moves in the opposite direction with three times its initial speed. What is the kinetic energy now?

Homework Equations



k =1/2mv^2

The Attempt at a Solution



k = 1/2m(3v)^2
k=1/2m(9v^2(
k=(m9v^2)/2
2k=m9v^2
2/9k=mv^2
...
?

answer: 9k
 
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You've actually found the answer within the first few steps.

As you've said, K = 1/2*mv^2

And afterwards, K' = 1/2*m(-3v)^2

so that K' = 9/2*mv^2

What do you notice about K and K' when comparing them?

K' = 9K.
 
I'm not understanding how you've arrived to that last step...what happened to all the other variables?
 
genu said:
I'm not understanding how you've arrived to that last step...what happened to all the other variables?

You found that the final KE was 9 * (½mv²)

If you divide by the initial KE you get

Final KE = 9 * (Initial KE)

The suggestion was to encourage you to recognize the initial KE in the final KE expression.
 
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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